NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

Q.1

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}1& \mathrm{sin\theta }& 1\\ -\mathrm{\hspace{0.17em}}\mathrm{sin\theta }& 1& \mathrm{sin\theta }\\ -1& -\mathrm{sin\theta }& 1\end{array}\right],\mathrm{where}0\le \mathrm{\theta }\le 2\mathrm{\pi }.\mathrm{Then}\\ \left(\mathrm{A}\right)\mathrm{Det}\left(\mathrm{A}\right)=0\left(\mathrm{B}\right)\mathrm{Det}\left(\mathrm{A}\right)\in (2,\mathrm{\infty })\\ \left(\mathrm{C}\right)\mathrm{Det}\left(\mathrm{A}\right)\in (2,4)\left(\mathrm{D}\right)\mathrm{Det}\left(\mathrm{A}\right)\in [2,4]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}\mathrm{1}& \mathrm{sin\theta }& \mathrm{1}\\ -\mathrm{\hspace{0.17em}}\mathrm{sin\theta }& \mathrm{1}& \mathrm{sin\theta }\\ -\mathrm{1}& -\mathrm{sin\theta }& \mathrm{1}\end{array}\right],\mathrm{where}0\le \mathrm{\theta }\le 2\mathrm{\pi }.\mathrm{Then}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\mathrm{A}\right|=\left|\begin{array}{ccc}\mathrm{1}& \mathrm{sin\theta }& \mathrm{1}\\ -\mathrm{\hspace{0.17em}}\mathrm{sin\theta }& \mathrm{1}& \mathrm{sin\theta }\\ -\mathrm{1}& -\mathrm{sin\theta }& \mathrm{1}\end{array}\right|\\ =1(1+{\sin }^2\mathrm{\theta })-\mathrm{sin\theta }(-\mathrm{sin\theta }+\mathrm{sin\theta })+1({\sin }^2\mathrm{\theta }+1)\\ =2(1+{\sin }^2\mathrm{\theta })\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{0}\le \mathrm{\theta }\le 2\mathrm{\pi }\\ \Rightarrow \mathrm{0}\le \mathrm{sin\theta }\le 1\\ \Rightarrow \mathrm{0}\le {\sin }^2\mathrm{\theta }\le 1\\ \Rightarrow \mathrm{1}\le 1+{\sin }^2\mathrm{\theta }\le 2\\ \Rightarrow \mathrm{2}\le 2(1+{\sin }^2\mathrm{\theta })\le 4\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{Det}\left(\mathrm{A}\right)\in [2,4]\\ \mathrm{Thus},\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{If}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{non}\mathrm{zero}\mathrm{real}\mathrm{numbers},\mathrm{then}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{matrix}\\ \mathrm{A}=\left[\begin{array}{ccc}\mathrm{x}& 0& 0\\ 0& \mathrm{y}& 0\\ 0& 0& \mathrm{z}\end{array}\right]\mathrm{\hspace{0.17em}}\mathrm{is}\\ \left(\mathrm{A}\right)\left[\begin{array}{ccc}{\mathrm{x}}^{–1}& 0& 0\\ 0& {\mathrm{y}}^{–1}& 0\\ 0& 0& {\mathrm{z}}^{–1}\end{array}\right]\\ \left(\mathrm{B}\right)\mathrm{xyz}\left[\begin{array}{ccc}{\mathrm{x}}^{–1}& 0& 0\\ 0& {\mathrm{y}}^{–1}& 0\\ 0& 0& {\mathrm{z}}^{–1}\end{array}\right]\\ \left(\mathrm{C}\right)\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{x}& 0& 0\\ 0& \mathrm{y}& 0\\ 0& 0& \mathrm{z}\end{array}\right]\\ \left(\mathrm{D}\right)\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{matrix}\mathrm{\hspace{0.17em}}\mathrm{is}\\ \mathrm{A}=\left[\begin{array}{ccc}\mathrm{x}& \mathrm{0}& \mathrm{0}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{z}\end{array}\right]\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}\mathrm{x}& \mathrm{0}& \mathrm{0}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{z}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{x}(\mathrm{yz}-0)-0+0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{xyz}\ne 0\\ \mathrm{Thus},\mathrm{inverse}\mathrm{of}\mathrm{matrix}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now}\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=\mathrm{yz},\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{12}}=0,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{13}}=0\\ {\mathrm{A}}_{\mathrm{21}}=0,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{22}}=\mathrm{xz},{\mathrm{A}}_{\mathrm{23}}=0\\ \mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{31}}=0,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{32}}=0,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{33}}=\mathrm{xy}\\ \mathrm{Adj}\hspace{0.17em}\hspace{0.17em}\mathrm{A}={\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]}^{‘}=\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\hspace{0.17em}\hspace{0.17em}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}\frac{\mathrm{yz}}{\mathrm{xyz}}& 0& 0\\ 0& \frac{\mathrm{xz}}{\mathrm{xyz}}& 0\\ 0& 0& \frac{\mathrm{xy}}{\mathrm{xyz}}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}\frac{1}{\mathrm{x}}& 0& 0\\ 0& \frac{1}{\mathrm{y}}& 0\\ 0& 0& \frac{1}{\mathrm{z}}\end{array}\right]=\left[\begin{array}{ccc}{\mathrm{x}}^{-1}& 0& 0\\ 0& {\mathrm{y}}^{-1}& 0\\ 0& 0& {\mathrm{z}}^{-1}\end{array}\right]\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.3

$\begin{array}{l}If\text{}a,b,careinA.P.,thenthedeterminant\\ \left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|is\\ \left(A\right)0\left(B\right)1\left(C\right)x\left(D\right)2x\end{array}$

Ans

$\begin{array}{l}\left|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}\right|\\ \mathrm{If}\mathrm{}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.,\mathrm{so}2\mathrm{b}=\mathrm{a}+\mathrm{c}\\ \mathrm{Then},\\ \left|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}\right|=\left|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+(\mathrm{a}+\mathrm{c})\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{ccc}-1& -1& \mathrm{a}-\mathrm{c}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+(\mathrm{a}+\mathrm{c})\\ 1& 1& \mathrm{c}-\mathrm{a}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{ccc}0& 0& 0\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+(\mathrm{a}+\mathrm{c})\\ 1& 1& \mathrm{c}-\mathrm{a}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\mathrm{Since},\mathrm{\hspace{0.17em}}\mathrm{all}\mathrm{elements}\mathrm{of}{\mathrm{R}}_{\mathrm{1}}\mathrm{are}\mathrm{zero}\mathrm{.}]\\ \mathrm{Hence},\\ \left|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}\right|=0\\ \mathrm{Therefore},\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{equations}\\ \frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}+\frac{10}{\mathrm{z}}=4\\ \frac{4}{\mathrm{x}}-\frac{6}{\mathrm{y}}+\frac{5}{\mathrm{z}}=1\\ \frac{6}{\mathrm{x}}+\frac{9}{\mathrm{y}}-\frac{20}{\mathrm{z}}=1\end{array}$

Ans

$\begin{array}{l}\frac{\mathrm{2}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{3}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{10}}{\mathrm{z}}=\mathrm{4}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{4}}{\mathrm{x}}-\frac{\mathrm{6}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{5}}{\mathrm{z}}=\mathrm{1}\\ \frac{\mathrm{6}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{9}}{\mathrm{y}}-\frac{\mathrm{20}}{\mathrm{z}}=\mathrm{1}\\ \mathrm{Let}\mathrm{p}=\frac{1}{\mathrm{x}}\mathrm{q}=\frac{1}{\mathrm{y}}\mathrm{and}\mathrm{r}=\frac{1}{\mathrm{z}},\mathrm{then}\\ \mathrm{\hspace{0.17em}}2\mathrm{p}+3\mathrm{q}+10\mathrm{r}=\mathrm{4}\\ \hspace{0.17em}\hspace{0.17em}4\mathrm{p}-6\mathrm{q}+5\mathrm{r}=\mathrm{1}\\ 6\mathrm{p}+9\mathrm{q}-20\mathrm{r}=\mathrm{1}\\ \mathrm{The}\mathrm{system}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right],\mathrm{\hspace{0.17em}}\mathrm{X}=\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 1\\ 1\end{array}\right]\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(120-45)-3(-80-30)+10(36+36)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=150+330+720\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1200\\ \mathrm{Thus},\mathrm{}\mathrm{A}\mathrm{is}\mathrm{non}–\mathrm{singular}.\mathrm{Therefore},\mathrm{its}\mathrm{inverse}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}=75,\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{12}}=110,\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{13}}=72,\\ {\mathrm{A}}_{\mathrm{21}}=150,\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{22}}=-100,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{23}}=0,\mathrm{\hspace{0.17em}}\\ {\mathrm{A}}_{\mathrm{31}}=75,\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{32}}=30,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{33}}=-24\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}75& 110& 72\\ 150& -100& 0\\ 75& 30& -24\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \Rightarrow \left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\hspace{0.17em}\hspace{0.17em}=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\left[\begin{array}{l}4\\ 1\\ 2\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\hspace{0.17em}\hspace{0.17em}=\frac{1}{1200}\left[\begin{array}{l}300+150+150\\ 440-100+60\\ 288+0-48\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\hspace{0.17em}\hspace{0.17em}=\frac{1}{1200}\left[\begin{array}{l}600\\ 400\\ 240\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]\\ \Rightarrow \mathrm{p}=\frac{1}{2},\mathrm{\hspace{0.17em}}\mathrm{q}=\frac{1}{3}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{r}=\frac{1}{5}\\ \Rightarrow \mathrm{x}=\frac{1}{\mathrm{p}}=2,\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{\mathrm{q}}=3\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{z}=\frac{1}{\mathrm{r}}=5\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{is}\\ \mathrm{x}=\mathrm{2,}\mathrm{y}=3\mathrm{and}\mathrm{z}=5\mathrm{.}\end{array}$

Q.5 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{sin\alpha }& \mathrm{cos\alpha }& \cos (\mathrm{\alpha }+\mathrm{\beta })\\ \mathrm{sin\beta }& \mathrm{cos\beta }& \cos (\mathrm{\beta }+\mathrm{\delta })\\ \mathrm{sin\gamma }& \mathrm{cos\gamma }& \cos (\mathrm{\gamma }+\mathrm{\delta })\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{ccc}\mathrm{sin\alpha }& \mathrm{cos\alpha }& \cos (\mathrm{\alpha }+\mathrm{\delta })\\ \mathrm{sin\beta }& \mathrm{cos\beta }& \cos (\mathrm{\beta }+\mathrm{\delta })\\ \mathrm{sin\gamma }& \mathrm{cos\gamma }& \cos (\mathrm{\gamma }+\mathrm{\delta })\end{array}\right|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{sin\delta C}}_{\mathrm{1}}\mathrm{and}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{cos\delta C}}_{\mathrm{2}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{sin\delta cos\delta }}\left|\begin{array}{ccc}\mathrm{sin\alpha sin\delta }& \mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }-\mathrm{sin\alpha sin\delta }\\ \mathrm{sin\beta sin\delta }& \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }-\mathrm{sin\beta sin\delta }\\ \mathrm{sin\gamma sin\delta }& \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }-\mathrm{sin\gamma sin\delta }\end{array}\right|\\ {\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_3,\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{sin\delta cos\delta }}\left|\begin{array}{ccc}\mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }-\mathrm{sin\alpha sin\delta }\\ \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }-\mathrm{sin\beta sin\delta }\\ \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }-\mathrm{sin\gamma sin\delta }\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{sin\delta cos\delta }}\times 0\left[\begin{array}{l}\because \mathrm{\Delta }=0,\mathrm{if}\mathrm{any}\mathrm{two}\mathrm{coloumns}\\ \mathrm{are}\mathrm{identical}\mathrm{.}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.6 Using properties of determinants prove that:

$\left|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 2& 3+2\mathrm{p}& 4+3\mathrm{p}+2\mathrm{q}\\ 3& 6+3\mathrm{p}& 10+6\mathrm{p}+3\mathrm{q}\end{array}\right|=1$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 2& 3+2\mathrm{p}& 4+3\mathrm{p}+2\mathrm{q}\\ 3& 6+3\mathrm{p}& 10+6\mathrm{p}+3\mathrm{q}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-2{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-3{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 0& 1& 2+\mathrm{p}\\ 0& 3& 7+3\mathrm{p}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\left|\begin{array}{cc}1& 2+\mathrm{p}\\ 3& 7+3\mathrm{p}\end{array}\right|-0+0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=7+3\mathrm{p}-6-3\mathrm{p}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.7 Using properties of determinants prove that:

$\left|\begin{array}{ccc}3\mathrm{a}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ -\mathrm{b}+\mathrm{a}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ -\mathrm{c}+\mathrm{a}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}\right|=3(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{ccc}3\mathrm{a}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ -\mathrm{b}+\mathrm{a}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ -\mathrm{c}+\mathrm{a}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}\right|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}+{\mathrm{C}}_{\mathrm{2}}+{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}\mathrm{a}+\mathrm{b}+\mathrm{c}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}+\mathrm{c}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}+\mathrm{c}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}1& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ 1& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ 1& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}\right|\\ \mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1\mathrm{and}{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1,\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}1& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ 0& \mathrm{a}+2\mathrm{b}& \mathrm{a}-\mathrm{b}\\ 0& \mathrm{a}-\mathrm{c}& \mathrm{a}+2\mathrm{c}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\{(\mathrm{a}+2\mathrm{b})(\mathrm{a}+2\mathrm{c})-(\mathrm{a}-\mathrm{b})(\mathrm{a}-\mathrm{c})\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\{{\mathrm{a}}^2+2\mathrm{ac}+2\mathrm{ab}+4\mathrm{bc}-({\mathrm{a}}^2-\mathrm{ac}-\mathrm{ba}+\mathrm{bc})\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\{{\mathrm{a}}^2+2\mathrm{ac}+2\mathrm{ab}+4\mathrm{bc}-{\mathrm{a}}^2+\mathrm{ac}+\mathrm{ba}-\mathrm{bc}\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.8 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& 1+{\mathrm{px}}^3\\ \mathrm{y}& {\mathrm{y}}^2& 1+{\mathrm{py}}^3\\ \mathrm{z}& {\mathrm{z}}^2& 1+{\mathrm{pz}}^3\end{array}\right|=(1+\mathrm{pxyz})(\mathrm{x}–\mathrm{y})(\mathrm{y}–\mathrm{z})(\mathrm{z}–\mathrm{x})$

Ans

$\begin{array}{l}\text{L}\text{.H}\text{.S}\text{.}=\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ y& y^2& 1+p{y}^3\\ z& z^2& 1+p{z}^3\end{array}\right|\\ Applying{\text{R}}_{\text{3}}\to {\text{R}}_{\text{3}}-{\text{R}}_{\text{1}}and{\text{R}}_{\text{2}}\to {\text{R}}_{\text{2}}-{\text{R}}_{\text{1}},we\text{get}\\ =\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ y-x& y^2-x^2& (1+p{y}^3)-(1+p{x}^3)\\ z-x& z^2-x^2& (1+p{z}^3)-(1+p{x}^3)\end{array}\right|\\ =(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ 1& y+x& (p{y}^3-p{x}^3)\\ 1& z+x& (p{z}^3-p{x}^3)\end{array}\right|\\ =(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ 1& y+x& p(y^2+yx+x^2)\\ 1& z+x& p(z^2+zx+x^2)\end{array}\right|\\ Applying{\text{R}}_{\text{3}}\to {\text{R}}_{\text{3}}-{\text{R}}_{\text{2}},\text{we get}\\ =(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ 1& y+x& p(y^2+yx+x^2)\\ 0& z-y& p(z^2-y^2+zx-yx)\end{array}\right|\\ =(y-x)(z-x)(z-y)\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ 1& y+x& p(y^2+yx+x^2)\\ 0& 1& p(z+y+x)\end{array}\right|\\ Expanding{\text{along R}}_{\text{3}}\text{, we get}\\ =(y-x)(z-x)(z-y)\left\{\begin{array}{l}0-1\left|\begin{array}{cc}x& 1+p{x}^3\\ 1& p(y^2+yx+x^2)\end{array}\right|\\ +p(z+y+x)\left|\begin{array}{cc}x& x^2\\ 1& y+x\end{array}\right|\end{array}\right\}\\ =(y-x)(z-x)(z-y)\left[\begin{array}{l}-1\{xp(y^2+yx+x^2)-(1+p{x}^3)\}\\ +p(z+y+x)\{x(y+x)-x^2\}\end{array}\right]\\ =(y-x)(z-x)(z-y)\left\{\begin{array}{l}-px{y}^2-p{x}^{2}y-p{x}^3+1+p{x}^3\\ +p{x}^{2}y+px{y}^2+pxyz\end{array}\right\}\\ =(y-x)(z-x)(z-y)(1+pxyz)\\ =(1+pxyz)(x-y)(y-z)(z-x)=R.H.S.\\ Therefore,\text{the given result is proved}\text{.}\end{array}$

Q.9 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^2& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }& {\mathrm{\beta }}^2& \mathrm{\gamma }+\mathrm{\alpha }\\ \mathrm{\gamma }& {\mathrm{\gamma }}^2& \mathrm{\alpha }+\mathrm{\beta }\end{array}\right|=(\mathrm{\beta }–\mathrm{\gamma })(\mathrm{\gamma }–\mathrm{\alpha })(\mathrm{\alpha }–\mathrm{\beta })(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^2& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }& {\mathrm{\beta }}^2& \mathrm{\gamma }+\mathrm{\alpha }\\ \mathrm{\gamma }& {\mathrm{\gamma }}^2& \mathrm{\alpha }+\mathrm{\beta }\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^2& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }-\mathrm{\alpha }& {\mathrm{\beta }}^2-{\mathrm{\alpha }}^2& \mathrm{\alpha }-\mathrm{\beta }\\ \mathrm{\gamma }-\mathrm{\alpha }& {\mathrm{\gamma }}^2-{\mathrm{\alpha }}^2& \mathrm{\alpha }-\mathrm{\gamma }\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{\beta }-\mathrm{\alpha })(\mathrm{\gamma }-\mathrm{\alpha })\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^2& \mathrm{\beta }+\mathrm{\gamma }\\ 1& \mathrm{\beta }+\mathrm{\alpha }& -1\\ 1& \mathrm{\gamma }+\mathrm{\alpha }& -1\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{2}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{\beta }-\mathrm{\alpha })(\mathrm{\gamma }-\mathrm{\alpha })\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^2& \mathrm{\beta }+\mathrm{\gamma }\\ 1& \mathrm{\beta }+\mathrm{\alpha }& -1\\ 0& \mathrm{\gamma }-\mathrm{\beta }& 0\end{array}\right|\\ \mathrm{Expanding}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{\beta }-\mathrm{\alpha })(\mathrm{\gamma }-\mathrm{\alpha })\{-(\mathrm{\gamma }-\mathrm{\beta })(-\mathrm{\alpha }-\mathrm{\beta }-\mathrm{\gamma })\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{\beta }-\mathrm{\gamma })(\mathrm{\gamma }-\mathrm{\alpha })(\mathrm{\alpha }-\mathrm{\beta })(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.10

$\left|\mathbf{\text{Evaluate}}\begin{array}{ccc}1& \mathrm{x}& \mathrm{y}\\ 1& \mathrm{x}+\mathrm{y}& \mathrm{y}\\ 1& \mathrm{x}& \mathrm{x}+\mathrm{y}\end{array}\right|.$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{1}& \mathrm{x}& \mathrm{y}\\ \mathrm{1}& \mathrm{x}+\mathrm{y}& \mathrm{y}\\ \mathrm{1}& \mathrm{x}& \mathrm{x}+\mathrm{y}\end{array}\right|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}\mathrm{1}& \mathrm{x}& \mathrm{y}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{x}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1(\mathrm{xy}-0)-0+0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{xy}\end{array}$

