# NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 (Ex 3.2)

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## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 (Ex 3.2)

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 help the students in scoring good marks in their examination from this particular chapter. The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 are provided for the benefit of the students. Students tend to get confused while solving questions, and Extramarks provides the students with access to get the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 in one place. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 provided by the Extramarks experts develop in students a basic understanding of the chapter and boost their confidence when they solve the questions. These solutions are provided in a step-by-step manner so that the students understand the concept and steps correctly.

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## Access NCERT Solutions for Class 12 Maths Chapter 3 Matrice

Here is an example of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2:

Question: The bookshop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books, and 10 dozen Economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer: The bookshop has 10 dozen Chemistry books, 8 dozen Physics books, and 10 dozen Economics books.

The selling prices of

a Chemistry book is Rs 80,

a Physics book is Rs 60,

and an Economics book is Rs 40

The total amount of money that will be received from the sale of all three books can be represented in the form of a matrix as:

Thus, the amount received by the bookshop is Rs 20160 from the sale of these three books.

### Topics Covered In Class 12 Maths Chapter 3 Exercise 3.2

NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.2 include topics such as

• Multiplication of matrices
• Properties of scalar multiplication of a matrix
• Properties of multiplication of matrices
• Multiplication of a matrix by a scalar

The properties of multiplications included in this chapter are ranging from existence, associative, and distributive multiplicative identity. This is also provided and covered in the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2. The exercise has 22 questions that can be easily solved if the students grasp the concept firmly.

The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 provided by Extramarks helps in redefining the students’ clarity as far as the exercises are concerned. The questions and topics covered in the NCERT Solutions help the students if they will be appearing in any kind of competitive examinations.

### What is a Matrix?

As per the NCERT definition of a matrix, “A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix”.

### What are the Operations of Matrices?

The chapter outline as provided by NCERT contains certain operations on matrices, namely, the addition of matrices, multiplication of a matrix by a scalar, difference, and multiplication of matrices. Combining any two or more matrices in a single matrix is also one of the operations.

### NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2

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### NCERT Solutions for Class 12 Maths Chapter 3 Other Exercises

 Chapter 3 Matrix Exercises Exercise 3.1 10 Questions & Solutions (5 Short Answers, 5 Long Answers) Exercise 3.3 12 Questions & Solutions (4 Short Answers, 8 Long Answers) Exercise 3.4 18 Questions & Solutions (18 Short Answers)

Q.1 In the matrix A= [2 43 2],B=[  1 32 5],C=[2 5  3 4]Find each of the following:(i) A+B (ii) AB (iii) 3AC (iv) AB(v) BA

Ans

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right], \mathrm{C}=\left[\begin{array}{l}-\mathrm{2}\mathrm{5}\\ 3\mathrm{4}\end{array}\right]\\ \left(\mathrm{i}\right)\mathrm{A}+\mathrm{B}=\left[\begin{array}{l}\mathrm{2}\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]+\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}\mathrm{2}+\mathrm{1}\mathrm{4}+\mathrm{3}\\ \mathrm{3}+\left(-2\right)\mathrm{2}+\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}37\\ 17\end{array}\right]\\ \left(\mathrm{ii}\right)\mathrm{A}-\mathrm{B}=\left[\begin{array}{l}\mathrm{2}\mathrm{4}\\ 3\mathrm{2}\end{array}\right]-\left[\begin{array}{l} \mathrm{ }1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}\mathrm{2}-1\mathrm{4}-\mathrm{3}\\ \mathrm{3}-\left(-2\right)\mathrm{2}-5\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1 \mathrm{ }1\\ 5-3\end{array}\right]\\ \left(\mathrm{iii}\right)3\mathrm{A}-\mathrm{C}=3\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]-\left[\begin{array}{l}-25\\ \mathrm{ }34\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{6}\mathrm{12}\\ 9 \mathrm{ }6\end{array}\right]-\left[\begin{array}{l}-25\\ \mathrm{ }34\end{array}\right]\\ =\left[\begin{array}{l}6-\left(-2\right)12-5\\ 9-36-4\end{array}\right]\\ =\left[\begin{array}{l}87\\ 62\end{array}\right]\\ \left(\mathrm{iv}\right) \mathrm{AB}=\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]\left[\begin{array}{l} 1\mathrm{3}\\ -2\mathrm{5}\end{array}\right]\\ =\left[\begin{array}{l}2-86+20\\ 3-49+10\end{array}\right]\\ =\left[\begin{array}{l}-626\\ -119\end{array}\right]\\ \left(\mathrm{v}\right) \mathrm{BA}=\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\left[\begin{array}{l}2\mathrm{4}\\ 3\mathrm{2}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }2+9\mathrm{ } 4+6\\ -4+15-8+10\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1110\\ 11 2\end{array}\right]\end{array}$

Q.2

$\begin{array}{l}\mathrm{Compute}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ –\mathrm{b}\mathrm{a}\end{array}\right]+\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{b}\mathrm{a}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}{\mathrm{a}}^{2}+{\mathrm{b}}^{2}{\mathrm{b}}^{2}+{\mathrm{c}}^{2}\\ {\mathrm{a}}^{2}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}–14–6\\ 8516\\ 28\mathrm{ } 5\end{array}\right]+\left[\begin{array}{l}1276\\ 805\\ 324\end{array}\right]\mathrm{ }\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}{\mathrm{cos}}^{2}\mathrm{x}{\mathrm{sin}}^{2}\mathrm{x}\\ {\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\end{array}\right]+\left[\begin{array}{l}{\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\\ {\mathrm{cos}}^{2}\mathrm{x}{\mathrm{sin}}^{2}\mathrm{x}\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ -\mathrm{b}\mathrm{a}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{b}\mathrm{a}\end{array}\right]=\left[\begin{array}{l} \mathrm{a}+\mathrm{a}\mathrm{b}+\mathrm{b}\\ -\mathrm{ }\mathrm{b}+\mathrm{b}\mathrm{a}+\mathrm{a}\end{array}\right]\\ =\left[\begin{array}{l}2\mathrm{a}2\mathrm{b}\\ 02\mathrm{a}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{2}{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}\\ {\mathrm{a}}^{\mathrm{2}}+{\mathrm{c}}^{2}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\end{array}\right]+\left[\begin{array}{l} \mathrm{ }2\mathrm{ab}2\mathrm{bc}\\ -2\mathrm{ac} -2\mathrm{ab}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}+2\mathrm{ab}{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}+2\mathrm{bc}\\ {\mathrm{a}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}-2\mathrm{ac}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}-2\mathrm{ab}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\left(\mathrm{a}+\mathrm{b}\right)}^{2}{\left(\mathrm{b}+\mathrm{c}\right)}^{2}\\ {\left(\mathrm{a}-\mathrm{c}\right)}^{2}{\left(\mathrm{a}-\mathrm{b}\right)}^{2}\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}-1\mathrm{4}- 6\\ \mathrm{8}5\mathrm{16}\\ 28 \mathrm{ }5\end{array}\right]\mathrm{+}\left[\begin{array}{l}127\mathrm{6}\\ 80\mathrm{5}\\ 32\mathrm{4}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-1+12\mathrm{4}+7 - 6+6\\ \mathrm{8}+\mathrm{8}\mathrm{5}+0\mathrm{16}+5\\ \mathrm{2}+3\mathrm{8}+2 \mathrm{5}+4\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}11110\\ 16\mathrm{ } 521\\ 510 9\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}+{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}+{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{sin}}^{\mathrm{2}}\mathrm{x}+{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}+{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}11\\ 11\end{array}\right]\end{array}$