Q.11

$\mathbf{\text{Evaluate}}\left|\begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}\right|.$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{x}+\mathrm{y}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}2\mathrm{x}+2\mathrm{y}& 2\mathrm{x}+2\mathrm{y}& 2\mathrm{x}+2\mathrm{y}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{2}(\mathrm{x}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\mathrm{y})\left|\begin{array}{ccc}\mathrm{1}& \mathrm{1}& \mathrm{1}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}\right|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{1}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_{\mathrm{3}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{2}(\mathrm{x}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\mathrm{y})\left|\begin{array}{ccc}\mathrm{1}& \mathrm{0}& \mathrm{0}\\ \mathrm{y}& \mathrm{x}& \mathrm{x}-\mathrm{y}\\ \mathrm{x}+\mathrm{y}& -\mathrm{y}& -\mathrm{x}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{2}(\mathrm{x}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\mathrm{y})\left|\begin{array}{cc}\mathrm{x}& \mathrm{x}-\mathrm{y}\\ -\mathrm{y}& -\mathrm{x}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{2}(\mathrm{x}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\mathrm{y})(-{\mathrm{x}}^2+\mathrm{xy}-{\mathrm{y}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-\mathrm{2}(\mathrm{x}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\mathrm{y})({\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-\mathrm{2}({\mathrm{x}}^3+{\mathrm{y}}^3)\\ \end{array}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right].\mathrm{Verify}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}{\left[\mathrm{adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right]}^{–1}=\mathrm{adj}\mathrm{\hspace{0.17em}}\left({\mathrm{A}}^{–1}\right)\left(\mathrm{ii}\right){\left({\mathrm{A}}^{–1}\right)}^{–1}=\mathrm{A}\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1(15-1)+2(-10-1)+1(-2-3)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=14-22-5=-13\ne 0\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=14,{\mathrm{A}}_{\mathrm{12}}=11,{\mathrm{A}}_{\mathrm{13}}=-5\\ {\mathrm{A}}_{\mathrm{21}}=11,{\mathrm{A}}_{\mathrm{22}}=4,{\mathrm{A}}_{\mathrm{23}}=-3\\ {\mathrm{A}}_{\mathrm{31}}=-5,{\mathrm{A}}_{\mathrm{32}}=-3,{\mathrm{A}}_{\mathrm{33}}=-1\\ \therefore \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]}^{‘}=\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{-13}\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]\\ \left(\mathrm{i}\right)\left|\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right|=\left|\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=14(-4-9)-11(-11-15)-5(-33+20)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-182+286+65\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=169\\ \mathrm{Cofactors}\mathrm{of}\mathrm{Adj}\mathrm{A}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=-13,{\mathrm{A}}_{\mathrm{12}}=26,{\mathrm{A}}_{\mathrm{13}}=-13\\ {\mathrm{A}}_{\mathrm{21}}=26,{\mathrm{A}}_{\mathrm{22}}=-39,{\mathrm{A}}_{\mathrm{23}}=-13\\ {\mathrm{A}}_{\mathrm{31}}=-13,{\mathrm{A}}_{\mathrm{32}}=-13,{\mathrm{A}}_{\mathrm{33}}=-65\\ \mathrm{Adj}\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)={\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ \therefore {\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)}^{-1}=\frac{1}{\left|\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right|}\mathrm{Adj}\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{169}\times 13\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}\mathrm{\hspace{0.17em}}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right]\end{array}$ \begin{array}{l}Adj{A}^{-1}={\left[\begin{array}{ccc}\frac{-4}{169}-\frac{9}{169}& -\left(-\frac{11}{169}-\frac{15}{169}\right)& -\frac{33}{169}-\frac{20}{169}\\ -\left(\frac{-11}{169}-\frac{15}{169}\right)& -\frac{14}{169}-\frac{25}{169}& -\left(-\frac{42}{169}+\frac{55}{169}\right)\\ -\frac{33}{169}+\frac{20}{169}& -\left(-\frac{42}{169}+\frac{55}{169}\right)& \frac{56}{169}-\frac{121}{169}\end{array}\right]}^{‘}\\ =\frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ =\frac{13}{169}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ Adj{A}^{-1}={\left(AdjA\right)}^{-1}\\ Hence,\text{it is proved}\text{.}\\ \left(ii\right)Since,\\ A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right]\\ andAdjA=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ \left|A^{-1}\right|=\left|\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right|\\ ={\left(\frac{1}{13}\right)}^3\left|\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right|\\ ={\left(\frac{1}{13}\right)}^3\left\{-14\left(-4-9\right)+11\left(-11-15\right)+5\left(-33+20\right)\right\}\\ ={\left(\frac{1}{13}\right)}^3\left(182-286-65\right)\\ ={\left(\frac{1}{13}\right)}^3\left(-169\right)=-\frac{1}{13}\\ \therefore {\left(A^{-1}\right)}^{-1}=\frac{Adj{A}^{-1}}{\left|A^{-1}\right|}\\ =\frac{1}{\left(-\frac{1}{13}\right)}\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ =\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]\Rightarrow {\left(A^{-1}\right)}^{-1}=A\\ Hence,\text{it is proved}\text{.}\end{array}

Q.13

${\text{If A}}^{–1}=\left|\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right|\mathrm{and}\mathrm{B}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right],\mathrm{find}{\left(\mathrm{AB}\right)}^{–1}.$

Ans

$\begin{array}{l}\mathrm{Since},\hspace{0.17em}\hspace{0.17em}{\left(\mathrm{AB}\right)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{B}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\left|\mathrm{B}\right|=\left|\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right|\\ =1(3-0)-2(-1-0)-2(2-0)\\ =3+2-4=1\ne 0\\ \mathrm{Thus},{\mathrm{B}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=3,{\mathrm{A}}_{\mathrm{12}}=1,{\mathrm{A}}_{\mathrm{13}}=2\\ {\mathrm{A}}_{\mathrm{21}}=2,{\mathrm{A}}_{\mathrm{22}}=1,{\mathrm{A}}_{\mathrm{23}}=2\\ {\mathrm{A}}_{\mathrm{31}}=6,{\mathrm{A}}_{\mathrm{32}}=2,{\mathrm{A}}_{\mathrm{33}}=5\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{B}={\left[\begin{array}{ccc}3& 1& 2\\ 2& 1& 2\\ 6& 2& 5\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\\ {\mathrm{B}}^{-1}=\frac{1}{\left|\mathrm{B}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{B}\\ \hspace{0.17em}\hspace{0.17em}=\frac{1}{1}\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}{\left(\mathrm{AB}\right)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}9-30+30& -3+12-12& 3-10+12\\ 3-15+10& -1+6-4& 1-5+4\\ 6-30+25& 2-12+10& 2-10+10\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right]\\ \mathrm{Thus},\mathrm{required}\mathrm{solution}\mathrm{is}\\ {\left(\mathrm{AB}\right)}^{-1}=\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right]\end{array}$

Q.14

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\\ \left|\begin{array}{ccc}{\mathrm{a}}^2& \mathrm{bc}& \mathrm{ac}+{\mathrm{c}}^2\\ {\mathrm{a}}^2+\mathrm{ab}& {\mathrm{b}}^2& \mathrm{ac}\\ \mathrm{ab}& {\mathrm{b}}^2+\mathrm{bc}& {\mathrm{c}}^2\end{array}\right|=4{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& \mathrm{bc}& \mathrm{ac}+{\mathrm{c}}^{\mathrm{2}}\\ {\mathrm{a}}^{\mathrm{2}}+\mathrm{ab}& {\mathrm{b}}^{\mathrm{2}}& \mathrm{ac}\\ \mathrm{ab}& {\mathrm{b}}^{\mathrm{2}}+\mathrm{bc}& {\mathrm{c}}^{\mathrm{2}}\end{array}\right|\\ \mathrm{Taking}\mathrm{out}\mathrm{common}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{from}{\mathrm{C}}_{\mathrm{1}},{\mathrm{C}}_{\mathrm{2}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\mathrm{respectively}\mathrm{.}\\ \mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{a}+\mathrm{\hspace{0.17em}}\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{\hspace{0.17em}}\mathrm{b}& \mathrm{b}-\mathrm{c}& -\mathrm{c}\\ \mathrm{b}-\mathrm{a}& \mathrm{b}& -\mathrm{a}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{b}-\mathrm{a}& \mathrm{b}& -\mathrm{a}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}+{\mathrm{R}}_{\mathrm{2}},\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ 2\mathrm{b}& 2\mathrm{b}& 0\end{array}\right|\\ \mathrm{\hspace{0.17em}}=2{\mathrm{ab}}^2\mathrm{c}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{1}& \mathrm{1}& 0\end{array}\right|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}=2{\mathrm{ab}}^2\mathrm{c}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{c}-\mathrm{a}& \mathrm{a}+\mathrm{c}\\ \mathrm{\hspace{0.17em}}\mathrm{a}+\mathrm{b}& -\mathrm{a}& \mathrm{a}\\ \mathrm{1}& \mathrm{0}& 0\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}=2{\mathrm{ab}}^2\mathrm{c}\{1(\mathrm{ac}-{\mathrm{a}}^2+{\mathrm{a}}^2+\mathrm{ac})-0+0\}\\ \mathrm{\hspace{0.17em}}=2{\mathrm{ab}}^2\mathrm{c}\left(2\mathrm{ac}\right)\\ \mathrm{\hspace{0.17em}}=4{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.15

$\begin{array}{l}\mathrm{Solve}\mathrm{thee}\mathrm{quation}\\ \left|\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=\mathbf{0},\mathrm{a}\ne \mathbf{0}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\left|\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=0,\mathrm{a}\ne 0.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=\mathrm{0}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{have}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{ccc}3\mathrm{x}+\mathrm{a}& 3\mathrm{x}+\mathrm{a}& 3\mathrm{x}+\mathrm{a}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=\mathrm{0}\\ \Rightarrow (3\mathrm{x}+\mathrm{a})\left|\begin{array}{ccc}1& 1& 1\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=\mathrm{0}\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_2-{\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_{\mathrm{3}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{}\mathrm{we}\mathrm{get}\\ \Rightarrow (3\mathrm{x}+\mathrm{a})\left|\begin{array}{ccc}1& 0& 0\\ \mathrm{x}& \mathrm{a}& 0\\ \mathrm{x}& 0& \mathrm{a}\end{array}\right|=\mathrm{0}\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \Rightarrow (3\mathrm{x}+\mathrm{a})\{1({\mathrm{a}}^2-0)-0+0\}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{a}}^2(3\mathrm{x}+\mathrm{a})=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(3\mathrm{x}+\mathrm{a})=0\Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{x}=-\frac{\mathrm{a}}{3}\\ \mathrm{Therefore},\mathrm{x}=-\frac{\mathrm{a}}{3}.\end{array}$

Q.16

$\begin{array}{l}\mathrm{If}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{real}\mathrm{numbers},\mathrm{and}\\ \mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}\right|=0,\\ \mathrm{Show}\mathrm{that}\mathrm{either}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\mathrm{or}\mathrm{a}=\mathrm{b}=\mathrm{c}.\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}\right|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{ccc}2(\mathrm{a}+\mathrm{b}+\mathrm{c})& 2(\mathrm{a}+\mathrm{b}+\mathrm{c})& 2(\mathrm{a}+\mathrm{b}+\mathrm{c})\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}1& 1& 1\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}\right|\\ \mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1,\mathrm{\hspace{0.17em}}{\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}1& 0& 0\\ \mathrm{c}+\mathrm{a}& \mathrm{b}-\mathrm{c}& \mathrm{b}-\mathrm{a}\\ \mathrm{a}+\mathrm{b}& \mathrm{c}-\mathrm{a}& \mathrm{c}-\mathrm{b}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(\mathrm{a}+\mathrm{b}+\mathrm{c}).[1\{(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{b})-(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\}-0+0]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(\mathrm{a}+\mathrm{b}+\mathrm{c})(\cancel{\mathrm{bc}}-{\mathrm{b}}^2-{\mathrm{c}}^2+\mathrm{cb}-\cancel{\mathrm{bc}}+\mathrm{ba}+\mathrm{ac}-{\mathrm{a}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(\mathrm{a}+\mathrm{b}+\mathrm{c})(-{\mathrm{a}}^2-{\mathrm{b}}^2-{\mathrm{c}}^2+\mathrm{cb}+\mathrm{ba}+\mathrm{ac})\\ \mathrm{But}\mathrm{\hspace{0.17em}}\mathrm{\Delta }=0\\ \mathrm{So},\\ 2(\mathrm{a}+\mathrm{b}+\mathrm{c})(-{\mathrm{a}}^2-{\mathrm{b}}^2-{\mathrm{c}}^2+\mathrm{cb}+\mathrm{ba}+\mathrm{ac})=0\\ \mathrm{Either}\hspace{0.17em}\hspace{0.17em}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-{\mathrm{a}}^2-{\mathrm{b}}^2-{\mathrm{c}}^2+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}=0\\ \mathrm{If}\mathrm{\hspace{0.17em}}-{\mathrm{a}}^2-{\mathrm{b}}^2-{\mathrm{c}}^2+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}=0\\ \mathrm{Then},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-2{\mathrm{a}}^2-2{\mathrm{b}}^2-2{\mathrm{c}}^2+2\mathrm{cb}+2\mathrm{ba}+2\mathrm{ac}=0\\ \Rightarrow 2{\mathrm{a}}^2+2{\mathrm{b}}^2+2{\mathrm{c}}^2-2\mathrm{cb}-2\mathrm{ba}-2\mathrm{ac}=0\\ \Rightarrow {(\mathrm{a}-\mathrm{b})}^2+{(\mathrm{b}-\mathrm{c})}^2+{(\mathrm{c}-\mathrm{a})}^2=0\\ \Rightarrow {(\mathrm{a}-\mathrm{b})}^2={(\mathrm{b}-\mathrm{c})}^2={(\mathrm{c}-\mathrm{a})}^2=0\left[\begin{array}{l}{(\mathrm{a}-\mathrm{b})}^2,{(\mathrm{b}-\mathrm{c})}^2,{(\mathrm{c}-\mathrm{a})}^2\mathrm{}\\ \mathrm{are}\mathrm{non}–\mathrm{negative}\mathrm{.}\end{array}\right]\\ \Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}\\ \mathrm{Thus},\mathrm{either}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\mathrm{or}\mathrm{a}=\mathrm{b}=\mathrm{c}.\end{array}$

Q.17

$\begin{array}{l}\mathrm{Evaluate}:\\ \mathrm{}\left|\begin{array}{ccc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }& -\mathrm{sin\alpha }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }& 0\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }& \mathrm{cos\alpha }\end{array}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }& -\mathrm{sin\alpha }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }& \mathrm{0}\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }& \mathrm{cos\alpha }\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}=-\mathrm{sin\alpha }\left|\begin{array}{l}-\mathrm{sin\beta }\mathrm{cos\beta }\\ \mathrm{sin\alpha cos\beta }\mathrm{sin\alpha sin\beta }\end{array}\right|-0\left|\begin{array}{cc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }\end{array}\right|\\ +\mathrm{cos\alpha }\left|\begin{array}{cc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }\end{array}\right|\\ =-\mathrm{sin\alpha }(-{\mathrm{sin\alpha sin}}^{\mathrm{2}}\mathrm{\beta }-{\mathrm{sin\alpha cos}}^2\mathrm{\beta })-0+\mathrm{cos\alpha }({\mathrm{cos\alpha cos}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{cos\alpha sin}}^{\mathrm{2}}\mathrm{\beta })\\ ={\sin }^{\mathrm{2}}{\mathrm{\alpha sin}}^{\mathrm{2}}\mathrm{\beta }+{\sin }^{\mathrm{2}}{\mathrm{\alpha cos}}^2\mathrm{\beta }+{\cos }^{\mathrm{2}}{\mathrm{\alpha cos}}^{\mathrm{2}}\mathrm{\beta }+{\cos }^{\mathrm{2}}{\mathrm{\alpha sin}}^{\mathrm{2}}\mathrm{\beta }\\ ={\sin }^{\mathrm{2}}\mathrm{\alpha }({\sin }^{\mathrm{2}}\mathrm{\beta }+{\cos }^2\mathrm{\beta })+{\cos }^{\mathrm{2}}\mathrm{\alpha }({\cos }^{\mathrm{2}}\mathrm{\beta }+{\sin }^{\mathrm{2}}\mathrm{\beta })[\because {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1]\\ ={\sin }^{\mathrm{2}}\mathrm{\alpha }\left(1\right)+{\cos }^{\mathrm{2}}\mathrm{\alpha }\left(1\right)\\ ={\sin }^{\mathrm{2}}\mathrm{\alpha }+{\cos }^{\mathrm{2}}\mathrm{\alpha }\\ =1=\mathrm{R}.\mathrm{H}.\mathrm{S}/\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.18

$\begin{array}{l}\mathrm{Without}\mathrm{expanding}\mathrm{the}\mathrm{determinant},\mathrm{prove}\mathrm{that}\\ \mathrm{}\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& \mathrm{bc}\\ \mathrm{b}& {\mathrm{b}}^2& \mathrm{ca}\\ \mathrm{c}& {\mathrm{c}}^2& \mathrm{ab}\end{array}\right|\mathrm{=}\left|\begin{array}{ccc}1& {\mathbf{a}}^2& {\mathbf{a}}^3\\ 1& {\mathbf{b}}^2& {\mathbf{b}}^3\\ 1& {\mathbf{c}}^2& {\mathbf{c}}^3\end{array}\right|\mathrm{.}\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^{\mathrm{2}}& \mathrm{bc}\\ \mathrm{b}& {\mathrm{b}}^{\mathrm{2}}& \mathrm{ca}\\ \mathrm{c}& {\mathrm{c}}^{\mathrm{2}}& \mathrm{ab}\end{array}\right|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{aR}}_{\mathrm{1}},{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{bR}}_{\mathrm{2}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{cR}}_{\mathrm{3}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ =\frac{1}{\mathrm{abc}}\left|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}& \mathrm{abc}\\ {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}& \mathrm{abc}\\ {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}& \mathrm{abc}\end{array}\right|\\ =\frac{1}{\mathrm{abc}}.\mathrm{abc}\left|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}& \mathrm{1}\\ {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}& \mathrm{1}\\ {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}& \mathrm{1}\end{array}\right|[\mathrm{Taking}\mathrm{out}\mathrm{abc}\mathrm{from}{\mathrm{C}}_{\mathrm{3}}.]\\ =(-1)\left|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& 1& {\mathrm{a}}^{\mathrm{3}}\\ {\mathrm{b}}^{\mathrm{2}}& 1& {\mathrm{b}}^{\mathrm{3}}\\ {\mathrm{c}}^{\mathrm{2}}& 1& {\mathrm{c}}^{\mathrm{3}}\end{array}\right|\left[{\mathrm{C}}_2\leftrightarrow {\mathrm{C}}_3\right]\\ ={(-1)}^2\left|\begin{array}{ccc}1& {\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}\\ 1& {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}\\ 1& {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}\end{array}\right|\left[{\mathrm{C}}_1\leftrightarrow {\mathrm{C}}_2\right]\\ =\left|\begin{array}{ccc}1& {\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}\\ 1& {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}\\ 1& {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}\end{array}\right|=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{determinant}\mathrm{}\left|\begin{array}{ccc}\mathrm{x}& \mathrm{sin\theta }& \mathrm{cos\theta }\\ -\mathrm{sin\theta }& -\mathrm{x}& 1\\ \mathrm{cos\theta }& 1& \mathrm{x}\end{array}\right|\mathrm{}\mathrm{is}\\ \mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\end{array}$

Ans

$\begin{array}{l}\left|\begin{array}{ccc}\mathrm{x}& \mathrm{sin\theta }& \mathrm{cos\theta }\\ -\mathrm{sin\theta }& -\mathrm{x}& \mathrm{1}\\ \mathrm{cos\theta }& \mathrm{1}& \mathrm{x}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\mathrm{x}(-{\mathrm{x}}^2-1)-\mathrm{sin\theta }(-\mathrm{xsin\theta }-\mathrm{cos\theta })+\mathrm{cos\theta }(-\mathrm{sin\theta }+\mathrm{xcos\theta })\\ =-{\mathrm{x}}^3-\mathrm{x}+{\mathrm{xsin}}^2\mathrm{\theta }+\mathrm{sin\theta cos\theta }-\mathrm{cos\theta sin\theta }+{\mathrm{xcos}}^2\mathrm{\theta }\\ =-{\mathrm{x}}^3-\mathrm{x}+\mathrm{x}({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta })\\ =-{\mathrm{x}}^3-\mathrm{x}+\mathrm{x}\left(1\right)[{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1]\\ =-{\mathrm{x}}^3,\mathrm{which}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\\ \mathrm{Therefore},\mathrm{determinant}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\end{array}$

Q.20 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Ans

$\begin{array}{l}\text{Let cost of 1 kg onions be ₹ x, 1 kg wheat be ₹ y and}\\ \text{1 kg rice be ₹z}\text{. Then, the given situation can be represented}\\ \text{by a system of equations as:}\\ \text{4x + 3y + 2z = 60}\\ \text{2x + 4y + 6z = 90}\\ \text{6x + 2y + 3z = 70}\\ \text{This system of equations can be written in the form of}\\ \text{AX = B as given below:}\\ \text{A}=\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]\text{and B=}\left[\begin{array}{l}60\\ 90\\ 70\end{array}\right]\\ \left|A\right|=\left|\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right|\\ =4\left(12-12\right)-3\left(6-36\right)+2\left(4-24\right)\\ =0+90-40=50\ne 0\\ Now,\\ A_{11}=0,A_{12}=30,A_{13}=-20\\ A_{21}=-5,A_{22}=0,A_{23}=10\\ A_{31}=10,A_{32}=-20,A_{33}=10\\ \therefore AdjA={\left[\begin{array}{ccc}0& 30& -20\\ -5& 0& 10\\ 10& -20& 10\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ \therefore A^{-1}=\frac{1}{\left|A\right|}AdjA\\ =\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ Now,\\ X=A^{-1}B\\ =\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{l}60\\ 90\\ 70\end{array}\right]\\ =\frac{1}{50}\left[\begin{array}{l}0-450+700\\ 1800+0-1400\\ -1200+900+700\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}250\\ 400\\ 400\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}5\\ 8\\ 8\end{array}\right]\\ \Rightarrow x=5,y=8andz=8.\\ \text{Thus, the cost of 1 kg onions is ₹5, 1 kg wheat is ₹ 8}\\ \text{and that of 1 kg rice is ₹8}\text{.}\end{array}$

Q.21

$\begin{array}{l}\mathbf{If}\hspace{0.17em}\hspace{0.17em}\mathbf{A}=\left|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right|,\mathrm{\hspace{0.17em}}\mathrm{find}{{\mathrm{A}}^{–}}^1.\mathrm{\hspace{0.17em}}{\mathrm{UsingA}}^{–1}\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\\ \mathrm{equations}:\\ 2\mathrm{x}-3\mathrm{y}+5\mathrm{z}=11\\ 3\mathrm{x}+2\mathrm{y}-4\mathrm{z}=-5\\ \mathrm{x}+\mathrm{y}-2\mathrm{z}=-3\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2(-4+4)+3(-6+4)+5(3-2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0-6+5\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-1\ne 0\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ {\mathrm{A}}_{11}=0,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{12}=2,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=-1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{22}=-9,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{23}=-5\\ {\mathrm{A}}_{31}=2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{32}=23,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{33}=13\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}0& 2& 1\\ -1& \hspace{0.17em}\hspace{0.17em}-9& -5\\ 2& 23& \hspace{0.17em}\hspace{0.17em}13\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \mathrm{\hspace{0.17em}}\therefore \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\\ \mathrm{of}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],\mathrm{\hspace{0.17em}}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}11\\ -5\\ -3\end{array}\right]\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{given}\mathrm{by}\\ \therefore \mathrm{X}=\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}\mathrm{B}\\ =\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{l}11\\ -5\\ -3\end{array}\right][\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)]\\ =\left[\begin{array}{l}0-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{x}=1,\mathrm{\hspace{0.17em}}\mathrm{y}=2\mathrm{and}\mathrm{z}=3.\end{array}$

Q.22 Solve system of linear equations, using matrix method,

x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}-\mathrm{y}+2\mathrm{z}=7\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\mathrm{x}+4\mathrm{y}-5\mathrm{z}=-5\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}-\mathrm{y}+3\mathrm{z}=12\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& -1& \mathrm{\hspace{0.17em}}2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}7\\ -5\\ \mathrm{\hspace{0.17em}}12\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& -1& \hspace{0.17em}\hspace{0.17em}2\\ 3& 4& -5\\ 2& -1& \hspace{0.17em}\hspace{0.17em}3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1(12-5)+1(9+10)+2(-3-8)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=7+19-22=4\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ {\mathrm{A}}_{11}=7,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{12}=-19,{\mathrm{A}}_{13}=-11\\ {\mathrm{A}}_{21}=1,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{22}=-1,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{23}=-1\\ {\mathrm{A}}_{31}=-3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{32}=11,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{33}=7\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}7& -19& -11\\ 1& \hspace{0.17em}\hspace{0.17em}-1& -1\\ -3& 11& \hspace{0.17em}\hspace{0.17em}7\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}7\\ -5\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}12\end{array}\right]\\ =\frac{1}{4}\left[\begin{array}{l}49-5-36\\ -133+5+132\\ -77+5+84\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}8\\ 4\\ 12\end{array}\right]\mathrm{\hspace{0.17em}}=\left[\begin{array}{l}2\\ 1\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{x}=2,\mathrm{\hspace{0.17em}}\mathrm{y}=1\mathrm{and}\mathrm{z}=3.\end{array}$

Q.23 Solve system of linear equations, using matrix method,

2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}+3\mathrm{y}+3\mathrm{z}=5\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}-2\mathrm{y}+\mathrm{z}=-4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\mathrm{x}-\mathrm{y}-2\mathrm{z}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& 3& \mathrm{\hspace{0.17em}}3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\\ -4\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}2& 3& \mathrm{\hspace{0.17em}}3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right|=2(4+1)-3(-2-3)+3(-1+6)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=10+15+15=40\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ {\mathrm{A}}_{11}=5,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{12}=5,{\mathrm{A}}_{13}=5\\ {\mathrm{A}}_{21}=3,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{22}=-13,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{23}=11\\ {\mathrm{A}}_{31}=9,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{32}=1,\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{33}=-7\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}5& 5& 5\\ 3& \hspace{0.17em}\hspace{0.17em}-13& 11\\ 9& 1& -7\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\\ -4\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right]\\ =\frac{1}{40}\left[\begin{array}{l}25-12+27\\ 25+52+3\\ 25-44-21\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{40}\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}40\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}80\\ -40\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}1\\ 2\\ -1\end{array}\right]\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{x}=1,\mathrm{\hspace{0.17em}}\mathrm{y}=2\mathrm{and}\mathrm{z}=-1.\end{array}$

Q.24 Solve system of linear equations, using matrix method,

x – y + z = 4
2x + y – 3z = 0
x + y + z = 2

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}-\mathrm{y}+\mathrm{z}=4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}+\mathrm{y}-3\mathrm{z}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y}+\mathrm{z}=2\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& -1& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 0\\ 2\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& -1& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1(1+3)+1(2+3)+1(2-1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4+5+1=10\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ {\mathrm{A}}_{11}=4,{\mathrm{A}}_{12}=-5,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=2,\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{22}=0,{\mathrm{A}}_{23}=-2\\ {\mathrm{A}}_{31}=2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{32}=5,{\mathrm{A}}_{33}=3\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{ccc}4& -5& 1\\ 2& \hspace{0.17em}\hspace{0.17em}0& -2\\ 2& 5& 3\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\left[\begin{array}{l}4\\ 0\\ 2\end{array}\right]\\ =\frac{1}{10}\left[\begin{array}{l}16+0+4\\ -20+0+10\\ 4+0+6\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{10}\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}20\\ -10\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}10\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}2\\ -1\\ 1\end{array}\right]\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{x}=2,\mathrm{\hspace{0.17em}}\mathrm{y}=-1\mathrm{and}\mathrm{z}=1.\end{array}$

Q.25 Solve system of linear equations, using matrix method,

2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9

Ans

\begin{array}{l}The\text{given system of equations is:}\\ 2x+y+z=1\\ x-2y-z=\frac{3}{2}\\ 3y-5z=9\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{ccc}2& 1& 1\\ 1& -2& -1\\ 0& 3& -5\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]\text{and B=}\left[\begin{array}{l}1\\ \frac{3}{2}\\ 9\end{array}\right]\\ Now,\\ \left|A\right|=\left|\begin{array}{ccc}2& 1& 1\\ 1& -2& -1\\ 0& 3& -5\end{array}\right|\\ =2\left(10+3\right)-1\left(-5-0\right)+1\left(3-0\right)\\ =26+5+3=34\ne 0\\ \therefore \text{A is a non-sigular}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ Now,\\ A_{11}=13,A_{12}=5,A_{13}=3\\ A_{21}=8,A_{22}=-10,A_{23}=-6\\ A_{31}=1,A_{32}=3,A_{33}=-5\\ AdjA={\left[\begin{array}{ccc}13& 5& 3\\ 8& -10& -6\\ 1& 3& -5\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ A^{-1}=\frac{1}{\left|A\right|}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ =\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \therefore X=A^{-1}B\\ =\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\left[\begin{array}{l}1\\ \frac{3}{\begin{array}{l}2\\ 9\end{array}}\end{array}\right]\\ =\frac{1}{34}\left[\begin{array}{l}13+12+9\\ 5-15+27\\ 3-9-45\end{array}\right]\\ \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{l}34\\ 17\\ -51\end{array}\right]=\left[\begin{array}{l}1\\ \frac{1}{2}\\ -\frac{3}{2}\end{array}\right]\\ Hence,x=1,y=\frac{1}{2}\text{and z}=-\frac{3}{2}.\end{array}

Q.26 Solve system of linear equations, using matrix method,

5x + 2y = 3
3x + 2y = 5

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\mathrm{x}+2\mathrm{y}=3\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\mathrm{x}+2\mathrm{y}=5\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}3\\ 5\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}5& 2\\ 3& 2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=10-6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{cc}\hspace{0.17em}\hspace{0.17em}2& -3\\ -2& \hspace{0.17em}\hspace{0.17em}5\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{–1}=\frac{1}{\left|\mathrm{A}\right|}(\mathrm{Adj}\mathrm{A})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\left[\begin{array}{l}3\\ 5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{4}\left[\begin{array}{l}6-10\\ -9+25\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}-4\\ 16\end{array}\right]\Rightarrow \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{l}-1\\ 4\end{array}\right]\\ \therefore \mathrm{x}=-1\hspace{0.17em}\hspace{0.17em}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{y}=4.\end{array}$

Q.27 Solve system of linear equations, using matrix method,

4x – 3y = 3
3x – 5y = 7

Ans

$\begin{array}{l}The\text{given system of equations is:}\\ 4x-3y=3\\ 3x-5y=7\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right],X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}3\\ 7\end{array}\right]\\ Now,\\ \left|A\right|=\left|\begin{array}{cc}4& -3\\ 3& -5\end{array}\right|\\ =-20+9\\ =-11\ne 0\\ \therefore \text{A is a non-sigular}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}-5& -3\\ 3& 4\end{array}\right]}^{‘}\\ =\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ {\text{A}}^{\text{-1}}=\frac{1}{\left|A\right|}\left(\text{Adj A}\right)\\ =\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ =\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\\ \therefore X=A^{-1}B\\ =\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{l}3\\ 7\end{array}\right]\\ =\frac{1}{11}\left[\begin{array}{l}15-21\\ 9-28\end{array}\right]\\ \left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-6\\ -19\end{array}\right]\\ \left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{-6}{11}\\ \frac{-19}{11}\end{array}\right]\\ \therefore x=\frac{-6}{11}andy=\frac{-19}{11}.\end{array}$

Q.29 Solve system of linear equations, using matrix method,

2x – y = –2
3x + 4y = 3

Ans

\begin{array}{l}The\text{given system of equations is:}\\ 2x-y=-2\\ 3x+4y=3\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}-2\\ 3\end{array}\right]\\ Now,\\ \left|A\right|=\left|\begin{array}{cc}2& -1\\ 3& 4\end{array}\right|\\ =8+3\\ =11\ne 0\\ \therefore \text{A is a non-sigular}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}4& -3\\ 1& 2\end{array}\right]}^{‘}=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ {\text{A}}^{\text{-1}}=\frac{1}{\left|A\right|}\left(\text{Adj A}\right)\\ =\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ \therefore X=A^{-1}B\\ =\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{l}-2\\ 3\end{array}\right]\\ =\frac{1}{11}\left[\begin{array}{l}-8+3\\ 6+6\end{array}\right]\\ \left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-5\\ 12\end{array}\right]\\ \left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{-5}{11}\\ \frac{12}{11}\end{array}\right]\\ \therefore x=\frac{-5}{11}andy=\frac{12}{11}.\end{array}

Q.30 Solve system of linear equations, using matrix method,

5x + 2y = 4
7x + 3y = 5

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}5\mathrm{x}+2\mathrm{y}=4\\ \hspace{0.17em}\hspace{0.17em}7\mathrm{x}+3\mathrm{y}=5\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}5& 2\\ 7& 3\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 5\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}5& 2\\ 7& 3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=15-14\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{cc}3& -7\\ -2& 5\end{array}\right]}^{‘}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{–1}=\frac{1}{\left|\mathrm{A}\right|}(\mathrm{Adj}\mathrm{A})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{1}\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\left[\begin{array}{l}4\\ 5\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}12-10\\ -28+25\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{l}2\\ -3\end{array}\right]\\ \therefore \mathrm{x}=2\hspace{0.17em}\hspace{0.17em}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{y}=-3.\end{array}$

Q.31 Examine the consistency of the system of equations.

5x – y + 4z = 5
2x + 3y + 5z = 2
5x –2y + 6z = –1

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5\mathrm{x}-\mathrm{y}+4\mathrm{z}=5\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{x}+3\mathrm{y}+5\mathrm{z}=2\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\mathrm{x}-2\mathrm{y}+6\mathrm{z}=-1\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}5& -1& 4\\ 2& 3& 5\\ 5& -2& 6\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 2\\ -1\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}5& -1& 4\\ 2& 3& 5\\ 5& -2& 6\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=5(18+10)+1(12-25)+4(-4-15)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=140-13-76=51\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.32 Examine the consistency of the system of equations.

3x – y – 2z = 2
2y – z = – 1
3x –5y = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\mathrm{x}-\mathrm{y}-2\mathrm{z}=2\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{y}-\mathrm{z}=-1\\ 3\mathrm{x}-5\mathrm{y}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}2\\ -1\\ 3\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3(0-5)+1(0+3)-2(0-6)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-15+3+12=0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{sigular}\mathrm{matrix}\mathrm{.}\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}\left|\begin{array}{cc}2& -1\\ -5& 0\end{array}\right|=-5\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}\left|\begin{array}{cc}0& -1\\ 3& 0\end{array}\right|=-3\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}\left|\begin{array}{cc}0& 2\\ 3& -5\end{array}\right|=-6\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left|\begin{array}{cc}-1& -2\\ -5& 0\end{array}\right|=10\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}\left|\begin{array}{cc}3& -2\\ 3& 0\end{array}\right|=6\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}\left|\begin{array}{cc}3& -1\\ 3& -5\end{array}\right|=12\\ {\mathrm{A}}_{\mathrm{31}}={(-1)}^{3+1}\left|\begin{array}{cc}-1& -2\\ \hspace{0.17em}\hspace{0.17em}2& -1\end{array}\right|=5\\ {\mathrm{A}}_{\mathrm{32}}={(-1)}^{3+2}\left|\begin{array}{cc}3& -2\\ 0& -1\end{array}\right|=3\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}\left|\begin{array}{cc}3& -1\\ 0& \hspace{0.17em}\hspace{0.17em}2\end{array}\right|=6\\ (\mathrm{Adj}\mathrm{A})={\left[\begin{array}{ccc}-5& -3& -6\\ 10& 6& 12\\ 5& 3& 6\end{array}\right]}^{‘}=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\\ (\mathrm{Adj}\mathrm{A})\mathrm{B}=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\left[\begin{array}{l}2\\ -1\\ 3\end{array}\right]\\ =\left[\begin{array}{l}-10-10+15\\ -6-6+9\\ -12-12+18\end{array}\right]\\ =\left[\begin{array}{l}-5\\ -3\\ -6\end{array}\right]\ne \mathrm{O}\\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{does}\\ \mathrm{not}\mathrm{exist}.\mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{inconsistent}\mathrm{.}\end{array}$

Q.33 Examine the consistency of the system of equations.

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{y}+\mathrm{z}=1\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}+3\mathrm{y}+2\mathrm{z}=2\\ \mathrm{ax}+\mathrm{ay}+2\mathrm{az}=4\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ \mathrm{a}& \mathrm{a}& 2\mathrm{a}\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}1\\ 2\\ 4\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ \mathrm{a}& \mathrm{a}& 2\mathrm{a}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1(6\mathrm{a}-2\mathrm{a})-1(4\mathrm{a}-2\mathrm{a})+1(2\mathrm{a}-3\mathrm{a})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4\mathrm{a}-2\mathrm{a}-\mathrm{a}=\mathrm{a}\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.34 Examine the consistency of the system of equations.

x + 3y = 5
2x + 6y = 8

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+3\mathrm{y}=5\\ 2\mathrm{x}+6\mathrm{y}=8\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 8\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}1& 3\\ 2& 6\end{array}\right|\mathrm{\hspace{0.17em}}=6-6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{sigular}\mathrm{matrix}\mathrm{.}\\ (\mathrm{Adj}\mathrm{A})={\left[\begin{array}{cc}6& -2\\ -3& 1\end{array}\right]}^{‘}=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]\\ (\mathrm{Adj}\mathrm{A})\mathrm{B}=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]\left[\begin{array}{l}5\\ 8\end{array}\right]\\ =\left[\begin{array}{l}30-24\\ -10+8\end{array}\right]\\ =\left[\begin{array}{l}6\\ -2\end{array}\right]\ne \mathrm{O}\\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{does}\\ \mathrm{not}\mathrm{exist}.\mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{inconsistent}\mathrm{.}\end{array}$

Q.35 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}2\mathrm{x}-\mathrm{y}=5\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{y}=4\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 4\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}2& -1\\ 1& 1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2+1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\hspace{0.17em}\hspace{0.17em}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.36 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{x}+2\mathrm{y}=2\\ 2\mathrm{x}+3\mathrm{y}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right],\hspace{0.17em}\hspace{0.17em}\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}2\\ 3\end{array}\right]\\ \mathrm{Now},\\ \left|\mathrm{A}\right|=\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3-4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-1\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\mathrm{\hspace{0.17em}}\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \end{array}$

Hence, the given system of equations is consistent.