Q.3

$\begin{array}{l}\mathrm{Compute}\mathrm{the}\mathrm{indicated}\mathrm{products}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ -\mathrm{b}\mathrm{a}\end{array}\right]\left[\begin{array}{l}\mathrm{a}–\mathrm{b}\\ \mathrm{b} \mathrm{ }\mathrm{a}\end{array}\right] \left(\mathrm{ii}\right)\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\left[234\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}1-2\\ 2 \mathrm{ }3\end{array}\right]\left[\begin{array}{l}123\\ 23\mathrm{}1\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}234\\ 345\\ 456\end{array}\right]\left[\begin{array}{l}1\mathrm{}–35\\ 0 \mathrm{ }24\\ 3 05\end{array}\right]\end{array}$

$\begin{array}{l}\left(v\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\text{}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}2\\ -1\text{}1\end{array}\right]\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{}0\text{}1\\ -1\text{}2\text{}1\end{array}\right]\\ \left(vi\right)\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}-1\text{}3\\ -1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{}2\end{array}\right]\left[\begin{array}{l}2\text{}-3\\ 1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 3\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{ab}\\ -\mathrm{ba}\end{array}\right]\left[\begin{array}{l}\mathrm{a}-\mathrm{b}\\ \mathrm{b} \mathrm{ }\mathrm{a}\end{array}\right]=\left[\begin{array}{l} \mathrm{ }{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-\mathrm{ab}+\mathrm{ab}\\ -\mathrm{ba}+\mathrm{ab} \mathrm{ }{\mathrm{b}}^{2}+{\mathrm{a}}^{2}\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-\mathrm{ab}+\mathrm{ab}\\ -\mathrm{ba}+\mathrm{ab} \mathrm{ }{\mathrm{b}}^{2}+{\mathrm{a}}^{2}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{1}\\ \mathrm{2}\\ \mathrm{3}\end{array}\right]\left[234\right]=\left[\begin{array}{l}234\\ 468\\ 6912\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}\mathrm{1}-\mathrm{2}\\ 2 \mathrm{3}\end{array}\right]\left[\begin{array}{l}12 3\\ 23 1\end{array}\right]=\left[\begin{array}{l}1-42-63-2\\ 2+64+96+3\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-3-41\\ \mathrm{ }8\mathrm{ }139\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}234\\ 345\\ 456\end{array}\right]\left[\begin{array}{l}\mathrm{1}-35\\ 0 24\\ 3 05\end{array}\right]\\ =\left[\begin{array}{l}2+0+12-6+6+010+12+20\\ 3+0+15-9+8+015+16+25\\ 4+0+18-12+10+020+20+30\end{array}\right]\\ =\left[\begin{array}{l}14042\\ 18-156\\ 22-270\end{array}\right]\\ \left(\mathrm{v}\right) \left[\begin{array}{l} \mathrm{ }21\\ \mathrm{ }32\\ -11\end{array}\right]\left[\begin{array}{l} \mathrm{ }101\\ -121\end{array}\right]=\left[\begin{array}{l}2-10+22+1\\ 3-20+43+2\\ -1-10+2-1+1\end{array}\right]\\ =\left[\begin{array}{l}123\\ 145\\ -220\end{array}\right]\\ \left(\mathrm{vi}\right)\left[\begin{array}{l} 3-13\\ -1 02\end{array}\right]\left[\begin{array}{l}\mathrm{2}-\mathrm{3}\\ 1 0\\ 3 1\end{array}\right]=\left[\begin{array}{l} \mathrm{ }6-1+9-9+0+3\\ -2+0+6 \mathrm{ }3+0+2\end{array}\right]\\ =\left[\begin{array}{l}14-6\\ 45\end{array}\right]\end{array}$

Q.4

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left[\begin{array}{l}12-3\\ 50\mathrm{ } 2\\ 1 -1 1\end{array}\right],\mathrm{ }\mathrm{B}=\left[\begin{array}{l}3-12\\ 4\mathrm{ } 25\\ 2 03\end{array}\right]\mathrm{and}\mathrm{C}=\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right],\\ \mathrm{then}\mathrm{compute}\left(\mathrm{A}+\mathrm{B}\right)\mathrm{and}\left(\mathrm{B}-\mathrm{C}\right).\mathrm{ }\mathrm{Also},\mathrm{verify}\mathrm{that}\\ \mathrm{A}+\left(\mathrm{B}-\mathrm{C}\right)=\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{C}.\end{array}$