Q.37

$$\begin{gathered} \begin{array}{l}\mathrm{If}\mathrm{A}\mathrm{is}\mathrm{aninvertible}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\mathrm{then}\hspace{0.17em}\hspace{0.17em}\det \left({\mathrm{A}}^{–1}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{is}\\ \mathrm{equalto}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \\ \mathrm{\hspace{0.17em}}\left(\mathrm{A}\right)\mathrm{Det}\left(\mathrm{A}\right) \\ \mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\frac{1}{\mathrm{Det}\left(\mathrm{A}\right)} \\ \left(\mathrm{C}\right)1 \\ \left(\mathrm{D}\right)0\end{array} \end{gathered}$$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{invertible}\mathrm{matrix},\mathrm{then}{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{and}{\mathrm{A}}^{–1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{adjA}.\\ \mathrm{Let}\mathrm{}\mathrm{a}\mathrm{matrix}\mathrm{of}\mathrm{order}2\mathrm{be}\mathrm{as}\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right],\\ \mathrm{then}\mathrm{}\left|\mathrm{A}\right|=\mathrm{ad}-\mathrm{bc}\mathrm{and}\mathrm{Adj}\mathrm{A}=\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]\\ \mathrm{Now},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\left|\mathrm{A}\right|}\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{cc}\frac{\mathrm{d}}{\left|\mathrm{A}\right|}& -\frac{\mathrm{b}}{\left|\mathrm{A}\right|}\\ -\frac{\mathrm{c}}{\left|\mathrm{A}\right|}& \frac{\mathrm{a}}{\left|\mathrm{A}\right|}\end{array}\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|{\mathrm{A}}^{-1}\right|=\left|\begin{array}{cc}\frac{\mathrm{d}}{\left|\mathrm{A}\right|}& -\frac{\mathrm{b}}{\left|\mathrm{A}\right|}\\ -\frac{\mathrm{c}}{\left|\mathrm{A}\right|}& \frac{\mathrm{a}}{\left|\mathrm{A}\right|}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{{\left|\mathrm{A}\right|}^2}\left|\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{{\left|\mathrm{A}\right|}^2}(\mathrm{ad}-\mathrm{bc})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{{\left|\mathrm{A}\right|}^2}\left|\mathrm{A}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\left|\mathrm{A}\right|}\\ \therefore \det \left({\mathrm{A}}^{-1}\right)=\frac{1}{\det \left(\mathrm{A}\right)}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{B}\right).\end{array}$

Q.38 Let A be a nonsingular square matrix of order 3 x 3. Then |adj A| is equal to

(A) |A|
(B) |A|2
(C) |A|3
(D) 3|A|

Ans

$\begin{array}{l}\mathrm{Since},\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\mathrm{A}=\left|\mathrm{A}\right|\mathrm{I}\\ =\left[\begin{array}{ccc}\left|\mathrm{A}\right|& 0& 0\\ 0& \left|\mathrm{A}\right|& 0\\ 0& 0& \left|\mathrm{A}\right|\end{array}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\mathrm{A}\right|=\left|\begin{array}{ccc}\left|\mathrm{A}\right|& 0& 0\\ 0& \left|\mathrm{A}\right|& 0\\ 0& 0& \left|\mathrm{A}\right|\end{array}\right|\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\right|\left|\mathrm{A}\right|={\left|\mathrm{A}\right|}^3\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\right|\left|\mathrm{A}\right|\mathrm{\hspace{0.17em}}={\left|\mathrm{A}\right|}^3\left(\mathrm{I}\right)\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\left(\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\right)\right|={\left|\mathrm{A}\right|}^2\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{B}\right).\end{array}$

Q.39

$\begin{array}{l}\mathrm{If}\mathrm{A}=\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathbf{2}& -\mathbf{1}& 1\\ -\mathbf{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathbf{2}& -\mathbf{1}\mathbf{}\mathbf{}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathbf{1}& -\mathbf{1}& \hspace{0.17em}\hspace{0.17em}\mathbf{2}\end{array}\right|,\mathrm{verify}\mathrm{that}\\ {\mathrm{A}}^3-6{\mathrm{A}}^2+9\mathrm{A}-4\mathrm{I}=\mathrm{O}\mathrm{and}\mathrm{hence}\mathrm{find}{\mathrm{A}}^{–1}.\end{array}$

Ans

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}]\\ {\mathrm{A}}^{\mathrm{2}}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}][\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}4+1+1& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-2-2-1& 2+1+2\\ -2-2-1& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1+4+1& -1-2-2\\ 2+1+2& -1-2-2& \hspace{0.17em}\hspace{0.17em}1+1+4\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}6& \hspace{0.17em}\hspace{0.17em}-5& 5\\ -5& \hspace{0.17em}\hspace{0.17em}6& -5\\ \hspace{0.17em}\hspace{0.17em}5& -5& 6\end{array}\right]\\ {\mathrm{A}}^{\mathrm{3}}={\mathrm{A}}^{\mathrm{2}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}6& \hspace{0.17em}\hspace{0.17em}-5& 5\\ -5& \hspace{0.17em}\hspace{0.17em}6& -5\\ \hspace{0.17em}\hspace{0.17em}5& -5& 6\end{array}\right][\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}12+5+5& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-6-10-5& \mathrm{6}+\mathrm{5}+\mathrm{10}\\ -\mathrm{10}-\mathrm{6}-\mathrm{5}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5+12+5& -\mathrm{5}-6-10\\ \mathrm{10}+5+6& -5-\mathrm{10}-6& \hspace{0.17em}\hspace{0.17em}\mathrm{5}+5+12\end{array}]=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{22}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\mathrm{22}\end{array}]\\ \therefore {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^2+9\mathrm{A}-4\mathrm{I}\\ =\mathrm{}[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{22}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\mathrm{22}\end{array}]-6\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}6& \hspace{0.17em}\hspace{0.17em}-5& 5\\ -5& \hspace{0.17em}\hspace{0.17em}6& -5\\ \hspace{0.17em}\hspace{0.17em}5& -5& 6\end{array}\right]+9[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}]\\ -4\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right]\\ =\mathrm{}[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{22}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \hspace{0.17em}\hspace{0.17em}\mathrm{22}\end{array}]-\left[\begin{array}{ccc}36& -30& 30\\ -30& 36& -30\\ 30& -30& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}36\end{array}\right]+\left[\begin{array}{ccc}18& -9& \hspace{0.17em}\hspace{0.17em}9\\ -9& 18& -9\\ 9& -9& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}18\end{array}\right]\\ -\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4\end{array}\right]\\ =[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{40}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{30}& \mathrm{30}\\ -\mathrm{30}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}40& -\mathrm{30}\\ \mathrm{30}& -\mathrm{30}& \hspace{0.17em}\hspace{0.17em}\mathrm{40}\end{array}]-\left[\begin{array}{ccc}40& -30& 30\\ -30& 40& -30\\ 30& -30& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}40\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\mathrm{O}\\ \mathrm{Thus},\mathrm{\hspace{0.17em}}{\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^2+5\mathrm{A}+11\mathrm{I}=0.\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{verified}\mathrm{.}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^3-6{\mathrm{A}}^2+9\mathrm{A}-4\mathrm{I}=\mathrm{O}\\ \Rightarrow \left(\mathrm{AAA}\right){\mathrm{A}}^{-1}-6\left(\mathrm{AA}\right){\mathrm{A}}^{-1}+9{\mathrm{AA}}^{-1}-4\mathrm{\hspace{0.17em}}{\mathrm{IA}}^{-1}=\mathrm{O}\\ [\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}\left|\mathrm{A}\right|\ne 0]\\ \Rightarrow \mathrm{AA}\left({\mathrm{AA}}^{-1}\right)-6\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+9\mathrm{I}=4{\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^2\mathrm{I}-6\mathrm{AI}+9\mathrm{I}=4{\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1}{4}\mathrm{\hspace{0.17em}}({\mathrm{A}}^2-6\mathrm{A}+9\mathrm{I})={\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\frac{1}{4}\mathrm{\hspace{0.17em}}({\mathrm{A}}^2-6\mathrm{A}+9\mathrm{I})...\left(\mathrm{i}\right)\\ \mathrm{Now},\\ {\mathrm{A}}^2-6\mathrm{A}+9\mathrm{I}=\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}6& \hspace{0.17em}\hspace{0.17em}-5& 5\\ -5& \hspace{0.17em}\hspace{0.17em}6& -5\\ \hspace{0.17em}\hspace{0.17em}5& -5& 6\end{array}\right]-6[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{1}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{2}\end{array}]+9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$ \begin{array}{l}=\left[\begin{array}{ccc}6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]-\left[\begin{array}{ccc}12& -6& 6\\ -6& 12& -6\\ 6& -6& 12\end{array}\right]+\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\\ =\left[\begin{array}{ccc}15& -5& 5\\ -5& 15& -5\\ 5& -5& 15\end{array}\right]-\left[\begin{array}{ccc}12& -6& 6\\ -6& 12& -6\\ 6& -6& 12\end{array}\right]=\left[\begin{array}{ccc}3& 1& -1\\ 1& 3& 1\\ -1& 1& 3\end{array}\right]\\ From\text{equation}\left(i\right),\text{we have}\\ {\text{A}}^{\text{-1}}=\left[\begin{array}{ccc}3& 1& -1\\ 1& 3& 1\\ -1& 1& 3\end{array}\right]\end{array}

Q.40

$\begin{array}{l}\mathrm{For}\mathrm{the}\mathrm{matrix}\mathrm{A}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}1& 1& 1\\ 1& 2& -\mathbf{3}\\ 2& -\mathbf{1}& \hspace{0.17em}\hspace{0.17em}\mathbf{3}\end{array}],\mathrm{show}\mathrm{that}\\ {\mathrm{A}}^3-6{\mathrm{A}}^2+5\mathrm{A}+11\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}.\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1& \mathrm{1}\\ \mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{3}\end{array}]\\ {\mathrm{A}}^{\mathrm{2}}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1& \mathrm{1}\\ \mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{3}\end{array}][\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1& \mathrm{1}\\ \mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{3}\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}1+1+2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1+2-\mathrm{1}& \mathrm{1}-3+3\\ 1+2-\mathrm{6}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1+4+3& 1-\mathrm{6}-\mathrm{9}\\ \mathrm{2}-1+6& 2-2-3& \hspace{0.17em}\hspace{0.17em}2+3+9\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}4& \hspace{0.17em}\hspace{0.17em}2& 1\\ -3& \hspace{0.17em}\hspace{0.17em}8& -14\\ 7& -3& 14\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{\mathrm{3}}={\mathrm{A}}^{\mathrm{2}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}4& \hspace{0.17em}\hspace{0.17em}2& 1\\ -3& \hspace{0.17em}\hspace{0.17em}8& -14\\ 7& -3& 14\end{array}\right][\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1& \mathrm{1}\\ \mathrm{1}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \hspace{0.17em}\hspace{0.17em}\mathrm{3}\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}4+2+2& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4+4-1& \mathrm{4}-\mathrm{6}+\mathrm{3}\\ -\mathrm{3}+\mathrm{8}-\mathrm{28}& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-3+16+14& -\mathrm{3}-24-42\\ \mathrm{7}-3+28& 7-\mathrm{6}-14& \hspace{0.17em}\hspace{0.17em}\mathrm{7}+9+42\end{array}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=[\hspace{0.17em}\hspace{0.17em}\begin{array}{ccc}\mathrm{8}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7& \mathrm{1}\\ -\mathrm{23}& \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}27& -\mathrm{69}\\ \mathrm{32}& -\mathrm{13}& \hspace{0.17em}\hspace{0.17em}\mathrm{58}\end{array}]\\ \therefore {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^2+5\mathrm{A}+11\mathrm{I}\\ =\mathrm{}\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]-6\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}14\end{array}\right]+5\left[\begin{array}{ccc}1& 1& \hspace{0.17em}\hspace{0.17em}1\\ 1& 2& -3\\ 2& -1& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right]\\ +11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]-\left[\begin{array}{ccc}24& 12& 6\\ -18& 48& -84\\ 42& -18& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}84\end{array}\right]+\left[\begin{array}{ccc}5& 5& \hspace{0.17em}\hspace{0.17em}5\\ 5& 10& -15\\ 10& -5& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}15\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}11\end{array}\right]\\ =\left[\begin{array}{ccc}8-24+5+11& 7-12+5+0& 1-6+5+0\\ -23+18+5+0& 27-48+10+11& -69+84-15+0\\ 32-42+10+0& -13+18-5+0& 58-84+15+11\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\mathrm{O}\\ \mathrm{Thus},\mathrm{\hspace{0.17em}}{\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^2+5\mathrm{A}+11\mathrm{I}=0.\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^3-6{\mathrm{A}}^2+5\mathrm{A}+11\mathrm{I}=\mathrm{O}\\ \Rightarrow \left(\mathrm{AAA}\right){\mathrm{A}}^{-1}-6\left(\mathrm{AA}\right){\mathrm{A}}^{-1}+5{\mathrm{AA}}^{-1}+11\mathrm{\hspace{0.17em}}{\mathrm{IA}}^{-1}=\mathrm{O}\\ [\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}\left|\mathrm{A}\right|\ne 0]\\ \Rightarrow \mathrm{AA}\left({\mathrm{AA}}^{-1}\right)-6\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+5\mathrm{I}=-11{\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^2\mathrm{I}-6\mathrm{AI}+5\mathrm{I}=-11{\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{1}{-11}\mathrm{\hspace{0.17em}}({\mathrm{A}}^2-6\mathrm{A}+5\mathrm{I})={\mathrm{A}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\frac{1}{-11}\mathrm{\hspace{0.17em}}({\mathrm{A}}^2-6\mathrm{A}+5\mathrm{I})...\left(\mathrm{i}\right)\\ \mathrm{Now},\\ {\mathrm{A}}^2-6\mathrm{A}+5\mathrm{I}=\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]-6\left[\begin{array}{ccc}1& 1& \hspace{0.17em}\hspace{0.17em}1\\ 1& 2& -3\\ 2& -1& \hspace{0.17em}\hspace{0.17em}3\end{array}\right]+5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]-\left[\begin{array}{ccc}6& 6& \hspace{0.17em}\hspace{0.17em}6\\ 6& 12& -18\\ 12& -6& \hspace{0.17em}\hspace{0.17em}18\end{array}\right]+\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]\\ \end{array}$ \begin{array}{l}=\left[\begin{array}{ccc}9& 2& 1\\ -3& 13& -14\\ 7& -3& 19\end{array}\right]-\left[\begin{array}{ccc}6& 6& 6\\ 6& 12& -18\\ 12& -6& 18\end{array}\right]\\ =\left[\begin{array}{ccc}3& -4& -5\\ -9& 1& 4\\ -5& 3& 1\end{array}\right]\\ From\text{equation}\left(i\right),\text{we have:}\\ A^{-1}=\frac{1}{-11}\left(A^2-6A+5I\right)\\ =\frac{1}{-11}\left[\begin{array}{ccc}3& -4& -5\\ -9& 1& 4\\ -5& 3& 1\end{array}\right]=\frac{1}{11}\left[\begin{array}{ccc}-3& 4& 5\\ 9& -1& -4\\ 5& -3& -1\end{array}\right]\end{array}

Q.41

$\begin{array}{l}\mathrm{For}\mathrm{matrix}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}32\\ 11\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{such}\mathrm{that}\\ {\mathrm{A}}^2+\mathrm{aA}+\mathrm{bI}=\mathrm{O}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{matrix}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}\mathrm{3}\mathrm{2}\\ \mathrm{1}\mathrm{1}\end{array}\right],\\ {\mathrm{A}}^2=\left[\begin{array}{l}3\mathrm{2}\\ \mathrm{1}\mathrm{1}\end{array}\right]\left[\begin{array}{l}\mathrm{3}\mathrm{2}\\ 1\mathrm{1}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}9+26+2\\ 3+12+1\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{\mathrm{2}}+\mathrm{aA}+\mathrm{bI}=\mathrm{O}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{AA}\left({\mathrm{A}}^{-1}\right)+{\mathrm{aAA}}^{-1}=-{\mathrm{bIA}}^{-1}[\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}\left|\mathrm{A}\right|\ne 0]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+\mathrm{aI}=-{\mathrm{bIA}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{A}\left(\mathrm{I}\right)+\mathrm{aI}=-{\mathrm{bIA}}^{-1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{-\mathrm{b}}(\mathrm{A}+\mathrm{aI})\\ \mathrm{Now},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\mathrm{A}\right|=\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|\\ =3-2=1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{AdjA}\\ =\frac{1}{1}\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}^{-1}=\frac{1}{-\mathrm{b}}(\left[\begin{array}{l}3\mathrm{2}\\ 1\mathrm{1}\end{array}\right]+\mathrm{a}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]=\frac{1}{-\mathrm{b}}\left[\begin{array}{cc}3+\mathrm{a}& 2\\ 1& 1+\mathrm{a}\end{array}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]=\left[\begin{array}{cc}\frac{3+\mathrm{a}}{-\mathrm{b}}& \frac{2}{-\mathrm{b}}\\ \frac{1}{-\mathrm{b}}& \frac{1+\mathrm{a}}{-\mathrm{b}}\end{array}\right]\\ \mathrm{Comparing}\mathrm{the}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{matrices},\mathrm{we}\mathrm{have}\\ -1=\frac{1}{-\mathrm{b}}\Rightarrow \mathrm{b}=1\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1=\frac{3+\mathrm{a}}{-\mathrm{b}}\Rightarrow -\mathrm{b}=3+\mathrm{a}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-1=3+\mathrm{a}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{a}=-4\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}-4\mathrm{and}1\mathrm{.}\end{array}$

Q.42

$\text{If\hspace{0.17em}\hspace{0.17em}A =}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}3\mathrm{\hspace{0.17em}}1\\ -12\end{array}\right|,\mathrm{show}\mathrm{that}{\mathrm{A}}^2-5\mathrm{A}+7\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}.$

Ans

$\begin{array}{l}\mathrm{Given},\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{3}\hspace{0.17em}1\\ -\mathrm{1}\mathrm{2}\end{array}\right],\mathrm{show}\mathrm{that}{\mathrm{A}}^{\mathrm{2}}–5\mathrm{A}+7\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}\mathrm{.}\\ {\mathrm{A}}^{\mathrm{2}}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{3}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}\mathrm{2}\end{array}\right]\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{3}\hspace{0.17em}\mathrm{1}\\ -\mathrm{1}\mathrm{2}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{9}-1\hspace{0.17em}3+\mathrm{2}\\ -\mathrm{3}-2-\mathrm{1}+\mathrm{4}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}85\\ -53\end{array}\right]\\ \therefore {\mathrm{A}}^{\mathrm{2}}-5\mathrm{A}+7\mathrm{I}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}85\\ -53\end{array}\right]-5\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{3}\hspace{0.17em}\mathrm{1}\\ -1\mathrm{2}\end{array}\right]+7\left[\begin{array}{l}10\\ 01\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}85\\ -53\end{array}\right]-\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{15\hspace{0.17em}}5\\ -\mathrm{5}\mathrm{10}\end{array}\right]+\left[\begin{array}{l}70\\ 07\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}8-15+75-5+0\\ -5+5+03-10+7\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\mathrm{\hspace{0.17em}}00\\ 00\end{array}\right]\\ \mathrm{Hence},\\ \mathrm{\hspace{0.17em}}{\mathrm{A}}^{\mathrm{2}}-5\mathrm{A}+7\mathrm{I}=\mathrm{O}\\ \therefore \mathrm{A}.\mathrm{A}-5\mathrm{A}=-7\mathrm{I}\\ \Rightarrow \mathrm{A}.\mathrm{A}\left({\mathrm{A}}^{-1}\right)-5{\mathrm{AA}}^{-1}=-7\mathrm{I}\left({\mathrm{A}}^{-1}\right)\\ [\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}\left|\mathrm{A}\right|\ne 0]\\ \Rightarrow \mathrm{A}\left({\mathrm{AA}}^{-1}\right)-5\mathrm{I}=-7{\mathrm{A}}^{-1}\\ \Rightarrow \mathrm{AI}-5\mathrm{I}=-7{\mathrm{A}}^{-1}\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{-7}(\mathrm{A}-5\mathrm{I})\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{7}(5\mathrm{I}-\mathrm{A})\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{7}(5\left[\begin{array}{l}10\\ 01\end{array}\right]-\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{3\hspace{0.17em}}1\\ -1\mathrm{2}\end{array}\right])\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{7}(\left[\begin{array}{l}50\\ 05\end{array}\right]-\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}3\hspace{0.17em}1\\ -1\mathrm{2}\end{array}\right])\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{7}\left[\begin{array}{cc}5-3& 0-1\\ 0+1& 5-2\end{array}\right]\\ \Rightarrow {\mathrm{A}}^{-1}=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\end{array}$

Q.43

$\text{Let\hspace{0.17em}\hspace{0.17em}A=}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}3\mathrm{\hspace{0.17em}}1\\ -12\end{array}\right|\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{B}=\left|\begin{array}{l}68\\ 79\end{array}\right|.\mathrm{\hspace{0.17em}}\mathrm{Verify}\mathrm{that}{\left(\mathrm{AB}\right)}^{–1}={\mathrm{B}}^{–1}{\mathrm{A}}^{–1}.$

Ans

$\begin{array}{l}Let\text{A}=\left[\begin{array}{cc}3& 7\\ 2& 5\end{array}\right]\\ \left|A\right|=\left|\begin{array}{cc}3& 7\\ 2& 5\end{array}\right|\\ =15-14=1\\ Now,\\ A_{11}={(-1)}^{1+1}5=5\\ A_{12}={(-1)}^{1+2}2=-2\\ A_{21}={(-1)}^{2+1}7=-7\\ A_{22}={(-1)}^{2+2}3=3\\ AdjA={\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]}^{‘}\\ =\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ A^{-1}=\frac{1}{\left|A\right|}.AdjA\\ =1\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ =\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ Let\text{B}=\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\right]\\ \left|B\right|=\left|\begin{array}{cc}6& 8\\ 7& 9\end{array}\right|\\ =54-56=-2\\ Now,\\ B_{11}={(-1)}^{1+1}9=9\\ B_{12}={(-1)}^{1+2}7=-7\\ B_{21}={(-1)}^{2+1}8=-8\\ B_{22}={(-1)}^{2+2}6=6\\ AdjB={\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]}^{‘}\\ =\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]\\ B^{-1}=\frac{1}{\left|B\right|}.AdjB\\ =-\frac{1}{2}\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{9}{2}& 4\\ \frac{7}{2}& -3\end{array}\right]\\ B^{-1}{A}^{-1}=\left[\begin{array}{cc}-\frac{9}{2}& 4\\ \frac{7}{2}& -3\end{array}\right]\left[\begin{array}{l}5-7\\ -23\end{array}\right]\\ =\left[\begin{array}{l}-\frac{45}{2}-8\frac{63}{2}+12\\ \frac{35}{2}+6-\frac{49}{2}-9\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right]\mathrm{\dots }\left(1\right)\\ AB=\left[\begin{array}{cc}3& 7\\ 2& 5\end{array}\right]\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\right]\\ =\left[\begin{array}{cc}18+49& 24+63\\ 12+35& 16+45\end{array}\right]\\ =\left[\begin{array}{cc}67& 87\\ 47& 61\end{array}\right]\\ \left|AB\right|=\left|\begin{array}{cc}67& 87\\ 47& 61\end{array}\right|\\ =4087-4089=-2\\ Adj\left(AB\right)={\left[\begin{array}{cc}61& -47\\ -87& 67\end{array}\right]}^{‘}\\ =\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]\\ {\left(AB\right)}^{-1}=-\frac{1}{2}\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right]\mathrm{\dots }\left(2\right)\\ Thus,\text{from equation}\left(1\right)\text{and equation}\left(2\right),\text{we have}\\ {\left(AB\right)}^{-1}=B^{-1}{A}^{-1}\\ Hence\text{proved}\text{.}\end{array}$