Ans

$\begin{array}{l}\mathrm{A}+\mathrm{B}=\left[\begin{array}{l}12-\mathrm{3}\\ 50 \mathrm{2}\\ 1 -1 1\end{array}\right]+\left[\begin{array}{l}\mathrm{3}-12\\ \mathrm{4} \mathrm{ }25\\ \mathrm{2} \mathrm{ }03\end{array}\right]\\ =\left[\begin{array}{l}1+32+\left(-1\right)-\mathrm{3}+\mathrm{2}\\ \mathrm{5}+40+22+5\\ \mathrm{1}+2 -\mathrm{1}+0 \mathrm{1}+3\end{array}\right]\\ =\left[\begin{array}{l}4\mathrm{ } 1-1\\ 9 2\mathrm{ } 7\\ 3-1 4\end{array}\right]\\ \mathrm{B}-\mathrm{C}=\left[\begin{array}{l}3-12\\ \mathrm{4} \mathrm{ }25\\ 2 03\end{array}\right]-\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right]\\ =\left[\begin{array}{l}3-\mathrm{4}-\mathrm{1}-12-2\\ 4-0 2-35-2\\ \mathrm{2}-1 0-\left(-2\right)\mathrm{3}-3\end{array}\right]\\ =\left[\begin{array}{l}-\mathrm{1}-20\\ \mathrm{4}-13\\ 1 20\end{array}\right]\\ \mathrm{A}+\left(\mathrm{B}-\mathrm{C}\right)=\left[\begin{array}{l}12-3\\ 50 \mathrm{ }2\\ 1 \mathrm{ }-1 1\end{array}\right]+\left(\left[\begin{array}{l}3-12\\ 4 \mathrm{ }25\\ 2 \mathrm{ }03\end{array}\right]-\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right]\right)\\ \mathrm{ }=\left[\begin{array}{l}12-3\\ 50 \mathrm{ }2\\ 1 \mathrm{ }-1 1\end{array}\right]+\left(\left[\begin{array}{l}3-4 -1-12-2\\ 4-0 \mathrm{ }2-35-2\\ 2-1 \mathrm{ }0+33-3\end{array}\right]\right)\\ \mathrm{ }=\left[\begin{array}{l}12-3\\ 50 \mathrm{ }2\\ 1 \mathrm{ }-1 1\end{array}\right]+\left[\begin{array}{l}-1-20\\ 4-13\\ 1 \mathrm{ }30\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1-12-2 \mathrm{ }-3+0\\ 5+40-1 \mathrm{ }2+3\\ 1+1 -1+3 1+0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}00 \mathrm{ }-3\\ 9 -1 \mathrm{ }5\\ 2 2 \mathrm{ }1\end{array}\right]\\ \left(\mathrm{A}+\mathrm{B}\right)-\mathrm{C}=\left(\left[\begin{array}{l}12-3\\ 50 \mathrm{ }2\\ 1 \mathrm{ }-1 1\end{array}\right]+\left[\begin{array}{l}3-12\\ 4 \mathrm{ }25\\ 2 \mathrm{ }03\end{array}\right]\right)-\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1+32-1 -3+2\\ 5+40+22+5\\ 1+2 \mathrm{ }-1+0 1+3\end{array}\right]-\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}41 \mathrm{ }-1\\ 927\\ 3 \mathrm{ }-14\end{array}\right]-\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}4-41-1 \mathrm{ }-1-2\\ 9-0 2-3 \mathrm{ }7-2\\ 3-1 \mathrm{ }-1+2 4-3\end{array}\right]\\ =\left[\begin{array}{l}00 \mathrm{ }-3\\ 9 \mathrm{ }-1\mathrm{ }5\\ 2 \mathrm{ }1\mathrm{ }1\end{array}\right]\\ \mathrm{Hence}, \mathrm{A}+\left(\mathrm{B}-\mathrm{C}\right)=\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{C}.\end{array}$

Q.5

$\text{If\hspace{0.17em} A=}\text{[}\begin{array}{l}\frac{2}{3}1\frac{5}{3}\\ \frac{1}{3}\frac{2}{3}\frac{4}{3}\\ \frac{7}{3}2\frac{2}{3}\end{array}\right]&\mathrm{B}=\left[\begin{array}{l}\frac{2}{5}\frac{3}{5}1\\ \frac{1}{5}\frac{2}{5}\frac{4}{5}\\ \frac{7}{5}\frac{6}{5}\frac{2}{5}\end{array}\right],\mathrm{then}\mathrm{compute}3\mathrm{A}-5\mathrm{B}.$

Ans

$\begin{array}{l}\mathrm{Given},\mathrm{A}=\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{3}}\mathrm{1}\frac{5}{\mathrm{3}}\\ \frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{2}}{\mathrm{3}}\frac{\mathrm{4}}{\mathrm{3}}\\ \frac{\mathrm{7}}{\mathrm{3}}2\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{3}}{\mathrm{5}}\mathrm{1}\\ \frac{\mathrm{1}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{4}}{\mathrm{5}}\\ \frac{\mathrm{7}}{\mathrm{5}}\frac{\mathrm{6}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\\ 3\mathrm{A}-5\mathrm{B}=3\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{3}}1\frac{\mathrm{5}}{\mathrm{3}}\\ \frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{2}}{\mathrm{3}}\frac{\mathrm{4}}{\mathrm{3}}\\ \frac{\mathrm{7}}{\mathrm{3}}2\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]-5\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{3}}{\mathrm{5}}\mathrm{1}\\ \frac{\mathrm{1}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{4}}{\mathrm{5}}\\ \frac{\mathrm{7}}{\mathrm{5}}\frac{\mathrm{6}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}3×\frac{\mathrm{2}}{\mathrm{3}}3×13×\frac{\mathrm{5}}{\mathrm{3}}\\ 3×\frac{\mathrm{1}}{\mathrm{3}}3×\frac{\mathrm{2}}{\mathrm{3}}3×\frac{\mathrm{4}}{\mathrm{3}}\\ 3×\frac{\mathrm{7}}{\mathrm{3}}3×23×\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]-\left[\begin{array}{l}5×\frac{\mathrm{2}}{\mathrm{5}}5×\frac{\mathrm{3}}{\mathrm{5}}5×\mathrm{1}\\ 5×\frac{\mathrm{1}}{\mathrm{5}}5×\frac{\mathrm{2}}{\mathrm{5}}5×\frac{\mathrm{4}}{\mathrm{5}}\\ 5×\frac{\mathrm{7}}{\mathrm{5}}5×\frac{\mathrm{6}}{\mathrm{5}}5×\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\mathrm{ }\\ =\left[\begin{array}{l}235\\ 124\\ 762\end{array}\right]-\left[\begin{array}{l}235\\ 124\\ 762\end{array}\right]\\ =\left[\begin{array}{l}2-23-35-5\\ 1-12-24-4\\ 7-76-62-2\end{array}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]\end{array}$

Q.6

$\text{Simplify\hspace{0.17em}cosθ}\text{[}\begin{array}{l}\mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }\end{array}\right]+\mathrm{sin\theta }\left[\begin{array}{l}\mathrm{sin\theta } -\mathrm{cos\theta }\\ \mathrm{cos\theta }\mathrm{sin\theta }\end{array}\right].$

Ans

$\begin{array}{l}\mathrm{cos\theta }\left[\begin{array}{l}\mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }\end{array}\right]+\mathrm{sin\theta }\left[\begin{array}{l}\mathrm{sin\theta } -\mathrm{cos\theta }\\ \mathrm{cos\theta }\mathrm{sin\theta }\end{array}\right]\\ =\left[\begin{array}{l} {\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta } \mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\mathrm{+}\left[\begin{array}{l}{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta } -\mathrm{sin\theta }\mathrm{cos\theta }\\ \mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\\ =\left[\begin{array}{l} {\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }+{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta } \mathrm{cos\theta }\mathrm{sin\theta }-\mathrm{sin\theta }\mathrm{cos\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }+\mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }+{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{ }10\\ 0\mathrm{1}\end{array}\right].\end{array}$