Q.44 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1\mathrm{\hspace{0.17em}}\hspace{0.17em}0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\\ 0\mathrm{cos\alpha }\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}s\mathrm{in\alpha }\\ 3\mathrm{sin\alpha }-\mathrm{cos\alpha }\end{array}\right|$

Ans

$\begin{array}{l}LetA=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right]\\ \left|A\right|=\left|\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right|\\ =1(-{\cos }^2\alpha -{\sin }^2\alpha )+0(0-0)+0(0-0)\\ =-1\ne 0\\ So,\text{the inverse of A exists}\text{.}\\ \text{Now,}\\ {\text{A}}_{\text{11}}={(-1)}^{1+1}\left|\begin{array}{l}\cos \alpha \sin \alpha \\ \sin \alpha -\cos \alpha \end{array}\right|\\ =-{\cos }^2\alpha -{\sin }^2\alpha \\ =-1\\ {\text{A}}_{\text{12}}={(-1)}^{1+2}\left|\begin{array}{l}0\sin \alpha \\ 0-\cos \alpha \end{array}\right|\\ =0\\ {\text{A}}_{\text{13}}={(-1)}^{1+3}\left|\begin{array}{l}0\cos \alpha \\ 0sin\alpha \end{array}\right|=0\\ {\text{A}}_{\text{21}}={(-1)}^{2+1}\left|\begin{array}{l}00\\ sin\alpha -\cos \alpha \end{array}\right|=0\\ A_{\text{22}}={(-1)}^{2+2}\left|\begin{array}{l}10\\ 0-\cos \alpha \end{array}\right|=-\cos \alpha \\ {\text{A}}_{\text{23}}={(-1)}^{2+3}\left|\begin{array}{l}10\\ 0\sin \alpha \end{array}\right|\\ =-\sin \alpha \\ {\text{A}}_{\text{31}}={(-1)}^{3+1}\left|\begin{array}{l}00\\ \cos \alpha \sin \alpha \end{array}\right|\\ =0\\ {\text{A}}_{\text{32}}={(-1)}^{3+2}\left|\begin{array}{l}10\\ 0\sin \alpha \end{array}\right|\\ =-sin\alpha \\ {\text{A}}_{\text{33}}={(-1)}^{3+3}\left|\begin{array}{l}10\\ 0\cos \alpha \end{array}\right|\\ =\cos \alpha \\ AdjA={\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\cos \alpha & -sin\alpha \\ 0& -sin\alpha & \cos \alpha \end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\cos \alpha & -sin\alpha \\ 0& -sin\alpha & \cos \alpha \end{array}\right]\\ A^{-1}=\frac{1}{\left|A\right|}AdjA\\ =-1\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\cos \alpha & -sin\alpha \\ 0& -sin\alpha & \cos \alpha \end{array}\right]\\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & sin\alpha \\ 0& sin\alpha & -\cos \alpha \end{array}\right]\end{array}$

Q.45 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1-1\mathrm{\hspace{0.17em}}2\\ 0\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}2-3\\ 3-2\hspace{0.17em}\hspace{0.17em}4\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=|\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\begin{array}{ccc}1& -1& 2\\ 0& \hspace{0.17em}\hspace{0.17em}2& -3\\ 3& -2& 4\end{array}|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=1(8-6)+1(0+9)+2(0-6)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2+9\mathrm{\hspace{0.17em}}-\mathrm{\hspace{0.17em}}12\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-1\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2 -\mathrm{3}\\ -\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}\left|\begin{array}{l}0-3\\ 3\hspace{0.17em}\hspace{0.17em}4\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\mathrm{\hspace{0.17em}}9\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}\left|\begin{array}{l}0\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\\ 3-\mathrm{\hspace{0.17em}}2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0-6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-6\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left|\begin{array}{l}-12\\ -24\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-(-4+4)\mathrm{\hspace{0.17em}}=0\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}\left|\begin{array}{l}12\\ 34\end{array}\right|\mathrm{\hspace{0.17em}}=-2\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}\left|\begin{array}{l}1-1\\ 3-2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-(-2+3)\hspace{0.17em}\hspace{0.17em}=-1\\ {\mathrm{A}}_{\mathrm{31}}={(-1)}^{3+1}\left|\begin{array}{l}-1\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2-3\end{array}\right|\hspace{0.17em}\hspace{0.17em}=-1\\ {\mathrm{A}}_{\mathrm{32}}={(-1)}^{3+2}\left|\begin{array}{l}1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\\ 0-3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}\left|\begin{array}{l}1-1\\ 0\hspace{0.17em}\hspace{0.17em}2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}=\mathrm{\hspace{0.17em}}{\left[\begin{array}{ccc}2& -9& -6\\ 0& -2& -1\\ -1& \mathrm{\hspace{0.17em}}3& \hspace{0.17em}\hspace{0.17em}2\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2& 0& -1\\ -9& -2& \hspace{0.17em}\hspace{0.17em}3\\ -6& -1& \hspace{0.17em}\hspace{0.17em}2\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}=-1\left[\begin{array}{ccc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2& 0& -1\\ -9& -2& \hspace{0.17em}\hspace{0.17em}3\\ -6& -1& \hspace{0.17em}\hspace{0.17em}2\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}-2& 0& \hspace{0.17em}\hspace{0.17em}1\\ \hspace{0.17em}\hspace{0.17em}9& 2& \hspace{0.17em}\hspace{0.17em}-3\\ \hspace{0.17em}\hspace{0.17em}6& 1& \hspace{0.17em}\hspace{0.17em}-2\end{array}\right]\end{array}$

Q.46 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}13\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4-10\\ -7\mathrm{\hspace{0.17em}}\hspace{0.17em}21\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}13\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4-1\mathrm{0}\\ -\mathrm{7\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\mathrm{2}\mathrm{1}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}13\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4-\mathrm{1}\mathrm{0}\\ -\mathrm{7\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}2\mathrm{1}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2(-1-0)-1(4-0)+3(8-7)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-2-4\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}3\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-3\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}\left|\begin{array}{l}-1 0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{1}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-1\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}40\\ -71\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\mathrm{\hspace{0.17em}}4\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}4-1\\ -7\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=8-7\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=1\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left|\begin{array}{l}13\\ 21\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-(1-6)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=5\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}23\\ -71\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=23\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2-1\\ -7\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-(4+7)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-11\\ {\mathrm{A}}_{\mathrm{31}}={(-1)}^{3+1}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}13\\ -10\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3\\ {\mathrm{A}}_{\mathrm{32}}={(-1)}^{3+2}\left|\begin{array}{l}23\\ 40\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=12\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}\left|\begin{array}{l}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1\\ 4-1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-6\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}=\mathrm{\hspace{0.17em}}{\left[\begin{array}{ccc}-1& -4& 1\\ 5& 23& -11\\ 3& 12& -6\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\\ \hspace{0.17em}\hspace{0.17em}=-\frac{1}{3}\left[\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right]\end{array}$

Q.47 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}10\hspace{0.17em}\hspace{0.17em}0\\ 3\mathrm{}3\mathrm{\hspace{0.17em}}\hspace{0.17em}0\\ 52-1\end{array}\right|$

Ans

$\begin{array}{l}LetA=\left[\begin{array}{l}\text{1}00\\ \text{3}\text{3}\text{0}\\ \text{5}\text{2}-\text{1}\end{array}\right]\\ \left|A\right|=\left|\begin{array}{l}\text{1}00\\ \text{3}\text{3}\text{0}\\ \text{5}\text{2}-\text{1}\end{array}\right|\\ =1(-3-0)-0(-3-0)+0(6-15)\\ =-3\ne 0\\ So,\text{the inverse of A exists}\text{.}\\ \text{Now,}\\ {\text{A}}_{\text{11}}={(-1)}^{1+1}\left|\begin{array}{l}\text{3}\text{0}\\ \text{2}-\text{1}\end{array}\right|=-3\\ {\text{A}}_{\text{12}}={(-1)}^{1+2}\left|\begin{array}{l}30\\ 5-1\end{array}\right|=3\\ {\text{A}}_{\text{13}}={(-1)}^{1+3}\left|\begin{array}{l}33\\ 52\end{array}\right|\\ =6-15=-9\\ {\text{A}}_{\text{21}}={(-1)}^{2+1}\left|\begin{array}{l}00\\ 2-1\end{array}\right|=0\\ {\text{A}}_{\text{22}}={(-1)}^{2+2}\left|\begin{array}{l}10\\ 5-1\end{array}\right|=-1\\ {\text{A}}_{\text{23}}={(-1)}^{2+3}\left|\begin{array}{l}10\\ 52\end{array}\right|\\ =-(2-0)=-2\\ {\text{A}}_{\text{31}}={(-1)}^{3+1}\left|\begin{array}{l}00\\ 30\end{array}\right|=0\\ {\text{A}}_{\text{32}}={(-1)}^{3+2}\left|\begin{array}{l}10\\ 30\end{array}\right|=0\\ {\text{A}}_{\text{33}}={(-1)}^{3+3}\left|\begin{array}{l}10\\ 33\end{array}\right|=3\\ AdjA={\left[\begin{array}{ccc}-3& 3& -9\\ 0& -1& -2\\ 0& 0& 3\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]\\ A^{-1}=\frac{1}{\left|A\right|}AdjA\\ =-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]\end{array}$

Q.48 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1\mathrm{}23\\ 024\\ 005\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}123\\ \mathrm{0}\mathrm{2}\mathrm{4}\\ \mathrm{0}\mathrm{0}\mathrm{5}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=\left|\begin{array}{l}\mathrm{1}23\\ \mathrm{0}2\mathrm{4}\\ 00\mathrm{5}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=1(10-0)-2(0-0)+3(0-0)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=10\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}\left|\begin{array}{l}\mathrm{2}\mathrm{4}\\ 0\mathrm{5}\end{array}\right|\mathrm{\hspace{0.17em}}=10\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}\left|\begin{array}{l}04\\ 05\end{array}\right|\mathrm{\hspace{0.17em}}=0\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}\left|\begin{array}{l}02\\ 00\end{array}\right|\mathrm{\hspace{0.17em}}=0\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left|\begin{array}{l}23\\ 05\end{array}\right|\mathrm{\hspace{0.17em}}=-10\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}\left|\begin{array}{l}13\\ 05\end{array}\right|\mathrm{\hspace{0.17em}}=5\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}\left|\begin{array}{l}12\\ 00\end{array}\right|\hspace{0.17em}\hspace{0.17em}=0\\ {\mathrm{A}}_{\mathrm{31}}={(-1)}^{3+1}\left|\begin{array}{l}12\\ 02\end{array}\right|\mathrm{\hspace{0.17em}}=2\\ {\mathrm{A}}_{\mathrm{32}}={(-1)}^{3+2}\left|\begin{array}{l}13\\ 04\end{array}\right|\mathrm{\hspace{0.17em}}=-4\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}\left|\begin{array}{l}12\\ 02\end{array}\right|\mathrm{\hspace{0.17em}}=2\end{array}$ \begin{array}{l}AdjA={\left[\begin{array}{ccc}10& 0& 0\\ -10& 5& 0\\ 2& -4& 2\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}10& -10& 2\\ 0& 5& -4\\ 0& 0& 2\end{array}\right]\\ A^{-1}=\frac{1}{\left|A\right|}AdjA\\ =\frac{1}{10}\left[\begin{array}{ccc}10& -10& 2\\ 0& 5& -4\\ 0& 0& 2\end{array}\right]\end{array}

Q.49 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}-15\\ -32\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}-\mathrm{1}5\\ -\mathrm{3}\mathrm{2}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=\left|\begin{array}{l}-\mathrm{1}5\\ -3\mathrm{2}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-2+15\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=13\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}2\hspace{0.17em}\hspace{0.17em}=2\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}(-3)\hspace{0.17em}\hspace{0.17em}=3\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left(5\right)\mathrm{\hspace{0.17em}}=-5\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}(-1)\mathrm{\hspace{0.17em}}=-1\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\\ -5-1\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}2-5\\ 3-1\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\mathrm{\hspace{0.17em}}=\frac{1}{13}\left[\begin{array}{l}2-5\\ 3-1\end{array}\right]\\ \end{array}$

Q.50 Find the inverse of each of the matrices (if it exists)

$\text{[}\begin{array}{l}2-2\\ 4\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}]$

Ans

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}2-\mathrm{2}\\ \mathrm{4\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}3\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=\left|\begin{array}{l}2-2\\ 4\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=6+8\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=14\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}3=3\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}4=-4\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}(-2)=2\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}2=2\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{l}3-4\\ 2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\end{array}\right]}^{‘}\\ \hspace{0.17em}\hspace{0.17em}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}32\\ -42\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}\mathrm{\hspace{0.17em}}=\frac{1}{14}\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}32\\ -42\end{array}\right]\end{array}$

Q.51 Verify A(adj A)=(adj A)A = |A|I

$\left|\begin{array}{l}1-1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\\ 3\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0-2\\ 1\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right|$

Ans

$\begin{array}{l}Let\text{}\text{A}=\left[\begin{array}{l}\text{1}-\text{1}\text{2}\\ \text{3}\text{0}-\text{2}\\ \text{1}\text{0}\text{3}\end{array}\right]\\ \left|A\right|=\left|\begin{array}{l}\text{1}-\text{1}\text{2}\\ \text{3}\text{0}-\text{2}\\ \text{1}\text{0}\text{3}\end{array}\right|\\ =1(0-0)+1(9+2)+2(0-0)\\ =11\\ A_{11}={(-1)}^{1+1}\left|\begin{array}{l}0-2\\ 03\end{array}\right|=0,\\ A_{12}={(-1)}^{1+2}\left|\begin{array}{l}3-2\\ 13\end{array}\right|=-11,\\ A_{13}={(-1)}^{1+3}\left|\begin{array}{l}30\\ 10\end{array}\right|=0,\\ A_{21}={(-1)}^{2+1}\left|\begin{array}{l}-12\\ 03\end{array}\right|=3\\ A_{22}={(-1)}^{2+2}\left|\begin{array}{l}12\\ 13\end{array}\right|=1\\ A_{23}={(-1)}^{2+3}\left|\begin{array}{l}1-1\\ 10\end{array}\right|=-1\\ A_{31}={(-1)}^{3+1}\left|\begin{array}{l}-12\\ 0-2\end{array}\right|=2\\ A_{32}={(-1)}^{3+2}\left|\begin{array}{l}12\\ 3-2\end{array}\right|=8\\ A_{33}={(-1)}^{3+3}\left|\begin{array}{l}1-1\\ 30\end{array}\right|=3\\ AdjA={\left[\begin{array}{l}0-110\\ 31-1\\ 283\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}032\\ -1118\\ 0-13\end{array}\right]\\ A\left(AdjA\right)=\left[\begin{array}{l}032\\ -1118\\ 0-13\end{array}\right]\left[\begin{array}{l}1-12\\ 30-2\\ 103\end{array}\right]\\ =\left[\begin{array}{l}0+9+20+0+00-6+6\\ -11+3+811+0+0-\mathrm{2\; 2}-2+24\\ 0-3+30+0+00+2+9\end{array}\right]\\ =\left[\begin{array}{l}1100\\ 0110\\ 0011\end{array}\right]\\ Also,\\ \left(AdjA\right)A=\left[\begin{array}{l}1-12\\ 30-2\\ 103\end{array}\right]\left[\begin{array}{l}032\\ -1118\\ 0-13\end{array}\right]\\ =\left[\begin{array}{l}0+11+03-1-22-8+6\\ 0+0+09+0+26+0-6\\ 0+0+03+0-32+0+9\end{array}\right]\\ =\left[\begin{array}{l}1100\\ 0110\\ 0011\end{array}\right]\\ \left|A\right|I=11\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]\\ =\left[\begin{array}{l}1100\\ 0110\\ 0011\end{array}\right]\Rightarrow A\left(AdjA\right)=\left(AdjA\right)A=\left|A\right|I\end{array}$

Q.52 Verify A(adj A) = (adj A)A = |A|I

$\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\\ -4-6\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{}\mathrm{A}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\\ -4-6\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\mathrm{A}\right|=\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\\ -4-6\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-12+12=0\\ {\mathrm{A}}_{11}={(-1)}^{1+1}(-6)=-6,\\ {\mathrm{A}}_{12}={(-1)}^{1+2}(-4)=4\\ {\mathrm{A}}_{21}={(-1)}^{2+1}3=-3\\ {\mathrm{A}}_{22}={(-1)}^{2+2}\left(2\right)=2\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{l}-64\\ -32\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}-6-3\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\end{array}\right]\\ \mathrm{A}.\left(\mathrm{AdjA}\right)=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\\ -4\mathrm{}-6\end{array}\right]\left[\begin{array}{l}-6\hspace{0.17em}-3\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{l}-12+12-6+6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}24-24\hspace{0.17em}\hspace{0.17em}12-12\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{l}00\\ 00\end{array}\right]=0\\ \left(\mathrm{AdjA}\right).\mathrm{A}=\left[\begin{array}{l}-6\mathrm{\hspace{0.17em}}-3\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\end{array}\right]\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\\ -4-6\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{l}-12+12-18+18\\ 12-1212-12\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left[\begin{array}{l}00\\ 00\end{array}\right]=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|\mathrm{I}=0\left[\begin{array}{l}10\\ 01\end{array}\right]=0\\ \mathrm{Thus},\\ \mathrm{A}.\left(\mathrm{AdjA}\right)=\left(\mathrm{AdjA}\right).\mathrm{A}=\left|\mathrm{A}\right|\mathrm{I}\end{array}$

Q.53 Find adjoint of each of the matrices

$\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}1-12\\ \hspace{0.17em}\mathrm{\hspace{0.17em}}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}35\\ -2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}01\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{1}-\mathrm{1}\mathrm{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{2}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\mathrm{5}\\ -\mathrm{2\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}0\mathrm{1}\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have}\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}\left|\begin{array}{l}35\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}25\\ -2\mathrm{}1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-(2+10)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-12\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}23\\ -20\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(0+6)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=6\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}\left|\begin{array}{l}-12\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}12\\ -21\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=5\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1-1\\ -2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-1(0-2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2\\ {\mathrm{A}}_{\mathrm{31}}={(-1)}^{3+1}\left|\begin{array}{l}-12\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}35\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-5-6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-11\\ {\mathrm{A}}_{\mathrm{32}}={(-1)}^{3+2}\left|\begin{array}{l}12\\ 25\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-(5-4)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-1\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}\left|\begin{array}{l}1-1\\ 2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3+2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=5\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{3}-12\mathrm{6}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{1\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}5\mathrm{2}\\ -11-1\mathrm{5}\end{array}\right]}^{‘}=\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1-11\\ -12\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}6\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\end{array}\right]\end{array}$

Q.54 Find adjoint of each of the matrices

$\left|\begin{array}{l}12\\ 34\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}1\mathrm{2}\\ \mathrm{3}\mathrm{4}\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have}\\ {\mathrm{A}}_{\mathrm{11}}={(-1)}^{1+1}4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4\\ {\mathrm{A}}_{\mathrm{12}}={(-1)}^{1+2}3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-3\\ {\mathrm{A}}_{\mathrm{21}}={(-1)}^{2+1}2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-2\\ {\mathrm{A}}_{\mathrm{22}}={(-1)}^{2+2}1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=1\\ \mathrm{Adj}\mathrm{\hspace{0.17em}}\mathrm{A}={\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4-3\\ -2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4-2\\ -3\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1\end{array}\right]\end{array}$