Q.7

$\begin{array}{l}\mathrm{Find}\mathrm{X}\mathrm{and}\mathrm{Y},\mathrm{if}\\ \left(\mathrm{i}\right)\mathrm{ }\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right] \mathrm{and} \mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}30\\ 03\end{array}\right]\\ \left(\mathrm{ii}\right)2\mathrm{X}+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right] \mathrm{and} 3\mathrm{X}+2\mathrm{Y}=\left[\begin{array}{l} \mathrm{ }2-2\\ -1 \mathrm{ }5\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) \mathrm{ }\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }...\left(\mathrm{i}\right) \\ \mathrm{and} \mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}30\\ 03\end{array}\right]\mathrm{ }...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\mathrm{both}\mathrm{equations},\mathrm{we}\mathrm{get}\\ \mathrm{X}+\mathrm{Y}+\mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]+\left[\begin{array}{l}30\\ 03\end{array}\right]\\ 2\mathrm{X}=\left[\begin{array}{l}\mathrm{7}+30+\mathrm{0}\\ \mathrm{2}+05+\mathrm{3}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}100\\ 28\end{array}\right]\\ \mathrm{X}=\frac{1}{2}\left[\begin{array}{l}100\\ 28\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}50\\ 14\end{array}\right]\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{X}\mathrm{in}\mathrm{equatin}\left(\mathrm{i}\right), \mathrm{we}\mathrm{ }\mathrm{get}\\ \left[\begin{array}{l}50\\ 14\end{array}\right]+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }\\ \mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }-\left[\begin{array}{l}50\\ 14\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{7}-50-0\\ \mathrm{2}-15-4\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}20\\ 11\end{array}\right]\\ \mathrm{Thus},\mathrm{ }\mathrm{X}=\left[\begin{array}{l}50\\ 14\end{array}\right]\mathrm{and}\mathrm{Y}=\left[\begin{array}{l}20\\ 11\end{array}\right].\\ \left(\mathrm{ii}\right)2\mathrm{X}+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]\mathrm{ }...\left(\mathrm{i}\right)\mathrm{ }\\ \mathrm{and} 3\mathrm{X}+2\mathrm{Y}=\left[\begin{array}{l} 2-\mathrm{2}\\ -1 \mathrm{5}\end{array}\right]\mathrm{ }...\left(\mathrm{ii}\right)\mathrm{ }\\ \mathrm{Multiplying}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{by}2\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}3,\mathrm{we}\mathrm{get}\\ 2\left(2\mathrm{X}+3\mathrm{Y}\right)=2\left[\begin{array}{l}23\\ 40\end{array}\right]\\ ⇒ 4\mathrm{X}+6\mathrm{Y}=\left[\begin{array}{l}46\\ 80\end{array}\right]...\left(\mathrm{iii}\right)\\ 3\left(3\mathrm{X}+2\mathrm{Y}\right)=3\left[\begin{array}{l} 2-\mathrm{2}\\ -1 \mathrm{5}\end{array}\right]\\ ⇒9\mathrm{X}+6\mathrm{Y}=\left[\begin{array}{l} 6-\mathrm{6}\\ -3 \mathrm{15}\end{array}\right]...\left(\mathrm{iv}\right)\\ \mathrm{Subtracting}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{from}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ 5\mathrm{X}=\left[\begin{array}{l} 6-\mathrm{6}\\ -3 \mathrm{15}\end{array}\right]-\left[\begin{array}{l}46\\ 80\end{array}\right]\\ \mathrm{ }\mathrm{X}=\frac{1}{5}\left[\begin{array}{l} 2-\mathrm{12}\\ -11 \mathrm{15}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{X}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{have}\\ \mathrm{2}\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]\\ \mathrm{ }3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]-\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{4}}{5}-\frac{\mathrm{24}}{5}\\ -\frac{\mathrm{22}}{5} 6\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{2}-\frac{\mathrm{4}}{5}\mathrm{3}-\left(-\frac{\mathrm{24}}{5}\right)\\ \mathrm{4}-\left(-\frac{\mathrm{22}}{5}\right)\mathrm{0}-6\end{array}\right]\\ \mathrm{ }\mathrm{Y}=\frac{1}{3}\left[\begin{array}{l}\frac{\mathrm{6}}{5}\frac{\mathrm{39}}{5}\\ \frac{\mathrm{42}}{5}-6\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{2}}{5}\frac{\mathrm{13}}{5}\\ \frac{\mathrm{14}}{5}-2\end{array}\right]\\ \mathrm{Thus},\mathrm{X}=\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]\mathrm{and} \mathrm{Y}=\left[\begin{array}{l}\frac{\mathrm{2}}{5}\frac{\mathrm{13}}{5}\\ \frac{\mathrm{14}}{5}-2\end{array}\right].\end{array}$

Q.8

$\text{Find X, if Y=}\text{[}\begin{array}{l}32\\ 40\end{array}\right]\mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l} 2-2\\ -1 \mathrm{ }5\end{array}\right].$

Ans

$\begin{array}{l}\mathrm{Given},\mathrm{Y}=\left[\begin{array}{l}32\\ 14\end{array}\right]\mathrm{}\\ \mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l} 10\\ -32\end{array}\right]...\left(\mathrm{i}\right)\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{Y}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ 2\mathrm{X}+\left[\begin{array}{l}32\\ 14\end{array}\right]=\left[\begin{array}{l} 10\\ -32\end{array}\right]\\ \mathrm{ }2\mathrm{X}=\left[\begin{array}{l} 10\\ -32\end{array}\right]-\left[\begin{array}{l}32\\ 14\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }1-30-2\\ -3-12-4\end{array}\right]\\ \mathrm{X}\mathrm{ }=\frac{1}{2}\left[\begin{array}{l}-2-2\\ -4-2\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-1-1\\ -2 -1\end{array}\right]\end{array}$

Q.9

$\text{If\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}and y, if 2}\text{[}\begin{array}{l}13\\ 0\mathrm{x}\end{array}\right]+\left[\begin{array}{l}\mathrm{y}0\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right].$

Ans

$\begin{array}{l}\mathrm{Given}: 2\left[\begin{array}{l}13\\ \mathrm{0}\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{y}\mathrm{0}\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒\left[\begin{array}{l}26\\ 02\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{y}\mathrm{0}\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{l}2+\mathrm{y} \mathrm{ }6+0\\ 0+12\mathrm{x}+2\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒ 2+\mathrm{y}=5 \mathrm{and} 2\mathrm{x}+2=8\\ ⇒ \mathrm{ }\mathrm{y}=3 \mathrm{and}\mathrm{ }\mathrm{x}=\frac{8-2}{2}=3\\ \mathrm{Thus},\mathrm{x}=3\mathrm{and}\mathrm{y}=3.\\ \end{array}$

Q.10

$\begin{array}{l}\mathrm{Solve}\mathrm{thee}\mathrm{quation}\mathrm{for}\mathrm{x}, \mathrm{y},\mathrm{ }\mathrm{z}\mathrm{and}\mathrm{t},\mathrm{if}\\ 2\left[\begin{array}{l}\mathrm{xz}\\ \mathrm{yt}\end{array}\right]+3\left[\begin{array}{l}1-1\\ 0 \mathrm{ }2\end{array}\right]=3\left[\begin{array}{l}35\\ 46\end{array}\right].\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{2}\left[\begin{array}{l}\mathrm{x}\mathrm{z}\\ \mathrm{y}\mathrm{t}\end{array}\right]+3\left[\begin{array}{l}1-\mathrm{1}\\ \mathrm{0} 2\end{array}\right]=\mathrm{3}\left[\begin{array}{l}35\\ 46\end{array}\right]\\ ⇒ \left[\begin{array}{l}2\mathrm{x}2\mathrm{z}\\ 2\mathrm{y}2\mathrm{t}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{3}-\mathrm{3}\\ \mathrm{0} 6\end{array}\right]=\left[\begin{array}{l}91 5\\ 121 8\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{l}2\mathrm{x}+32\mathrm{z}-\mathrm{3}\\ 2\mathrm{y}+02\mathrm{t}+\mathrm{6}\end{array}\right]=\left[\begin{array}{l}91 5\\ 121 8\end{array}\right]\\ ⇒2\mathrm{x}+\mathrm{3}=9, 2\mathrm{y}=12, 2\mathrm{z}-\mathrm{3}=15 \mathrm{and} 2\mathrm{t}+\mathrm{6}=18\\ ⇒\mathrm{x}=3,\mathrm{y}=6, \mathrm{z}=9\mathrm{â€‹}\mathrm{and}\mathrm{t}=6.\end{array}$

Q.11

$\text{If\hspace{0.17em} x}\text{[}\begin{array}{l}2\\ 3\end{array}\right]+\mathrm{y}\left[\begin{array}{l}-1\\ \mathrm{ }1\end{array}\right]=\left[\begin{array}{l}10\\ 5\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.$

Ans

$\begin{array}{l}\mathrm{Goven}:\\ \mathrm{x}\left[\begin{array}{l}\mathrm{2}\\ \mathrm{3}\end{array}\right]+\mathrm{y}\left[\begin{array}{l}-\mathrm{1}\\ 1\end{array}\right]\mathrm{=}\left[\begin{array}{l}\mathrm{10}\\ \mathrm{5}\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{l}2\mathrm{x}\\ 3\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}-\mathrm{y}\\ \mathrm{ }\mathrm{y}\end{array}\right]\mathrm{=}\left[\begin{array}{l}\mathrm{10}\\ \mathrm{5}\end{array}\right]\\ ⇒ \left[\begin{array}{l}2\mathrm{x}-\mathrm{y}\\ 3\mathrm{x}+\mathrm{y}\end{array}\right]\mathrm{=}\left[\begin{array}{l}\mathrm{10}\\ \mathrm{5}\end{array}\right]\\ ⇒2\mathrm{x}-\mathrm{y}=10, 3\mathrm{x}+\mathrm{y}=5\\ ⇒\mathrm{x}=3 \mathrm{and} \mathrm{y}=-4\end{array}$

Q.12

$\begin{array}{l}\mathrm{Given}\mathrm{ }3\left[\begin{array}{l}\mathrm{xy}\\ \mathrm{zw}\end{array}\right]=\left[\begin{array}{l}\mathrm{x}6\\ -12\mathrm{w}\end{array}\right]+\left[\begin{array}{l}4 \mathrm{ }\mathrm{x}+\mathrm{y}\\ \mathrm{z}+\mathrm{w}3\end{array}\right],\mathrm{f}\mathrm{in}\mathrm{d}\mathrm{the}\\ \mathrm{value}\mathrm{of}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{and}\mathrm{w}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{ }:\\ 3\left[\begin{array}{l}\mathrm{xy}\\ \mathrm{zw}\end{array}\right]=\left[\begin{array}{l}\mathrm{x}6\\ -12\mathrm{w}\end{array}\right]+\left[\begin{array}{l}4 \mathrm{ }\mathrm{x}+\mathrm{y}\\ \mathrm{z}+\mathrm{w}3\end{array}\right]\\ ⇒\left[\begin{array}{l}3\mathrm{x}3\mathrm{y}\\ 3\mathrm{z}3\mathrm{w}\end{array}\right]=\left[\begin{array}{l}\mathrm{x}+4 6+\mathrm{x}+\mathrm{y}\\ -1+\mathrm{z}+\mathrm{w}2\mathrm{w}+3\end{array}\right]\\ ⇒ 3\mathrm{x}=\mathrm{x}+4⇒\mathrm{x}=2\\ 3\mathrm{y}=6+\mathrm{x}+\mathrm{y}⇒\mathrm{y}=4\\ \mathrm{ }3\mathrm{w}=2\mathrm{w}+3⇒\mathrm{w}=3\\ \mathrm{ }\mathrm{and} \mathrm{ }3\mathrm{z}=-1+\mathrm{z}+\mathrm{w}⇒\mathrm{z}=1\\ \mathrm{Therefore},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=4,\mathrm{ }\mathrm{z}=1\mathrm{ }\mathrm{and} \mathrm{w}=3.\end{array}$

Q.13

$\text{If F(x) =}\left[\begin{array}{l}\mathrm{cosx}-\mathrm{sinx}0\\ \mathrm{sinx}\mathrm{cosx}0\\ 0 \mathrm{ }01\end{array}\right], \mathrm{show}\mathrm{that}\mathrm{F}\left(\mathrm{x}\right)\mathrm{F}\left(\mathrm{y}\right)=\mathrm{F}\left(\mathrm{x}+\mathrm{y}\right).$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{F}\left(\mathrm{x}\right)=\left[\begin{array}{l}\mathrm{cosx}-\mathrm{ }\mathrm{sinx}\mathrm{0}\\ \mathrm{sinx}\mathrm{cosx}\mathrm{0}\\ 0 01\end{array}\right]\\ \mathrm{F}\left(\mathrm{y}\right)=\left[\begin{array}{l}\mathrm{cosy} -\mathrm{ }\mathrm{siny} \mathrm{0}\\ \mathrm{siny}\mathrm{cosy}\mathrm{0}\\ 0 01\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \mathrm{F}\left(\mathrm{x}\right).\mathrm{F}\left(\mathrm{y}\right)=\left[\begin{array}{l}\mathrm{cosx}-\mathrm{ }\mathrm{sinx}\mathrm{0}\\ \mathrm{sinx}\mathrm{cosx}\mathrm{0}\\ 0 01\end{array}\right]\left[\begin{array}{l}\mathrm{cosy} \mathrm{ }-\mathrm{ }\mathrm{siny} \mathrm{0}\\ \mathrm{siny}\mathrm{cosy}\mathrm{0}\\ 0 01\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{cosx}\mathrm{cosy}-\mathrm{sinx}\mathrm{siny}-\mathrm{ }\mathrm{siny}\mathrm{cosx}-\mathrm{sinx}\mathrm{cosy}\mathrm{0}\\ \mathrm{sinx}\mathrm{cosy}\mathrm{ }+\mathrm{ }\mathrm{cosx}\mathrm{siny}-\mathrm{sinx}\mathrm{siny}+\mathrm{cosx}\mathrm{cosy}\mathrm{0}\\ 0 01\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{ }\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{0}\\ \mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{ }\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{0}\\ 0 01\end{array}\right]\\ =\mathrm{F}\left(\mathrm{x}+\mathrm{y}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.14