Q.55

$\begin{array}{l}\mathrm{If}\mathrm{\Delta }=\left|\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}{\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}{\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}{\mathrm{a}}_{32}{\mathrm{a}}_{33}\end{array}\right|\mathrm{}\mathrm{and}{\mathrm{A}}_{\mathrm{ij}}\mathrm{is}\mathrm{Co}\mathrm{factor}\mathrm{so}{\mathrm{fa}}_{\mathrm{ij}},\mathrm{then}\mathrm{value}\\ \mathrm{of}\mathrm{\Delta }\mathrm{is}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right){\mathrm{a}}_{11}{\mathrm{A}}_{31}+{\mathrm{a}}_{12}{\mathrm{A}}_{32}+{\mathrm{a}}_{13}{\mathrm{A}}_{33}\left(\mathrm{B}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{12}{\mathrm{A}}_{21}+{\mathrm{a}}_{13}{\mathrm{A}}_{31}\\ \left(\mathrm{C}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{a}}_{21}{\mathrm{A}}_{11}+{\mathrm{a}}_{22}{\mathrm{A}}_{12}+{\mathrm{a}}_{23}{\mathrm{A}}_{13}\left(\mathrm{D}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{31}{\mathrm{A}}_{31}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{we}\mathrm{know}\mathrm{that}\\ \mathrm{\Delta }=\mathrm{Sum}\mathrm{of}\mathrm{product}\mathrm{of}\mathrm{elements}\mathrm{of}{\mathrm{C}}_{\mathrm{1}}\mathrm{with}\mathrm{their}\mathrm{corresponding}\mathrm{cofactors}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{31}{\mathrm{A}}_{31}\\ \mathrm{Thus},\mathrm{the}\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{correct}<mi\; mathvariant="normal">.</mi>\end{array}$

Q.56

$\begin{array}{l}\mathrm{Using}\mathrm{Co}\mathrm{factors}\mathrm{of}\mathrm{element}\mathrm{sof}\mathrm{third}\mathrm{column},\mathrm{evaluate}\\ \mathrm{\Delta }=\left|\begin{array}{l}1\mathrm{x}\mathrm{yz}\\ 1\mathrm{y}\mathrm{zx}\\ 1\mathrm{z}\mathrm{xy}\end{array}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{determinant}\mathrm{is}\\ \mathrm{\Delta }=\left|\begin{array}{l}\mathrm{1}\mathrm{x}\mathrm{yz}\\ 1\mathrm{y}\mathrm{zx}\\ 1\mathrm{z}\mathrm{xy}\end{array}\right|\\ \mathrm{Minors}\mathrm{of}\mathrm{elements}\mathrm{of}\mathrm{third}\mathrm{column}\mathrm{are}\\ {\mathrm{M}}_{\mathrm{13}}=\left|\begin{array}{l}1\mathrm{y}\\ 1\mathrm{z}\end{array}\right|=\mathrm{z}-\mathrm{y}\\ {\mathrm{M}}_{\mathrm{23}}=\left|\begin{array}{l}1\mathrm{x}\\ 1\mathrm{z}\end{array}\right|=\mathrm{z}-\mathrm{x}\\ {\mathrm{M}}_{\mathrm{33}}=\left|\begin{array}{l}1\mathrm{x}\\ 1\mathrm{y}\end{array}\right|=\mathrm{y}-\mathrm{x}\\ \mathrm{Cofactors}\mathrm{of}\mathrm{elements}\mathrm{of}\mathrm{third}\mathrm{column}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{13}}={(-1)}^{1+3}{\mathrm{M}}_{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{z}-\mathrm{y}\\ {\mathrm{A}}_{\mathrm{23}}={(-1)}^{2+3}{\mathrm{M}}_{23}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-(\mathrm{z}-\mathrm{x})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{x}-\mathrm{z}\\ {\mathrm{A}}_{\mathrm{33}}={(-1)}^{3+3}{\mathrm{M}}_{33}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{y}-\mathrm{x}\\ \mathrm{Expanding}\mathrm{the}\mathrm{determinant}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }={\mathrm{a}}_{13}{\mathrm{A}}_{13}+{\mathrm{a}}_{23}{\mathrm{A}}_{23}+{\mathrm{a}}_{33}{\mathrm{A}}_{33}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{yz}(\mathrm{z}-\mathrm{y})+\mathrm{zx}(\mathrm{x}-\mathrm{z})+\mathrm{xy}(\mathrm{y}-\mathrm{x})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{yz}}^2-{\mathrm{y}}^2\mathrm{z}+{\mathrm{zx}}^2-{\mathrm{z}}^2\mathrm{x}+{\mathrm{xy}}^2-{\mathrm{x}}^2\mathrm{y}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{zx}}^2-{\mathrm{y}}^2\mathrm{z}-{\mathrm{x}}^2\mathrm{y}+{\mathrm{xy}}^2+{\mathrm{yz}}^2-{\mathrm{z}}^2\mathrm{x}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{z}({\mathrm{x}}^2-{\mathrm{y}}^2)+\mathrm{xy}(\mathrm{y}-\mathrm{x})+{\mathrm{z}}^2(\mathrm{y}-\mathrm{x})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{x}-\mathrm{y})\{\mathrm{z}(\mathrm{x}+\mathrm{y})-\mathrm{xy}-{\mathrm{z}}^2\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{x}-\mathrm{y})(\mathrm{zx}+\mathrm{zy}-\mathrm{xy}-{\mathrm{z}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{x}-\mathrm{y})(\mathrm{zx}-\mathrm{xy}+\mathrm{zy}-{\mathrm{z}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{x}-\mathrm{y})\{-\mathrm{x}(\mathrm{y}-\mathrm{z})+\mathrm{z}(\mathrm{y}-\mathrm{z})\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\end{array}$

Q.57

$\begin{array}{l}\mathrm{Using}\mathrm{Co}\mathrm{factors}\mathrm{of}\mathrm{element}\mathrm{so}\mathrm{f}\mathrm{second}\mathrm{row},\mathrm{evaluate}\\ \mathrm{\Delta }=\left|\begin{array}{l}538\\ 201\\ 123\end{array}\right|\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l}\mathrm{5}\mathrm{3}\mathrm{8}\\ 2\mathrm{0}\mathrm{1}\\ \mathrm{1}2\mathrm{3}\end{array}\right|\\ \mathrm{Minors}\mathrm{corresponding}\mathrm{to}\mathrm{second}\mathrm{row}:\\ {\mathrm{M}}_{\mathrm{21}}=\left|\begin{array}{l}38\\ 23\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=9-16\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-7\\ \because \mathrm{Cofactor}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{ij}}={(-1)}^{\mathrm{i}+\mathrm{j}}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore {\mathrm{A}}_{21}={(-1)}^{2+1}{\mathrm{M}}_{21}\\ =(-1)(-7)\\ =7\\ {\mathrm{M}}_{\mathrm{22}}=\left|\begin{array}{l}58\\ 13\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=15-8\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=7\\ {\mathrm{A}}_{22}={(-1)}^{2+2}{\mathrm{M}}_{22}\\ =7\\ {\mathrm{M}}_{\mathrm{23}}=\left|\begin{array}{l}53\\ 12\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=10-3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=7\\ \mathrm{\hspace{0.17em}}{\mathrm{A}}_{23}={(-1)}^{2+3}{\mathrm{M}}_{23}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(-1)\left(7\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-7\\ \mathrm{Expanding}\mathrm{the}\mathrm{determinant}\mathrm{along}{\mathrm{R}}_{\mathrm{2}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }={\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{22}{\mathrm{A}}_{22}+{\mathrm{a}}_{23}{\mathrm{A}}_{23}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=2\times 7+0\times 7+1\times -7\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=14+0-7\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=7\\ \end{array}$

Q.58 Write Minors and Cofactors of the elements of following determinants:

$\text{(i)}\left|\begin{array}{l}100\\ 010\\ 001\end{array}\right|$ $\text{(ii)}\left|\begin{array}{l}10\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}4\\ 35-1\\ 01\mathrm{\hspace{0.17em}}\mathrm{\hspace{0.17em}}2\end{array}\right|$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{l}1\mathrm{0}\mathrm{0}\\ 01\mathrm{0}\\ 00\mathrm{1}\end{array}\right|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}00\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{13}}={\mathrm{M}}_{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}01\\ 00\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}00\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{23}}={\mathrm{M}}_{23}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\\ 00\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{31}}={\mathrm{M}}_{31}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}00\\ 10\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{32}}={\mathrm{M}}_{32}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\\ 00\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{33}}={\mathrm{M}}_{33}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\\ \mathrm{Now},\mathrm{cofactor}\mathrm{of}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}},\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{ij}}={(-1)}^{\mathrm{i}+\mathrm{j}}{\mathrm{M}}_{\mathrm{ij}}\\ {\mathrm{A}}_{11}={(-1)}^{1+1}{\mathrm{M}}_{11}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1\\ {\mathrm{A}}_{12}={(-1)}^{1+2}{\mathrm{M}}_{12}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\\ {\mathrm{A}}_{13}={(-1)}^{1+3}{\mathrm{M}}_{13}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\\ {\mathrm{A}}_{21}={(-1)}^{2+1}{\mathrm{M}}_{21}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\end{array}$ \begin{array}{l}{A}_{22}={\left(-1\right)}^{2+2}{M}_{22}\\ =1\\ A_{23}={\left(-1\right)}^{2+3}{M}_{23}\\ =0\\ A_{31}={\left(-1\right)}^{3+1}{M}_{31}\\ =0\\ A_{32}={\left(-1\right)}^{3+2}{M}_{32}\\ =0\\ A_{33}={\left(-1\right)}^{3+3}{M}_{33}\\ =1\\ \left(\text{ii}\right)\left|\begin{array}{l}\text{1}\text{0}\text{4}\\ \text{3}\text{5}-\text{1}\\ \text{0}\text{1}\text{2}\end{array}\right|\\ Minor{\text{of the element a}}_{\text{ij}}{\text{is M}}_{\text{ij}}\\ \therefore Minor{\text{of element a}}_{\text{11}}=M_{11}\\ =\left|\begin{array}{l}5-1\\ 12\end{array}\right|\\ =10+1\\ =11\\ Minor{\text{of element a}}_{\text{12}}=M_{12}\\ =\left|\begin{array}{l}3-1\\ 02\end{array}\right|\\ =6\\ Minor{\text{of element a}}_{\text{13}}=M_{13}\\ =\left|\begin{array}{l}35\\ 01\end{array}\right|\\ =3\\ Minor{\text{of element a}}_{\text{21}}=M_{21}\\ =\left|\begin{array}{l}04\\ 12\end{array}\right|\\ =-4\\ Minor{\text{of element a}}_{\text{22}}=M_{22}\\ =\left|\begin{array}{l}14\\ 02\end{array}\right|\\ =2\\ Minor{\text{of element a}}_{\text{23}}=M_{23}\\ =\left|\begin{array}{l}10\\ 01\end{array}\right|\\ =1\\ Minor{\text{of element a}}_{\text{31}}=M_{31}\\ =\left|\begin{array}{l}04\\ 5-1\end{array}\right|\\ =-20\\ Minor{\text{of element a}}_{\text{32}}=M_{32}\\ =\left|\begin{array}{l}14\\ 3-1\end{array}\right|\\ =-13\\ Minor{\text{of element a}}_{\text{33}}=M_{33}\\ =\left|\begin{array}{l}10\\ 35\end{array}\right|\\ =5\\ Now,{\text{cofactor of a}}_{\text{ij}}{\text{is A}}_{\text{ij}}\text{, so}\\ {\text{A}}_{\text{ij}}={\left(-1\right)}^{i+j}{M}_{ij}\\ A_{11}={\left(-1\right)}^{1+1}{M}_{11}\\ =11\\ A_{12}={\left(-1\right)}^{1+2}{M}_{12}\\ =-6\\ A_{13}={\left(-1\right)}^{1+3}{M}_{13}\\ =3\\ A_{21}={\left(-1\right)}^{2+1}{M}_{21}\\ =\left(-1\right)\left(-4\right)\\ =4\\ A_{22}={\left(-1\right)}^{2+2}{M}_{22}\\ =2\\ \end{array} \begin{array}{l}{A}_{23}={\left(-1\right)}^{2+3}{M}_{23}\\ =-1\\ A_{31}={\left(-1\right)}^{3+1}{M}_{31}\\ =-20\\ A_{32}={\left(-1\right)}^{3+2}{M}_{32}\\ =\left(-1\right)\left(-1\right)\\ =13\\ A_{33}={\left(-1\right)}^{3+3}{M}_{33}\\ =5\end{array}

Q.59 Write Minors and Cofactors of the elements of following determinants:

$\left(\mathrm{i}\right)\left|\begin{array}{l}2-4\\ 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\end{array}\right|\left(\mathrm{ii}\right)\left|\begin{array}{l}\mathrm{a}\mathrm{c}\\ \mathrm{b}\mathrm{d}\end{array}\right|$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{l}2-4\\ 0\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}3\end{array}\right|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}=3,\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}=0,\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}=-4,\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}=2.\\ \mathrm{Cofactor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}}={(-1)}^{(\mathrm{i}+\mathrm{j})}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{11}}={(-1)}^{(1+1)}{\mathrm{M}}_{11}=3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{12}}={(-1)}^{(1+2)}{\mathrm{M}}_{12}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{21}}={(-1)}^{(2+1)}{\mathrm{M}}_{21}=4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{22}}={(-1)}^{(2+2)}{\mathrm{M}}_{22}=2\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{l}\mathrm{a}\mathrm{c}\\ \mathrm{b}\mathrm{d}\end{array}\right|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}=\mathrm{d},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}=\mathrm{b},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}=\mathrm{c},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Minor}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}=\mathrm{a}.\\ \mathrm{Cofactor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}}={(-1)}^{(\mathrm{i}+\mathrm{j})}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{11}}={(-1)}^{(1+1)}{\mathrm{M}}_{11}=\mathrm{d}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{\mathrm{12}}={(-1)}^{(1+2)}{\mathrm{M}}_{12}=-\mathrm{b}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{21}}={(-1)}^{(2+1)}{\mathrm{M}}_{21}=-\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{A}}_{\mathrm{22}}={(-1)}^{(2+2)}{\mathrm{M}}_{22}=\mathrm{a}\end{array}$

Q.60 If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=\pm \hspace{0.17em}35\hspace{0.17em}\hspace{0.17em}\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(2,-6),\\ \hspace{0.17em}\hspace{0.17em}(5,4),(\mathrm{k},4)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}2-61\\ 5\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}41\\ \mathrm{k}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}41\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 35=\frac{1}{2}\{2(4-4)-(-6)(5-\mathrm{k})+1(20-4\mathrm{k})\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 35=\frac{1}{2}(0+30-6\mathrm{k}+20-4\mathrm{k})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 70=-10\mathrm{k}+50\\ \Rightarrow \mathrm{\hspace{0.17em}}-10\mathrm{k}+50=70\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}-10\mathrm{k}+50=-70\\ \Rightarrow \mathrm{k}=\frac{20}{-10}=-2\mathrm{or}\mathrm{k}=\frac{-120}{-10}=12\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}12\mathrm{or}-2.\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{D}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.61 (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Ans

$\begin{array}{l}\left(i\right)\text{Point A}(1,2)\text{and B}(3,6)\text{are joining each other and a point P}(x,y)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Areaof\text{triangle ABP}=\text{0}\\ \left|\begin{array}{l}121\\ 361\\ xy1\end{array}\right|=0\\ \Rightarrow \frac{1}{2}\{1(6-y)-2(3-x)+1(3y-6x)\}=0\\ \Rightarrow 6-y-6+2x+3y-6x=0\\ \Rightarrow 2y-4x=0\\ \Rightarrow y=2x\\ \text{Thus, the equation of line joining two points (1, 2) and}(3,6)\\ is\text{}\text{y}=\text{2x}\text{.}\\ \left(ii\right)\text{Point A}(3,1)\text{and B}(9,3)\text{are joining each other and a point P}(x,y)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Areaof\text{triangle ABP}=\text{0}\\ \left|\begin{array}{l}311\\ 931\\ xy1\end{array}\right|=0\\ \Rightarrow \frac{1}{2}\{3(3-y)-1(9-x)+1(9y-3x)\}=0\\ \Rightarrow 9-3y-9+x+9y-3x=0\\ \Rightarrow 6y-2x=0\\ \Rightarrow 3y=x\\ \text{Thus, the equation of line joining two points (1, 2) and}(3,6)\\ is\text{}\text{x}-\text{3y}=\text{0}\text{.}\end{array}$

b

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=\pm \mathrm{4\hspace{0.17em}}\mathrm{\hspace{0.17em}}\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(\mathrm{k},0),\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(4,0),(0,2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}\mathrm{k}01\\ 401\\ 021\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 4=\frac{1}{2}\{\mathrm{k}(0-2)-0(4-0)+1(8-0)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 4=\frac{1}{2}(-2\mathrm{k}-0+8)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 8=-2\mathrm{k}+8\\ \Rightarrow \mathrm{\hspace{0.17em}}-2\mathrm{k}+8=8\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}-2\mathrm{k}+8=-8\\ \Rightarrow \mathrm{k}=0\mathrm{}\mathrm{or}\mathrm{k}=\frac{-16}{-2}=8\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}0\mathrm{or}8\mathrm{.}\\ \left(\mathrm{ii}\right)\hspace{0.17em}\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=\pm \mathrm{4\hspace{0.17em}}\mathrm{\hspace{0.17em}}\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(-2,0),\\ \hspace{0.17em}\hspace{0.17em}(0,4),(0,\mathrm{k})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}-201\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}041\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\mathrm{k}1\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 4=\frac{1}{2}\{-2(4-\mathrm{k})-0(0-0)+1(0-0)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 4=\frac{1}{2}(-8+2\mathrm{k}-0+0)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\pm 8=2\mathrm{k}-8\\ \Rightarrow \mathrm{\hspace{0.17em}}2\mathrm{k}-8=8\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}2\mathrm{k}-8=-8\\ \Rightarrow \mathrm{k}=\frac{16}{2}=\mathrm{\hspace{0.17em}}8\mathrm{}\mathrm{or}\mathrm{k}=\frac{0}{2}=0\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}8\mathrm{or}0\mathrm{.}\end{array}$

Q.63 Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Ans

$\begin{array}{l}\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\mathrm{A}(\mathrm{a},\mathrm{b}+\mathrm{c}),\\ \mathrm{B}(\mathrm{b},\mathrm{c}+\mathrm{a}),\mathrm{C}(\mathrm{c},\mathrm{a}+\mathrm{b})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{a}& \mathrm{b}+\mathrm{c}& 1\\ \mathrm{b}& \mathrm{c}+\mathrm{a}& 1\\ \mathrm{c}& \mathrm{a}+\mathrm{b}& 1\end{array}\right|\\ =\frac{1}{2}\{\mathrm{a}(\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b})-(\mathrm{b}+\mathrm{c})(\mathrm{b}-\mathrm{c})+1(\mathrm{ba}+{\mathrm{b}}^2-{\mathrm{c}}^2-\mathrm{ca})\}\\ =\frac{1}{2}\{\mathrm{ac}-\mathrm{ab}-({\mathrm{b}}^2-{\mathrm{c}}^2)+(\mathrm{ba}+{\mathrm{b}}^2-{\mathrm{c}}^2-\mathrm{ca})\}\\ =\frac{1}{2}(\cancel{\mathrm{ac}}-\bcancel{\mathrm{ab}}-\bcancel{{\mathrm{b}}^2}+\bcancel{{\mathrm{c}}^2}+\cancel{\mathrm{ba}}+\bcancel{{\mathrm{b}}^2}-\bcancel{{\mathrm{c}}^2}-\cancel{\mathrm{ca}})\\ =0\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{is}\mathrm{zero},\mathrm{so}\mathrm{these}\\ \mathrm{points}\mathrm{are}\mathrm{collinear}\mathrm{.}\end{array}$

Q.64 Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(1,0),(6,0),(4,3)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}101\\ 601\\ 431\end{array}\right|\\ =\frac{1}{2}\{1(0-3)-0(6-4)+1(18-0)\}\\ =\frac{1}{2}(-3-0+18)\\ =\frac{1}{2}\left(15\right)\\ =\frac{15}{2}\mathrm{\hspace{0.17em}}\mathrm{sq}.\mathrm{\hspace{0.17em}}\mathrm{units}\\ \left(\mathrm{ii}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(2,7),(1,1),(10,8)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}271\\ 111\\ 1081\end{array}\right|\\ =\frac{1}{2}\{2(1-8)-7(1-10)+1(8-10)\}\\ =\frac{1}{2}(-14+63-2)\\ =\frac{1}{2}\left(47\right)\\ =\frac{47}{2}\mathrm{\hspace{0.17em}}\mathrm{sq}.\mathrm{\hspace{0.17em}}\mathrm{units}\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}(-2,-3),\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(3,2),(-1,-8)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{l}-2-31\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}21\\ -1-81\end{array}\right|\\ =\frac{1}{2}\{-2(2+8)-(-3)(3+1)+1(-24+2)\}\\ =\frac{1}{2}(-20+12-22)\\ =\frac{1}{2}(-30)\\ =-15\mathrm{\hspace{0.17em}}\mathrm{sq}.\mathrm{\hspace{0.17em}}\mathrm{units}\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{can}\mathrm{never}\mathrm{be}\mathrm{negative}.\\ \mathrm{So},\mathrm{area}\mathrm{of}\mathrm{triangle}=\; 15\mathrm{square}\mathrm{units}\mathrm{.}\end{array}$

Q.65 Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these.