$\begin{array}{l}\mathrm{Show}\mathrm{that}\\ \mathrm{ }\left(\mathrm{i}\right)\left[\begin{array}{l}5-1\\ 6 \mathrm{ }7\end{array}\right]\left[\begin{array}{l}21\\ 34\end{array}\right]\ne \left[\begin{array}{l}21\\ 34\end{array}\right]\left[\begin{array}{l}5-1\\ 6 \mathrm{ }7\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]\left[\begin{array}{l}-110\\ 0 -11\\ 234\end{array}\right]\ne \left[\begin{array}{l}-110\\ 0 \mathrm{ }-11\\ 234\end{array}\right]\left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{ }\left(\mathrm{i}\right) \mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \left[\begin{array}{l}\mathrm{5}-\mathrm{1}\\ 6 \mathrm{7}\end{array}\right]\left[\begin{array}{l}21\\ 34\end{array}\right]=\left[\begin{array}{l}10-35-4\\ 12+216+28\end{array}\right]\\ =\left[\begin{array}{l} 7 \mathrm{ }1\\ 3334\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\\ \left[\begin{array}{l}21\\ 34\end{array}\right]\left[\begin{array}{l}\mathrm{5}-\mathrm{1}\\ 6 \mathrm{7}\end{array}\right]=\left[\begin{array}{l}\mathrm{10}+\mathrm{6}-\mathrm{2}+\mathrm{7}\\ \mathrm{15}+\mathrm{24}-\mathrm{3}+\mathrm{28}\end{array}\right]\\ =\left[\begin{array}{l}165\\ 3925\end{array}\right]\\ \mathrm{Therefore},\\ \left[\begin{array}{l}\mathrm{5}-\mathrm{1}\\ 6 \mathrm{7}\end{array}\right]\left[\begin{array}{l}21\\ 34\end{array}\right]\ne \left[\begin{array}{l}21\\ 34\end{array}\right]\left[\begin{array}{l}\mathrm{5}-\mathrm{1}\\ 6 \mathrm{7}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]\left[\begin{array}{l}-110\\ 0 \mathrm{ }-11\\ 234\end{array}\right]\ne \left[\begin{array}{l}-110\\ 0 -11\\ 234\end{array}\right]\left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}:\\ \left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]\left[\begin{array}{l}-110\\ 0 \mathrm{ }-11\\ 234\end{array}\right]=\left[\begin{array}{l}-\mathrm{1}+\mathrm{0}+61-\mathrm{2}+90+2+12\\ 0+0+00-1+0 \mathrm{0}+1+0\\ -1+0+01-1+0 0+1+0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }5814\\ \mathrm{ }0-11\\ -10 1\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}:\\ \left[\begin{array}{l}-110\\ 0 -11\\ 234\end{array}\right]\left[\begin{array}{l}123\\ 010\\ 110\end{array}\right]=\left[\begin{array}{l}-1+0+0-2+1+0-3+0+0\\ 0+0+10-1+10+0+0\\ 2+0+44+3+46+0+0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-1-1-3\\ \mathrm{ }1 \mathrm{ }0 0\\ 6\mathrm{ }11 6\end{array}\right]\\ \therefore \mathrm{ }\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$

Q.15

$Find\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{2}-5A+6I,ifA=\left[\begin{array}{l}2\text{}0\text{}1\\ 2\text{}1\text{}3\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1\text{}0\end{array}\right].$

Ans

$\begin{array}{l}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}-\text{5A+6I, A}=\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{-1}\text{}\text{0}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}=A×A\\ \text{}=\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\\ \text{}=\left[\begin{array}{l}4+0+1\text{}0+0-1\text{}2+0+0\\ 4+2+3\text{}0+1-3\text{}2+3+0\\ 2-2+0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0-1+0\text{}1-3+0\end{array}\right]\\ \text{}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]\\ Now,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}-\text{5A+6I}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]-\text{5}\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\text{+6}\left[\begin{array}{l}1\text{}0\text{}0\\ 0\text{}1\text{}0\\ 0\text{}0\text{}1\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]-\left[\begin{array}{l}\text{10}\text{}\text{0}\text{}\text{5}\\ \text{10}\text{}\text{5}\text{}\text{15}\\ \text{5}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{5}\text{}\text{0}\end{array}\right]\text{+}\left[\begin{array}{l}6\text{}0\text{}0\\ 0\text{}6\text{}0\\ 0\text{}0\text{}6\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}5-10+6\text{}\text{}-1\text{}\text{}2-5\\ 9-10\text{}-2-5+6\text{}5-15\\ 0-5\text{}\text{}-1+5\text{}\text{}-2+6\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}1\text{}-1\text{}-3\\ -1\text{}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}-10\\ -5\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\text{}\text{\hspace{0.17em}}4\end{array}\right]\end{array}$

Q.16

$\text{If A =}\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\mathrm{}, \mathrm{prove}\mathrm{that} {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+7\mathrm{A}+2\mathrm{I}=0.$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right], \\ \mathrm{ }{\mathrm{A}}^{\mathrm{2}}\mathrm{}=\mathrm{A}×\mathrm{A}\\ =\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\\ =\left[\begin{array}{l}1+0+4 0+0+02+0+6\\ 0+0+2 \mathrm{ }0+4+00+2+3\\ 2+0+6 \mathrm{ }0+0+04+0+9\end{array}\right]\\ =\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\\ 6{\mathrm{A}}^{2}=6\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\\ =\left[\begin{array}{l}30 \mathrm{ }048\\ 122430\\ 16 \mathrm{ }078\end{array}\right]\\ \\ {\mathrm{A}}^{3}={\mathrm{A}}^{2}.\mathrm{A}\\ =\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\left[\begin{array}{l}102\\ 021\\ 203\end{array}\right]\\ =\left[\begin{array}{l}5+0+160+0+010+0+24\\ 2+0+100+8+04+4+15\\ 8+0+260+0+016+0+39\end{array}\right]\\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]\\ 7\mathrm{A}=7\left[\begin{array}{l}102\\ 021\\ 203\end{array}\right]\\ =\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]\\ 2\mathrm{I}=2\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]\\ =\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}:\\ {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{\mathrm{2}}+7\mathrm{A}+2\mathrm{I}\\ \\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]-6\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]+\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]+\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]-\left[\begin{array}{l}30048\\ 122430\\ 48078\end{array}\right]+\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]+\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ =\left[\begin{array}{l}21-30+7+20-0+0+034-48+14+0\\ 12-12+0+08-24+14+223-30+7+0\\ 34-48+14+00+0+0+055-78+21+2\end{array}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]=0\end{array}$

Q.17

$\text{If A=}\text{[}\begin{array}{l}3-2\\ 4-2\end{array}\right], \mathrm{and}\mathrm{I}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\mathrm{f}\mathrm{indk}\mathrm{so}\mathrm{that}\mathrm{ }{\mathrm{A}}^{2}=\mathrm{kA}-2\mathrm{I}.$

Ans

$\begin{array}{l}âˆµ\mathrm{A}=\left[\begin{array}{l}\mathrm{3}-\mathrm{2}\\ \mathrm{4}-\mathrm{2}\end{array}\right]\\ {\mathrm{A}}^{2}=\left[\begin{array}{l}\mathrm{3}-\mathrm{2}\\ \mathrm{4}-\mathrm{2}\end{array}\right]\left[\begin{array}{l}\mathrm{3}-\mathrm{2}\\ \mathrm{4}-\mathrm{2}\end{array}\right]\\ =\left[\begin{array}{l}9-8-6+4\\ 12-8-8+4\end{array}\right]\\ =\left[\begin{array}{l}1-2\\ 4-4\end{array}\right]\\ \mathrm{Now}, {\mathrm{A}}^{\mathrm{2}}=\mathrm{kA}-2\mathrm{I}\\ \left[\begin{array}{l}1-2\\ 4-4\end{array}\right]=\mathrm{k}\left[\begin{array}{l}\mathrm{3}-\mathrm{2}\\ \mathrm{4}-\mathrm{2}\end{array}\right]-2\left[\begin{array}{l}10\\ 01\end{array}\right]\\ \left[\begin{array}{l}1-2\\ 4-4\end{array}\right]=\left[\begin{array}{l}3\mathrm{k}-\mathrm{2}\mathrm{k}\\ \mathrm{4}\mathrm{k}-\mathrm{2}\mathrm{k}\end{array}\right]-\left[\begin{array}{l}20\\ 02\end{array}\right]\\ \left[\begin{array}{l}1-2\\ 4-4\end{array}\right]=\left[\begin{array}{l}3\mathrm{k}-\mathrm{2}-\mathrm{2}\mathrm{k}\\ \mathrm{4}\mathrm{k} \mathrm{ }-\mathrm{2}\mathrm{k}-2\end{array}\right]\\ ⇒ \mathrm{ }4=4\mathrm{k}\\ ⇒ \mathrm{ }\mathrm{k}=1\\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}1\mathrm{.}\end{array}$

Q.18

$\begin{array}{l}\mathrm{If} \mathrm{A}=\left[\begin{array}{l}0-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}0\end{array}\right]\mathrm{and}\mathrm{I}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\\ \mathrm{show}\mathrm{that}\mathrm{I}+\mathrm{A}=\left(\mathrm{I}-\mathrm{A}\right)\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{ }\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{I}+\mathrm{ }\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{0}-\mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\mathrm{0}\end{array}\right]\\ =\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \mathrm{I}-\mathrm{ }\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]-\left[\begin{array}{l}\mathrm{0}-\mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\mathrm{0}\end{array}\right]\\ =\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\left(\mathrm{I}-\mathrm{A}\right)\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}1\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ -\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}1\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{cos\alpha }+\frac{\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}-\mathrm{sin\alpha }+\mathrm{cos\alpha }\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ -\frac{\mathrm{cos\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}+\mathrm{sin\alpha }\frac{\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}+\mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos\alpha }.\mathrm{cos}\frac{\mathrm{\alpha }}{2}+\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\left(\mathrm{sin\alpha }.\mathrm{cos}\frac{\mathrm{\alpha }}{2}-\mathrm{cos\alpha sin}\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin\alpha cos}\frac{\mathrm{\alpha }}{2}-\mathrm{cos\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos\alpha cos}\frac{\mathrm{\alpha }}{2}+\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\mathrm{sin}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\mathrm{sin}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]=\mathrm{L}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.19 A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide â‚¹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) Rs 1800 (b) Rs 2000.

Ans

$\begin{array}{l} \mathrm{Total}\mathrm{money}\mathrm{to}\mathrm{invest}=\mathrm{â€„} â‚¹ 30,000\\ \mathrm{Money}\mathrm{invested}\mathrm{in}{\mathrm{I}}^{\mathrm{st}}\mathrm{type}\mathrm{of}\mathrm{bond}= â€„â‚¹\mathrm{x}\\ \mathrm{Money}\mathrm{invested}\mathrm{in}{2}^{\mathrm{nd}}\mathrm{type}\mathrm{of}\mathrm{bond}=â€„â‚¹\left(30,000-\mathrm{x}\right)\\ \mathrm{Rate}\mathrm{of}\mathrm{interest}\mathrm{on}\mathrm{first}\mathrm{type}\mathrm{of}\mathrm{bond}=5%\\ \mathrm{Rate}\mathrm{of}\mathrm{interest}\mathrm{on}\mathrm{secon}\mathrm{type}\mathrm{of}\mathrm{bond}=7%\\ \left(\mathrm{i}\right) \mathrm{Total}\mathrm{obtained}\mathrm{interest}=1800\\ \left(\mathrm{ii}\right) \mathrm{Total}\mathrm{obtained}\mathrm{interest}=2000\\ \mathrm{ }\left(\begin{array}{cc}\mathrm{I}& \mathrm{II}\\ \mathrm{x}& \left(30,000-\mathrm{x}\right)\end{array}\right)\left(\begin{array}{l}5%\\ 7%\end{array}\right)\mathrm{=}\left(\begin{array}{l}\mathrm{1800}\\ \mathrm{2000}\end{array}\right)\\ \left(\mathrm{i}\right)5% \mathrm{of}\mathrm{ }\mathrm{x}+7%\left(30,000-\mathrm{x}\right)=1800\\ \frac{\mathrm{5}}{\mathrm{100}}\mathrm{x}+\frac{\mathrm{7}}{\mathrm{100}}\left(30,000-\mathrm{x}\right)=1800\\ \mathrm{ }5\mathrm{x}+210000-7\mathrm{x}=180000\\ 210000-180000=2\mathrm{x}\\ 30000=2\mathrm{x}\\ \mathrm{ }\mathrm{x}=\frac{\mathrm{30000}}{\mathrm{2}}\\ \mathrm{ }=15000\\ \mathrm{To}\mathrm{get}\mathrm{annual}\mathrm{interest}\mathrm{of}1800,\mathrm{he}\mathrm{should}\mathrm{invest}â‚¹\mathrm{â€„}15,000\\ \mathrm{in}\mathrm{I}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{and}â‚¹\mathrm{â€„}15,000\mathrm{in}\mathrm{II}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{.}\\ \left(\mathrm{ii}\right)5% \mathrm{of}\mathrm{ }\mathrm{x}+7%\left(30,000-\mathrm{x}\right)=2000\\ \frac{\mathrm{5}}{\mathrm{100}}\mathrm{x}+\frac{\mathrm{7}}{\mathrm{100}}\left(30,000-\mathrm{x}\right)=2000\\ \mathrm{ }5\mathrm{x}+210000-7\mathrm{x}=200000\\ 210000-200000=2\mathrm{x}\\ 10000=2\mathrm{x}\\ \mathrm{ }\mathrm{x}=\frac{\mathrm{10000}}{\mathrm{2}}\\ = 5000\\ \mathrm{To}\mathrm{get}\mathrm{annual}\mathrm{interest}\mathrm{of}â‚¹\mathrm{â€„}2000\mathrm{he}\mathrm{should}\mathrm{invest}â‚¹\mathrm{â€„}5000&\\ â‚¹â€„25000\mathrm{respectively}\mathrm{in}\mathrm{I}\mathrm{and}\mathrm{II}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{.}\end{array}$