Ans

To every square matrix A = [a_ij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where a_ij = (i,j)^th element of A.
Therefore, option (C) is correct

Q.66 Let A b a square matrix of order 3 x 3, then |kA| is equal to

(A) k|A| (B) k²|A| (C) k³|A| (D) 3k|A|

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{A}\mathrm{be}\mathrm{a}\mathrm{square}\mathrm{matrix}\mathrm{of}\mathrm{order}3\times 3,\\ \mathrm{then}\hspace{0.17em}\hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_1{\mathrm{a}}_2{\mathrm{a}}_3\\ {\mathrm{b}}_1{\mathrm{b}}_2{\mathrm{b}}_3\\ {\mathrm{c}}_1{\mathrm{c}}_2{\mathrm{c}}_3\end{array}\right]\\ \mathrm{and}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{kA}=\left[\begin{array}{l}{\mathrm{ka}}_1{\mathrm{ka}}_2{\mathrm{ka}}_3\\ {\mathrm{kb}}_1{\mathrm{kb}}_2{\mathrm{kb}}_3\\ {\mathrm{kc}}_1{\mathrm{kc}}_2{\mathrm{kc}}_3\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\mathrm{kA}\right|=\left|\begin{array}{l}{\mathrm{ka}}_1{\mathrm{ka}}_2{\mathrm{ka}}_3\\ {\mathrm{kb}}_1{\mathrm{kb}}_2{\mathrm{kb}}_3\\ {\mathrm{kc}}_1{\mathrm{kc}}_2{\mathrm{kc}}_3\end{array}\right|\\ ={\mathrm{k}}^3\left|\begin{array}{l}{\mathrm{a}}_1{\mathrm{a}}_2{\mathrm{a}}_3\\ {\mathrm{b}}_1{\mathrm{b}}_2{\mathrm{b}}_3\\ {\mathrm{c}}_1{\mathrm{c}}_2{\mathrm{c}}_3\end{array}\right|[\mathrm{Taking}\mathrm{k}\mathrm{common}\mathrm{from}{\mathrm{R}}_{\mathrm{1}},\hspace{0.17em}{\mathrm{R}}_{\mathrm{2}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}.]\\ ={\mathrm{k}}^3\left|\mathrm{A}\right|\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{C}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.67 By using properties of determinants, show that:

$\left|\begin{array}{l}1+{\mathrm{a}}^2\mathrm{ab}\mathrm{ab}\\ \mathrm{ab}{\mathrm{b}}^2+1\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^2+1\end{array}\right|=1+{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{1}+{\mathrm{a}}^2\mathrm{ab}\mathrm{ac}\\ \mathrm{ab}{\mathrm{b}}^{\mathrm{2}}+\mathrm{1}\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}\right|\\ \mathrm{Applying}\hspace{0.17em}\hspace{0.17em}{\mathrm{R}}_1\to \frac{1}{\mathrm{a}}{\mathrm{R}}_1,{\mathrm{R}}_2\to \frac{1}{\mathrm{b}}{\mathrm{R}}_2,{\mathrm{R}}_3\to \frac{1}{\mathrm{c}}{\mathrm{R}}_3\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{abc}\left|\begin{array}{l}\frac{\mathrm{1}}{\mathrm{a}}+\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}\mathrm{b}+\frac{\mathrm{1}}{\mathrm{b}}\mathrm{c}\\ \mathrm{a}\mathrm{b}\mathrm{c}+\frac{\mathrm{1}}{\mathrm{c}}\end{array}\right|\\ \mathrm{Applying}\hspace{0.17em}\hspace{0.17em}{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2,{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{abc}\left|\begin{array}{l}\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\mathrm{0}\\ \mathrm{0\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{1}}{\mathrm{b}}-\frac{1}{\mathrm{c}}\\ \mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{b}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{c}+\frac{\mathrm{1}}{\mathrm{c}}\end{array}\right|\\ \mathrm{Applying}\hspace{0.17em}\hspace{0.17em}{\mathrm{C}}_1\to {\mathrm{aC}}_1,{\mathrm{C}}_2\to {\mathrm{bC}}_2,{\mathrm{C}}_3\to {\mathrm{cC}}_3\\ \hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{abc}}{\mathrm{abc}}\left|\begin{array}{l}1-1\mathrm{0}\\ \mathrm{0}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1-1\\ {\mathrm{a}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}\right|\\ \mathrm{Applying}\hspace{0.17em}\hspace{0.17em}{\mathrm{C}}_2\to {\mathrm{C}}_2+{\mathrm{C}}_1\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}10\mathrm{0}\\ \mathrm{0}\mathrm{\hspace{0.17em}}1-1\\ {\mathrm{a}}^2{\mathrm{a}}^2+\mathrm{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}\right|\\ \mathrm{Expanding}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}1-1\\ {\mathrm{a}}^2+\mathrm{\hspace{0.17em}}{\mathrm{b}}^2{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=({\mathrm{c}}^{\mathrm{2}}+\mathrm{1})-(-1)({\mathrm{a}}^2+\mathrm{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{2}})\\ \hspace{0.17em}\hspace{0.17em}=1+{\mathrm{c}}^2+{\mathrm{a}}^2+\mathrm{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{2}}\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.68 By using properties of determinants, show that:

$\left|\begin{array}{l}1+{\mathrm{a}}^2-{\mathrm{b}}^{2}2\mathrm{ab}-2\mathrm{b}\\ 2\mathrm{ab}1-{\mathrm{a}}^2+{\mathrm{b}}^{2}2\mathrm{a}\\ 2\mathrm{b}-2\mathrm{a}1-{\mathrm{a}}^2-{\mathrm{b}}^2\end{array}\right|={(1+{\mathrm{a}}^2+{\mathrm{b}}^2)}^3$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left|\begin{array}{l}\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{ab}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-2\mathrm{b}\\ 2\mathrm{ab}\mathrm{1}-{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{a}\\ 2\mathrm{b}-2\mathrm{a}\mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}\right|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{bR}}_{\mathrm{3}},{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{aR}}_{\mathrm{3}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{ccc}\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}& \mathrm{0}& -\mathrm{b}(1+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})\\ \mathrm{0}& \mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}& \mathrm{\hspace{0.17em}}\mathrm{a}(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})\\ 2\mathrm{b}& -2\mathrm{a}& \mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})}^2\left|\begin{array}{ccc}\mathrm{1}& \mathrm{0}& -\mathrm{b}\\ \mathrm{0}& \mathrm{1}& \mathrm{\hspace{0.17em}}\mathrm{a}\\ 2\mathrm{b}& -2\mathrm{a}& \mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})}^2[1\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1\mathrm{a}\\ -2\mathrm{a}1-{\mathrm{a}}^2-{\mathrm{b}}^2\end{array}\right|+(-\mathrm{b})\left|\begin{array}{l}01\\ 2\mathrm{b}-1\mathrm{a}\end{array}\right|]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})}^2(1-{\mathrm{a}}^2-{\mathrm{b}}^2+2{\mathrm{a}}^2-\mathrm{b}\times -2\mathrm{b})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}})}^2(1-{\mathrm{a}}^2-{\mathrm{b}}^2+2{\mathrm{a}}^2+2{\mathrm{b}}^2)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(1+{\mathrm{a}}^2+{\mathrm{b}}^2)}^{\mathrm{3}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.69 By using properties of determinants, show that:

$\left|\begin{array}{l}1\mathrm{x}{\mathrm{x}}^2\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^{2}1\end{array}\right|={(1-{\mathrm{x}}^3)}^2$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{1}\mathrm{x}{\mathrm{x}}^{\mathrm{2}}\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^2\mathrm{1}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^2\mathrm{1}\end{array}\right|\left[{\mathrm{R}}_1\to {\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}=(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2)\left|\begin{array}{l}111\\ {\mathrm{x}}^2\mathrm{1}\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^2\mathrm{1}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2)\left|\begin{array}{l}1\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}00\\ {\mathrm{x}}^2\mathrm{1}-{\mathrm{x}}^2\mathrm{x}-{\mathrm{x}}^{\mathrm{2}}\\ \mathrm{x}{\mathrm{x}}^{\mathrm{2}}-\mathrm{x}\mathrm{1}-\mathrm{x}\end{array}\right|\left[\begin{array}{l}{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1,\\ {\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}=(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2)\left|\begin{array}{l}1\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}00\\ {\mathrm{x}}^2(\mathrm{1}-\mathrm{x})(\mathrm{1}+\mathrm{x})\mathrm{x}(1-\mathrm{x})\\ \mathrm{x}\mathrm{x}(\mathrm{x}-1)(\mathrm{1}-\mathrm{x})\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2)(1-\mathrm{x})(1-\mathrm{x})\left|\begin{array}{l}1\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}00\\ {\mathrm{x}}^2(\mathrm{1}+\mathrm{x})\mathrm{x}\\ \mathrm{x}-\mathrm{x}1\end{array}\right|\\ \left[\begin{array}{l}\mathrm{Taking}\mathrm{}(1-\mathrm{x})\hspace{0.17em}\hspace{0.17em}\mathrm{common}\mathrm{from}\\ {\mathrm{C}}_{\mathrm{2}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\mathrm{.}\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}=(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^2)(1-\mathrm{x})(1-\mathrm{x})(1+\mathrm{x}+{\mathrm{x}}^2)\\ \hspace{0.17em}\hspace{0.17em}=\left\{(1-\mathrm{x})(1+\mathrm{x}+{\mathrm{x}}^2)\right\}\left\{(1-\mathrm{x})(1+\mathrm{x}+{\mathrm{x}}^2)\right\}\\ \hspace{0.17em}\hspace{0.17em}=(1-{\mathrm{x}}^3)(1-{\mathrm{x}}^3)\\ \hspace{0.17em}\hspace{0.17em}={(1–{\mathrm{x}}^{\mathrm{3}})}^{\mathrm{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Hence}\mathrm{Proved}\mathrm{.}\end{array}$

Q.70 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|={(\mathrm{a}+\mathrm{b}+\mathrm{c})}^3\\ \left(\mathrm{ii}\right)\left|\begin{array}{l}\mathrm{x}+\mathrm{y}+2\mathrm{z}\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\\ \mathrm{z}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}+\mathrm{z}+2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\\ \mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{x}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}\right|=2{(\mathrm{x}+\mathrm{y}+\mathrm{z})}^3\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\mathrm{=}{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^{\mathrm{3}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\left|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\\ =\left|\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{a}+\mathrm{b}+\mathrm{c}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\left[\begin{array}{l}\mathrm{Apply}\\ {\mathrm{R}}_1\to {\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3\end{array}\right]\\ =(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{l}111\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\mathrm{\hspace{0.17em}}2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\\ =(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{l}1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}00\\ 2\mathrm{b}-\mathrm{b}-\mathrm{c}-\mathrm{a}\hspace{0.17em}0\\ 2\mathrm{c}0-\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[\begin{array}{l}{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1,\mathrm{\hspace{0.17em}}\\ {\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1\end{array}\right]\\ =(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{l}-\mathrm{b}-\mathrm{c}-\mathrm{a}\mathrm{\hspace{0.17em}}\mathrm{0}\\ \mathrm{0\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|\\ =(\mathrm{a}+\mathrm{b}+\mathrm{c}){(-\mathrm{b}-\mathrm{c}-\mathrm{a})}^2\\ ={(\mathrm{a}+\mathrm{b}+\mathrm{c})}^3\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{x}+\mathrm{y}+2\mathrm{z}\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ \mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}+\mathrm{z}+2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ \mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{x}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{x}\mathrm{y}\\ 2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{y}+\mathrm{z}+2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ 2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{x}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\left[{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})\left|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 1\mathrm{y}+\mathrm{z}+2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ 1\mathrm{x}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})\left|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 0\mathrm{y}+\mathrm{z}+\mathrm{x}\hspace{0.17em}\mathrm{0}\\ 00\mathrm{z}+\mathrm{x}+\mathrm{y}\end{array}\right|\left[\begin{array}{l}{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1\\ {\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})}^3\left|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 01\hspace{0.17em}0\\ 001\end{array}\right|\\ \mathrm{Expanding}\hspace{0.17em}\hspace{0.17em}\mathrm{along}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{C}}_1,\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})}^3\left|\begin{array}{l}10\\ 01\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})}^3(1-0)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={(2\mathrm{x}+2\mathrm{y}+2\mathrm{z})}^3=\mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\end{array}$

Q.71 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 2\mathrm{x}\mathrm{x}+42\mathrm{x}\\ 2\mathrm{x}2\mathrm{x}\mathrm{x}+4\end{array}\right|=(5\mathrm{x}+4){(4-\mathrm{x})}^2\\ \end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\left|\begin{array}{l}\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}\\ \mathrm{y}\mathrm{y}+4\mathrm{y}\\ \mathrm{y}\mathrm{y}\mathrm{y}+4\end{array}\right|={\mathrm{k}}^2(3\mathrm{y}+\mathrm{k})\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 2\mathrm{x}\mathrm{x}+42\mathrm{x}\\ 2\mathrm{x}2\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}5\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 5\mathrm{x}+\mathrm{5}\mathrm{x}+42\mathrm{x}\\ 5\mathrm{x}+\mathrm{5}2\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}\right|\mathrm{\hspace{0.17em}}\left[\mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(5\mathrm{x}+4)\left|\begin{array}{l}12\mathrm{x}2\mathrm{x}\\ 1\mathrm{x}+42\mathrm{x}\\ 12\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(5\mathrm{x}+4)\left|\begin{array}{l}12\mathrm{x}2\mathrm{x}\\ 0\mathrm{4}-\mathrm{x}\mathrm{0}\\ 0\mathrm{0}\mathrm{4}-\mathrm{x}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Apply}\\ {\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1,{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(5\mathrm{x}+4)\left|\begin{array}{l}4-\mathrm{x}0\\ 04-\mathrm{x}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(5\mathrm{x}+4){(4-\mathrm{x})}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left|\begin{array}{l}\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}\\ \mathrm{y}\mathrm{y}+\mathrm{k}\mathrm{y}\\ \mathrm{y}\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}3\mathrm{y}+\mathrm{k}\mathrm{y}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\\ 3\mathrm{y}+\mathrm{k}\mathrm{y}+\mathrm{k}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ 3\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}\right|\mathrm{\hspace{0.17em}}\left[\mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(3\mathrm{y}+\mathrm{k})\left|\begin{array}{l}1\mathrm{y}\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\\ 1\mathrm{y}+\mathrm{k}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\\ 1\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}\right|\\ \left[\mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(3\mathrm{y}+\mathrm{k})\left|\begin{array}{l}1\mathrm{y}\mathrm{y}\\ 0\mathrm{k}\mathrm{0}\\ 00\mathrm{k}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(3\mathrm{y}+\mathrm{k})\left|\begin{array}{l}\mathrm{k}0\\ 0\mathrm{k}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(3\mathrm{y}+\mathrm{k})\times {\mathrm{k}}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{k}}^{\mathrm{2}}(3\mathrm{y}+\mathrm{k})\end{array}$

Q.72 By using properties of determinants, show that:

$\left|\begin{array}{l}\mathrm{x}{\mathrm{x}}^2\mathrm{yz}\\ \mathrm{y}{\mathrm{y}}^2\mathrm{zx}\\ \mathrm{z}{\mathrm{z}}^2\mathrm{xy}\end{array}\right|=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{x}{\mathrm{x}}^2\mathrm{yz}\\ \mathrm{y}{\mathrm{y}}^2\mathrm{zx}\\ \mathrm{z}{\mathrm{z}}^2\mathrm{xy}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}\mathrm{x}-\mathrm{y}{\mathrm{x}}^{\mathrm{2}}-{\mathrm{y}}^2\mathrm{yz}-\mathrm{zx}\\ \mathrm{y}-\mathrm{z}{\mathrm{y}}^{\mathrm{2}}-{\mathrm{z}}^2\mathrm{zx}-\mathrm{xy}\\ \mathrm{z}{\mathrm{z}}^2\mathrm{xy}\end{array}\right|\\ \left[\mathrm{Applying}\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}},\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{3}}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}(\mathrm{x}-\mathrm{y})(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})-\mathrm{z}(\mathrm{x}-\mathrm{y})\\ (\mathrm{y}-\mathrm{z})(\mathrm{y}-\mathrm{z})(\mathrm{y}+\mathrm{z})-\mathrm{x}(\mathrm{y}-\mathrm{z})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{z}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{xy}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})\left|\begin{array}{l}1(\mathrm{x}+\mathrm{y})-\mathrm{z}\\ 1(\mathrm{y}+\mathrm{z})-\mathrm{x}\\ \mathrm{\hspace{0.17em}}\mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{z}}^{\mathrm{2}}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{xy}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Taking}\mathrm{common}(\mathrm{x}-\mathrm{y})\mathrm{\hspace{0.17em}}\\ \mathrm{and}\mathrm{\hspace{0.17em}}(\mathrm{y}-\mathrm{z})\mathrm{\hspace{0.17em}}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{2}}\\ \mathrm{respectively}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})\left|\begin{array}{l}0(\mathrm{x}-\mathrm{z})(\mathrm{x}-\mathrm{z})\\ 1(\mathrm{y}+\mathrm{z})-\mathrm{x}\\ \mathrm{\hspace{0.17em}}\mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{z}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\mathrm{xy}\end{array}\right|\\ \left[\mathrm{Applying}\mathrm{\hspace{0.17em}}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\left|\begin{array}{l}0-1-1\\ 1(\mathrm{y}+\mathrm{z})-\mathrm{x}\\ \mathrm{\hspace{0.17em}}\mathrm{z}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{z}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\mathrm{xy}\end{array}\right|\left[\begin{array}{l}\mathrm{Taking}\mathrm{common}\\ (\mathrm{z}-\mathrm{x})\mathrm{\hspace{0.17em}}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\left|\begin{array}{l}00-1\\ 1(\mathrm{x}+\mathrm{y}+\mathrm{z})-\mathrm{x}\\ \mathrm{\hspace{0.17em}}\mathrm{z}\mathrm{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}{\mathrm{z}}^2-\mathrm{xy}\hspace{0.17em}\hspace{0.17em}\mathrm{xy}\end{array}\right|\left[\begin{array}{l}\mathrm{Applying}\hspace{0.17em}\hspace{0.17em}\\ {\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_3\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\left\{(-1)\left|\begin{array}{l}1(\mathrm{x}+\mathrm{y}+\mathrm{z})\\ \mathrm{z}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{z}}^2-\mathrm{xy}\end{array}\right|\right\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\times (-1)\{\mathrm{\hspace{0.17em}}{\mathrm{z}}^2-\mathrm{xy}-\left(\mathrm{z}\right)(\mathrm{x}+\mathrm{y}+\mathrm{z})\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})\times (-1)\{\mathrm{\hspace{0.17em}}{\mathrm{z}}^2-\mathrm{xy}-\mathrm{zx}-\mathrm{zy}-{\mathrm{z}}^2\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}\mathrm{a}$