Q.20 The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are â‚¹80, â‚¹60 and â‚¹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans

$\begin{array}{l}\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{chemistry}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 10\mathrm{dozen}\end{array}\\ \begin{array}{l}=10×12\end{array}\\ \begin{array}{l}=120\end{array}\\ \begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{physics}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 8\mathrm{dozen}\end{array}\\ \begin{array}{l}=8×12\end{array}\\ =\mathrm{â€„}\begin{array}{l}\mathrm{96}\end{array}\\ \begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{economics}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 10\mathrm{dozen}\end{array}\\ \begin{array}{l}=10×12\end{array}\\ \begin{array}{l}=120\end{array}\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{chemistry}\mathrm{book}=\end{array}â€„â‚¹â€„80\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{physics}\mathrm{book}=\end{array}â€„â‚¹â€„60\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{economics}\mathrm{book}=\end{array}â€„â‚¹â€„40\\ \begin{array}{l}\mathrm{The}\mathrm{total}\mathrm{amount}\mathrm{the}\mathrm{bookshop}\mathrm{will}\mathrm{receive}\end{array}\\ \begin{array}{l}=\left[\mathrm{12096120}\right]\left[\begin{array}{l}\mathrm{80}\\ \mathrm{60}\\ \mathrm{40}\end{array}\right]\end{array}\\ \begin{array}{l}=â€„\left[9600+5760+4800\right]\end{array}\\ \begin{array}{l}=â€„\left[\mathrm{20160}\right]\end{array}\\ \begin{array}{l}\mathrm{Thus},\mathrm{the}\mathrm{total}\mathrm{amount}\mathrm{the}\mathrm{bookshop}\mathrm{will}\mathrm{receive}\mathrm{from}\mathrm{selling}\end{array}\mathrm{â€„}\begin{array}{l}\mathrm{all}\mathrm{the}\mathrm{books}\mathrm{ }\mathrm{is}\end{array}â€„â‚¹\mathrm{â€„}20,â€„160\mathrm{.}\end{array}$

Q.21 Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Exercises 21 and 22.

The restriction of n, k and p so that PY + WY will be defined are:

(A) k = 3, p = n

(B) k is arbitrary, p = 2

(C) p is arbitrary, k = 3

(D) k = 2, p = 3

Ans

 Matrix Order X 2 x n Y 3 x k Z 2 x p W n x 3

PY + WY=Order of P× Order of Y+ Order of W× Order of Y

= ( p×k )×( 3×k )+( n×3 )×( 3×k )

= ( p×k )+( n×k )

Multiplication of PY is possible if k=3 and sum of PY and WY is possible if p = n.

Thus, option ( A ) is correct.

Q.22 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k respectively. Choose the correct answer in Exercises 21 and 22.

If n = p, then the order of the matrix 7X – 5Z is:

(A) P × 2

(B) 2 × n

(C) n × 3

(D) p × n

Ans

$\begin{array}{l}\mathrm{The}\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{X}\mathrm{is}2×\mathrm{n}\\ \mathrm{The}\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{Z}\mathrm{is}2×\mathrm{p}\\ 7\mathrm{X}-5\mathrm{Z}\mathrm{is}\mathrm{defined}\mathrm{when}\mathrm{X}\mathrm{and}\mathrm{Z}\mathrm{an}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{order}\mathrm{.}\\ ⇒\mathrm{n}=\mathrm{p}\left(\mathrm{Given}\right)\\ \mathrm{Thus}\mathrm{the}\mathrm{order}\mathrm{of}7\mathrm{X}-5\mathrm{Z}\mathrm{is}2×\mathrm{n}\mathrm{.}\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{correct}\mathrm{answer}\mathrm{.}\end{array}$

## 1. What are the topics and sub-topics contained in Class 12 Maths Chapter 3?

• 3.1 Introduction
• 3.2 Matrix
• 3.2.1 Order of a matrix
• 3.3 Types of Matrices
• 3.3.1 Equality of Matrices
• 3.4 Operations on Matrices
• 3.4.2 Multiplication of a Matrix by a scalar
• 3.4.3 Properties of matrix addition
• 3.4.4 Properties of scalar multiplication of a Matrix
• 3.4.5 Multiplication of Matrices
• 3.4.6 Properties of multiplication of Matrices
• 3.5. Transpose of a Matrix
• 3.5.1 Properties of the transpose of the Matrices
• 3.6 Symmetric and Skew Symmetric Matrices
• 3.7 Elementary Operation (Transformation) of a Matrix
• 3.8 Invertible Matrices
• 3.8.1 Inverse of a Matrix by Elementary Operations

## 2. How many questions are there in Class 12 Maths Chapter 3 Exercise 3.2?

Exercise 3.2 of Chapter 3 named Matrices in the NCERT Class 12 book contains a total of 22 questions. The solutions provided by the Extramarks’ experts contain answers to clarify doubts. These solutions are carefully provided by our in-house subject matter experts.

## 3. How can I find NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2 to prepare for exams?

Students can download NCERT Solutions for Chapter 3 Class 12 Maths from Extramarks. Complete Solutions for Chapter 3 Matrices of Class 12 Maths are given in such a simple, easy, and comprehensive method so that conceptual clarity can be gained. They can utilize NCERT questions to revise the main concepts given in NCERT Solutions for Class 12 Maths Chapter 3 to score high marks. The NCERT Solutions are given by experts so that students can understand the main concepts more readily. Apart from NCERT solutions, students can also access multiple other learning tools that can help them learn in the comfort of their homes. Extramarks functions as a 24×7 learning guide for students. They can download multiple things from the website and the app that can be accessed later. Students can study and learn at any time, anywhere with the help of the Extramarks app.

NCERT Solutions Class-wise List for Class 1 to 12

There is a list of NCERT Solutions that is created by the subject matter experts to help students with the preparations for their board exams and other competitive exams. These solutions are provided in such a simple language that is easy to understand. These solutions provide students with assistance to prepare for the examination by using various ways and answering the questions by getting an overwhelming victory. Students can access other NCERT Solutions by clicking on the links provided below:

• NCERT Solutions Class 12
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