Q.73 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}1\mathrm{a}{\mathrm{a}}^2\\ 1\mathrm{b}{\mathbf{b}}^2\\ 1\mathrm{c}{\mathrm{c}}^2\end{array}\right|=(\mathrm{a}–\mathrm{b})(\mathrm{b}–\mathrm{c})(\mathrm{c}–\mathrm{a})\\ \left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{l}111\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ {\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3\end{array}\right|=(\mathrm{a}–\mathrm{b})(\mathrm{b}–\mathrm{c})(\mathrm{c}–\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}1\mathrm{a}{\mathrm{a}}^{\mathrm{2}}\\ \mathrm{1}\mathrm{b}{\mathrm{b}}^{\mathrm{2}}\\ \mathrm{1}\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}0\mathrm{a}-\mathrm{b}{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^2\\ \mathrm{0}\mathrm{b}-\mathrm{c}{\mathrm{b}}^{\mathrm{2}}-{\mathrm{c}}^2\\ \mathrm{1}\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}\right|\left[\mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{3}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{l}01\mathrm{a}+\mathrm{b}\\ 01\mathrm{b}+\mathrm{c}\\ 1\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{l}1\mathrm{a}+\mathrm{b}\\ 1\mathrm{b}+\mathrm{c}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c}-\mathrm{a}-\mathrm{b})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}11\mathrm{1}\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ {\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^{\mathrm{3}}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}0\mathrm{0}\mathrm{1}\\ \mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}\\ {\mathrm{a}}^{\mathrm{3}}-{\mathrm{b}}^3{\mathrm{b}}^{\mathrm{3}}-{\mathrm{c}}^3{\mathrm{c}}^{\mathrm{3}}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\\ \left[\mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}-{\mathrm{C}}_{\mathrm{2}}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{3}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}0\mathrm{0}\mathrm{1}\\ \mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}\\ (\mathrm{a}-\mathrm{b})({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2)(\mathrm{b}-\mathrm{c})({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2){\mathrm{c}}^{\mathrm{3}}\end{array}\right|\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{l}\mathrm{0}\mathrm{0}\mathrm{1}\\ 11\mathrm{c}\\ ({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2)({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2){\mathrm{c}}^{\mathrm{3}}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{l}11\\ ({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2)({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2)\end{array}\right|\mathrm{\hspace{0.17em}}[\mathrm{Expending}\mathrm{along}{\mathrm{R}}_{\mathrm{1}}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\{({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2)-({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2)\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2-{\mathrm{a}}^2-\mathrm{ab}-{\mathrm{b}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{bc}+{\mathrm{c}}^2-{\mathrm{a}}^2-\mathrm{ab})\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\{{\mathrm{c}}^2-{\mathrm{a}}^2+\mathrm{bc}-\mathrm{ab}\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\{(\mathrm{c}-\mathrm{a})(\mathrm{c}+\mathrm{a})-\mathrm{b}(\mathrm{c}-\mathrm{a})\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})\end{array}$

Q.74 By using properties of determinants, show that:

$\left|\begin{array}{l}-{\mathrm{a}}^2\mathrm{ab}\mathrm{ac}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathbf{b}\mathrm{a}-{\mathrm{b}}^2\mathbf{b}\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ca}\mathrm{cb}-{\mathrm{c}}^2\end{array}\right|=4{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2$

Ans

$\begin{array}{c}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\mathrm{\Delta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}-{\mathrm{a}}^{\mathrm{2}}\mathrm{abac}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ba}-{\mathrm{b}}^2\mathrm{bc}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{cacb}-{\mathrm{c}}^2\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{l}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}-\mathrm{b}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{c}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{c}-\mathrm{c}\end{array}\right|\left[\begin{array}{l}\mathrm{Applying}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{C}}_1\to \frac{1}{\mathrm{a}}{\mathrm{C}}_1,{\mathrm{C}}_2\to \frac{1}{\mathrm{b}}{\mathrm{C}}_2\mathrm{\hspace{0.17em}}\\ \mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{C}}_3\to \frac{1}{\mathrm{c}}{\mathrm{C}}_3\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{l}-\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}-\mathrm{b}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{c}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{c}-\mathrm{c}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\left|\begin{array}{l}0\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\\ 2\mathrm{b}-\mathrm{b}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{b}\\ \mathrm{0\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\mathrm{c}-\mathrm{c}\end{array}\right|\left[\mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}+{\mathrm{C}}_{\mathrm{3}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\times -2\mathrm{b}\left|\begin{array}{l}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{a}\\ \mathrm{c}-\mathrm{c}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\times -2\mathrm{b}\left(-\mathrm{ac}-\mathrm{ac}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{abc}\times -2\mathrm{b}\left(-2\mathrm{ac}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=4{\mathrm{a}}^{\mathrm{2}}{\mathrm{b}}^{\mathrm{2}}{\mathrm{c}}^{\mathrm{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}$

Q.75 By using properties of determinants, show that:

$\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0a-b\\ -a0-c\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bc\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|=\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\mathrm{a}-\mathrm{b}\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{bc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\mathrm{c}}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\mathrm{ac}-\mathrm{bc}\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{bc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\left[{\mathrm{R}}_1\to {\mathrm{cR}}_1\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\mathrm{c}}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{abac}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{bc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\left[{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{bR}}_2\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{a}}{\mathrm{c}}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{bc}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{bc}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\left[\mathrm{Taking}\mathrm{a}\mathrm{common}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\mathrm{.}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{a}}{\mathrm{c}}\times 0\left[\mathrm{Two}\mathrm{rows}\mathrm{are}\mathrm{identical},\mathrm{so}\mathrm{\Delta }=0.\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\end{array}$

Q.76

$\left|\begin{array}{l}\mathrm{b}+\mathrm{cq}+\mathrm{ry}+\mathrm{z}\\ \mathrm{c}+\mathrm{ar}+\mathrm{pz}+\mathrm{x}\\ \mathrm{a}+\mathrm{bp}+\mathrm{qx}+\mathrm{y}\end{array}\right|=2\left|\begin{array}{l}\\ \\ \end{array}\right|$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b}\mathrm{p}+\mathrm{q}\mathrm{x}+\mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}\right|+\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}={\mathrm{\Delta }}_1+{\mathrm{\Delta }}_2\left[\mathrm{Let}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{\Delta }}_1=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2\right]\\ \hspace{0.17em}\hspace{0.17em}=(-1)\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[{\mathrm{R}}_1\leftrightarrow {\mathrm{R}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{\Delta }}_2=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\left[{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\left[{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1\right]{(-1)}^2\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[{\mathrm{R}}_1\leftrightarrow {\mathrm{R}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}\right|\left[{\mathrm{R}}_1\leftrightarrow {\mathrm{R}}_2\right]\\ \hspace{0.17em}\hspace{0.17em}=\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}\right|\left[{\mathrm{R}}_2\leftrightarrow {\mathrm{R}}_3\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{\Delta }={\mathrm{\Delta }}_1+{\mathrm{\Delta }}_2\\ =\mathrm{2}\left|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}\right|\end{array}$

Q.77 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}1\mathrm{bc}\mathrm{a}(\mathrm{b}+\mathrm{c})\\ 1\mathrm{ca}\mathrm{b}(\mathrm{c}+\mathrm{a})\\ 1\mathrm{ab}\mathrm{c}(\mathrm{a}+\mathrm{b})\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{a}(\mathrm{b}+\mathrm{c})\\ \mathrm{1}\mathrm{ca}\mathrm{b}(\mathrm{c}+\mathrm{a})\\ \mathrm{1}\mathrm{ab}\mathrm{c}(\mathrm{a}+\mathrm{b})\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{ab}+\mathrm{ac}\\ 1\mathrm{ca}\mathrm{bc}+\mathrm{ba}\\ \mathrm{1}\mathrm{ab}\mathrm{ca}+\mathrm{cb}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\\ 1\mathrm{ca}\mathrm{bc}+\mathrm{ba}+\mathrm{ca}\\ \mathrm{1}\mathrm{ab}\mathrm{ca}+\mathrm{cb}+\mathrm{ab}\end{array}\right|\left[\mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{C}}_3\to {\mathrm{C}}_3+{\mathrm{C}}_2\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\left|\begin{array}{l}1\mathrm{bc}\mathrm{1}\\ \mathrm{1}\mathrm{ca}\mathrm{1}\\ 1\mathrm{ab}\mathrm{1}\end{array}\right|[\mathrm{Taking}\mathrm{common}(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\mathrm{from}{\mathrm{C}}_{\mathrm{3}}.]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\times 0[\mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\end{array}$

Q.78 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}2765\\ 3875\\ 5986\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l}2765\\ 3875\\ 5986\end{array}\right|\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_3+{\mathrm{C}}_1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}2763+\mathrm{2}\\ 3872+\mathrm{3}\\ 5981+\mathrm{5}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}2763\\ 3872\\ 5981\end{array}\right|+\left|\begin{array}{l}272\\ 383\\ 595\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}2763\\ 3872\\ 5981\end{array}\right|+\mathrm{0}\left[\because \mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}\right]\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{3}}\to \frac{1}{9}{\mathrm{C}}_3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=9\left|\begin{array}{l}277\\ 388\\ 599\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=9\times 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left[\because \mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\end{array}$

Q.79 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}\right|=0$

Ans

$\begin{array}{l}\left|\begin{array}{l}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}\right|=\left|\begin{array}{l}\mathrm{a}-\mathrm{b}+\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}+\mathrm{b}-\mathrm{c}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}\right|\\ \left[\mathrm{Apply}\mathrm{\hspace{0.17em}}{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left|\begin{array}{l}0\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ 0\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ 0\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=0\\ [\mathrm{Since},\mathrm{}\mathrm{all}\mathrm{the}\mathrm{elements}\mathrm{of}\mathrm{a}\mathrm{column}\mathrm{are}\mathrm{zero}\mathrm{.}]\end{array}$

Q.80 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}+\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{y}+\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{z}+\mathrm{c}\end{array}\right|=0$

Ans

$\begin{array}{l}\left|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}+\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{y}+\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{z}+\mathrm{c}\end{array}\right|=\left|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}\\ \mathrm{y}\mathrm{b}\mathrm{y}\\ \mathrm{z}\mathrm{c}\mathrm{z}\end{array}\right|+\left|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{c}\end{array}\right|\\ =0+0\left[\begin{array}{l}\mathrm{If}\mathrm{two}\mathrm{columns}\mathrm{of}\mathrm{a}\mathrm{determinant}\mathrm{is}\\ \mathrm{identical},\mathrm{then}\mathrm{their}\mathrm{value}\mathrm{is}\mathrm{zero}.\end{array}\right]\\ =0\end{array}$

Q.81

$\begin{array}{l}\mathrm{If}\hspace{0.17em}\hspace{0.17em}\left|\begin{array}{l}\mathrm{x}2\\ 18\mathrm{x}\end{array}\right|=\left|\begin{array}{l}62\\ 186\end{array}\right|,\mathrm{then}\mathrm{x}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)6\left(\mathrm{B}\right)\pm 6(\mathrm{C})-6\left(\mathrm{D}\right)0.\end{array}$

Ans

$\begin{array}{l}\because \mathrm{\hspace{0.17em}}\left|\begin{array}{l}\mathrm{x}\mathrm{2}\\ 18\mathrm{x}\end{array}\right|=\left|\begin{array}{l}6\mathrm{2}\\ \mathrm{\hspace{0.17em}}18\mathrm{6}\end{array}\right|\\ \therefore {\mathrm{x}}^2-36=36-36\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^2=36\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\pm 6\\ \mathrm{Therefore},\mathrm{option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.82

$\begin{array}{l}\mathrm{Find}\mathrm{value}\mathrm{so}\mathrm{fx},\mathrm{if}\\ \left(\mathrm{i}\right)\left|\begin{array}{l}24\\ 51\end{array}\right|=\left|\begin{array}{l}2\mathrm{x}4\\ \mathrm{\hspace{0.17em}}6\mathrm{x}\end{array}\right|\left(\mathrm{ii}\right)\left|\begin{array}{l}23\\ 45\end{array}\right|=\left|\begin{array}{l}\mathrm{x}3\\ 2\mathrm{x}5\end{array}\right|\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}2\mathrm{4}\\ \mathrm{5}\mathrm{1}\end{array}\right|=\left|\begin{array}{l}2\mathrm{x}\mathrm{4}\\ \mathrm{\hspace{0.17em}}6\mathrm{x}\end{array}\right|\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2-20=2{\mathrm{x}}^2-24\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-18=2{\mathrm{x}}^2-24\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}24-18=2{\mathrm{x}}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}^2=\frac{6}{2}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\pm \sqrt{3}\\ \left(\mathrm{ii}\right)\left|\begin{array}{l}2\mathrm{3}\\ \mathrm{4}\mathrm{5}\end{array}\right|\mathrm{=}\left|\begin{array}{l}\mathrm{x}\mathrm{3}\\ 2\mathrm{x}\mathrm{5}\end{array}\right|\\ \Rightarrow 10-12=5\mathrm{x}-6\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-2=-\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=2.\end{array}$

Q.83

$\text{If A=}\text{[}\begin{array}{l}11-2\\ 21-3\\ 54-9\end{array}],\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{find}\mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|.$

Ans

$\begin{array}{l}\mathrm{\hspace{0.17em}}\mathrm{A}=\left[\begin{array}{l}\mathrm{1}\mathrm{1}-\mathrm{2}\\ 2\mathrm{1}-\mathrm{3}\\ 5\mathrm{4}-\mathrm{9}\end{array}\right],\hspace{0.17em}\mathrm{\hspace{0.17em}}\\ \left|\mathrm{A}\right|=\left|\begin{array}{l}\mathrm{1}\mathrm{1}-\mathrm{2}\\ 2\mathrm{1}-\mathrm{3}\\ 5\mathrm{4}-\mathrm{9}\end{array}\right|[\mathrm{Expending}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{1}}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=1\left|\begin{array}{l}1-3\\ 4-9\end{array}\right|-1\left|\begin{array}{l}2-3\\ 5-9\end{array}\right|+(-2)\left|\begin{array}{l}21\\ 54\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(-9+12)-1(-18+15)-2(8-5)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3+3-6\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=0\end{array}$

Q.84

$\begin{array}{l}\mathrm{\hspace{0.17em}}\mathrm{Evaluate}\mathrm{the}\mathrm{determinants}\\ \\ \left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\begin{array}{l}3-1-2\\ 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0-1\\ 3-5\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{ii}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\begin{array}{l}3-4\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\\ 1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1-2\\ 2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1\end{array}\right|\\ \left(\mathrm{iii}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\mathrm{\hspace{0.17em}}1\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\mathrm{\hspace{0.17em}}0-3\\ -2\mathrm{\hspace{0.17em}}3\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\end{array}\right|\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{iv}\right)\mathrm{\hspace{0.17em}}\left|\begin{array}{l}2-1-2\\ 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2-1\\ 3-5\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\end{array}\right|\end{array}$

Ans

$\begin{array}{c}\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left|\begin{array}{l}\mathrm{3}-\mathrm{1}-\mathrm{2}\\ \mathrm{0\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0-\mathrm{1}\\ \mathrm{3}-\mathrm{5\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\mathrm{0}\end{array}\right|=3\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0-1\\ -5\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\end{array}\right|-(-1)\left|\begin{array}{l}0-1\\ 3\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\end{array}\right|+(-2)\left|\begin{array}{l}0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\\ 3-5\end{array}\right|\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\mathrm{Expanding}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{1}}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3(0-5)+1(0+3)-2(0-0)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-15+3-0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-12\\ \left(\mathrm{ii}\right)\left|\begin{array}{l}3-4\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}5\\ 1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1-2\\ 2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right|=3\left|\begin{array}{l}1-2\\ 3\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right|-(-4)\left|\begin{array}{l}1-2\\ 2\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\end{array}\right|+5\left|\begin{array}{l}11\\ 23\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3(1+6)+4(1+4)+5(3-2)\\ =21+20+5\\ =46\end{array}$ \begin{array}{c}\left(\text{iii}\right)\left|\begin{array}{l}\text{0}\text{1}\text{2}\\ -\text{1}\text{0}-\text{3}\\ -\text{2}3\text{0}\end{array}\right|=0\left|\begin{array}{l}0-3\\ 30\end{array}\right|-1\left|\begin{array}{l}-1-3\\ -20\end{array}\right|+2\left|\begin{array}{l}-10\\ -23\end{array}\right|\\ \left[Expanding\text{}\text{}\text{}{\text{along R}}_{\text{1}}\right]\\ =0\left(0+9\right)-1\left(0-6\right)+2\left(-3-0\right)\\ =0+6-6\\ =0\\ \left(iv\right)\left|\begin{array}{l}2-1-2\\ 02-1\\ 3-50\end{array}\right|=2\left|\begin{array}{l}2-1\\ -50\end{array}\right|-\left(-1\right)\left|\begin{array}{l}0-1\\ 30\end{array}\right|-2\left|\begin{array}{l}02\\ 3-5\end{array}\right|\\ \left[Expanding\text{}\text{}\text{}{\text{along R}}_{\text{1}}\right]\\ =2\left(0-5\right)+1\left(0+3\right)-2\left(0-6\right)\\ =-10+3+12\\ =5\end{array}

Q.85

$\text{If A=}\text{[}\begin{array}{l}101\\ 012\\ 004\end{array}],\mathrm{then}\mathrm{show}\mathrm{that}\left|3\mathrm{A}\right|=27\left|\mathrm{A}\right|.$

Ans

$\begin{array}{l}\mathrm{Given}\\ \hspace{0.17em}\mathrm{A}=\left[\begin{array}{l}1\mathrm{0}\mathrm{1}\\ \mathrm{0}1\mathrm{2}\\ \mathrm{0}0\mathrm{4}\end{array}\right]\\ \mathrm{\hspace{0.17em}}3\mathrm{A}=\left[\begin{array}{l}3\mathrm{0}\mathrm{3}\\ \mathrm{0}3\mathrm{6}\\ \mathrm{0}\mathrm{0}\mathrm{12}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\left|\mathrm{A}\right|=\left|\begin{array}{l}3\mathrm{0}\mathrm{3}\\ \mathrm{0}\mathrm{3}\mathrm{6}\\ \mathrm{0}\mathrm{0}\mathrm{12}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3(36-0)-0(0-0)+3(0-0)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=108+0+0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=108\\ \left|3\mathrm{A}\right|=\left|\begin{array}{l}\mathrm{9}\mathrm{0}\mathrm{9}\\ \mathrm{0}9\mathrm{18}\\ \mathrm{0}\mathrm{0}\mathrm{36}\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=9(324-0)-0(0-0)+9(0-0)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2916+0+0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=27\times 108\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=27\times \left|\mathrm{A}\right|\\ \left|3\mathrm{A}\right|=\mathrm{27}\left|\mathrm{A}\right|\\ \mathrm{Hence}\mathrm{Proved}\mathrm{.}\end{array}$

Q.86

$\text{If A=}\left|\begin{array}{l}12\\ 42\end{array}\right|,\mathrm{then}\mathrm{show}\mathrm{that}\left|2\mathrm{A}\right|=4\left|\mathrm{A}\right|.$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \left[\mathrm{A}\right]=\left[\begin{array}{l}12\\ 42\end{array}\right]\hspace{0.17em}\hspace{0.17em}\\ \mathrm{\hspace{0.17em}}\left|\mathrm{A}\right|=\left|\begin{array}{l}12\\ 42\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2-8\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-6\\ \hspace{0.17em}\hspace{0.17em}2\mathrm{A}=\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{8}\mathrm{4}\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\\ \left|2\mathrm{A}\right|=\left|\begin{array}{l}24\\ 84\end{array}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=8-32\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-24\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4\times (-6)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4\left|\mathrm{A}\right|=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}$

Q.87 Evaluate the determinants

$\text{|}\begin{array}{l}({\mathrm{x}}^2-\mathrm{x}+1)(\mathrm{x}-1)\\ (\mathrm{x}+1)(\mathrm{x}+1)\end{array}|$

Ans

$\begin{array}{l}\left|\begin{array}{l}{\mathrm{x}}^{\mathrm{2}}-\mathrm{x}+1\mathrm{x}-\mathrm{1}\\ \mathrm{x}+1\mathrm{x}+\mathrm{1}\end{array}\right|=(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)-(\mathrm{x}-1)(\mathrm{x}+1)\\ \hspace{0.17em}\hspace{0.17em}=({\mathrm{x}}^3-1)-({\mathrm{x}}^2-1)={\mathrm{x}}^3-1-{\mathrm{x}}^2+1\\ \hspace{0.17em}\hspace{0.17em}={\mathrm{x}}^3-{\mathrm{x}}^2\end{array}$

Q.88 Evaluate the determinants

$\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4\\ -5-1\end{array}\right|$

Ans

$\begin{array}{l}\left|\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{2\hspace{0.17em}}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{4}\\ -\mathrm{5}-\mathrm{1}\end{array}\right|=2\times (-1)-4(-5)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-2+20\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=18\end{array}$