# NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 (Ex 4.6)

Class 12 plays an important role in the life of a student as well as their parents. The hard work that a student puts in all his or her school life comes full circle. The hopes and dreams of millions of children and their parents depend upon the Class 12 board exams. Students spend countless hours to make sure they secure good marks in the examination.

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## NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 (Ex 4.6)

The Class 12 Mathematics syllabus is vast and one needs to have an immaculate plan from the beginning. Students are suggested to finish the whole syllabus in a few months to have proper time for revision. Early preparation helps, so it doesn’t feel like a mountain to climb in the end. The NCERT books are your one-stop solution to beat the odds and score stellar marks in Mathematics.

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Further, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 help you get a comprehensive idea regarding the application of Determinants and Matrices. It ensures that you are able to solve the questions with the utmost accuracy and confidence.

### Important Topics

 Sections Topics 4.1 Introduction 4.2 Determinant 4.3 Properties of Determinants 4.4 Area of a Triangle 4.5 Minors and Cofactors 4.6 Adjoint and Inverse of a Matrix 4.7 Applications of Determinants and Matrices

The introduction part gives you a brief idea of the various areas where determinants could be useful in real-life situations. The objective behind going through these chapters is provided in detail. Determinants are applicable in the study of Engineering, Science, Economics, and Social Science. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 discusses the condition for the existence of the inverse of a matrix.

Here, students study various properties of Determinants, Minors, Cofactors, and Application Of Determinants in Finding the Area Of a Triangle, Adjoint And Inverse of a Square Matrix. Solutions of Linear Equations in two or three variables using the inverse of a matrix. Consistency and inconsistency of system of linear equations. Four Properties of Determinants are essential for students to understand the topic and are useful to solve complex linear equations.

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### Important Points

NCERT Solutions help the student adapt to the style of writing the answers that give them an approach to perform well in exams. NCERT solutions help students to prepare for board exams. Extramarks provides NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1.

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### NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6

If a student wants to do well in the board exams, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 really helps the student. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 prepared by the experts at Extramarks accelerate their learning and growth. The biggest reason students fail to achieve good marks is the lack of authentic and detailed solutions.

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### Related Questions

Determinant is described as the numerical value of the square matrix. If A is a square matrix, which is a matrix that has an equal number of rows and columns, i.e A = aij of order n, then the determinant of this matrix is denoted by det A. The element aij signifies the entry in the ith row or jth column. In simple terms, a determinant is a special number that is calculated from the column.

Apart from the Class 12 Maths Chapter 4 Exercise 4.6, there are other questions for which complete and detailed NCERT Solutions are given, that are authentic and comprehensive.

Apart from NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6, Extramarks provides solutions for other exercises of determinants.

In exercise 4.1, the first exercise deals with evaluating the Determinants. Other than the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6. Extramarks provides complete solutions for this exercise as well.

Exercise 4.2 deals with the properties of Determinants. For questions 1 to 16, it is required to prove using the properties of determinants without expanding. Extramarks provides genuine solutions to these exercises apart from the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6.

Exercise 4.3 deals with the Area of a Triangle. Questions 1 to 5 deal with finding the area of a triangle, showing the points are collinear, finding the value of k, and finding the equation of the line joining using determinants, respectively. Extensive NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 and Exercise 4.3 would prepare students in less time. The questions in these exercises are vital when preparing for Determinants to attain terrific marks in these topics.

### Chapter 4 – Determinants of Other Exercises

 Chapter 4 – Determinants of Other Exercises Exercise 4.1 8 Questions & Solutions (3 Short Answers, 5 Long Answers) Exercise 4.2 10 Questions & Solutions (4 Short Answers, 10 Long Answers) Exercise 4.3 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.4 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.5 18 Questions & Solutions (4 Short Answers, 14 Long Answers)

Q.1 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3

Ans

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

Hence, the given system of equations is consistent.

Q.2 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

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

Q.3 Examine the consistency of the system of equations.

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

Ans

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

Q.4 Examine the consistency of the system of equations.

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

Ans

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

Q.5 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}:\\ \mathrm{ }3\mathrm{x}-\mathrm{y}-2\mathrm{z}=2\\ \mathrm{ }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], \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},\\ |\mathrm{A}|=|\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}|\\ =3\left(0-5\right)+1\left(0+3\right)-2\left(0-6\right)\\ =-15+3+12=0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{sigular}\mathrm{matrix}\mathrm{.}\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{cc}2& -1\\ -5& 0\end{array}|=-5\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{cc}0& -1\\ 3& 0\end{array}|=-3\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{cc}0& 2\\ 3& -5\end{array}|=-6\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{cc}-1& -2\\ -5& 0\end{array}|=10\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{cc}3& -2\\ 3& 0\end{array}|=6\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{cc}3& -1\\ 3& -5\end{array}|=12\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{cc}-1& -2\\ 2& -1\end{array}|=5\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{cc}3& -2\\ 0& -1\end{array}|=3\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{cc}3& -1\\ 0& 2\end{array}|=6\\ \left(\mathrm{Adj}\mathrm{A}\right)={\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]\\ \left(\mathrm{Adj}\mathrm{A}\right)\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.6 Examine the consistency of the system of equations.

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

Ans

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

Q.7 Solve system of linear equations, using matrix method,

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

Ans

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

Q.8 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:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x-y=-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}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],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ Now,\\ |A|=|\begin{array}{cc}2& -1\\ 3& 4\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8+3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}4& -3\\ 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}\right]}^{‘}=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{-1}}=\frac{1}{|A|}\left(\text{Adj A}\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ \therefore \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}}X={A}^{-1}B\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{l}-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{l}-8+3\\ 6+6\end{array}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-5\\ 12\end{array}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\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}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}y=\frac{12}{11}.\end{array}$

Q.9 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:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x-3y=3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}3\\ 7\end{array}\right]\\ Now,\\ |A|=|\begin{array}{cc}4& -3\\ 3& -5\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-20+9\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-11\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}-5& -3\\ 3& \text{\hspace{0.17em}}\text{\hspace{0.17em}}4\end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{-1}}=\frac{1}{|A|}\left(\text{Adj A}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\\ \therefore \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}}X={A}^{-1}B\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{l}3\\ 7\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{l}15-21\\ 9-28\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-6\\ -19\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\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}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}y=\frac{-19}{11}.\end{array}$

Q.10 Solve system of linear equations, using matrix method,

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

Ans

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

Q.11 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:}\\ \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}}\text{\hspace{0.17em}}2x+y+z=1\\ \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}}x-2y-z=\frac{3}{2}\\ \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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ 1& -2& -1\\ 0& 3& -5\end{array}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}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,\\ |A|=|\begin{array}{ccc}2& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ 1& -2& -1\\ 0& 3& -5\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(10+3\right)-1\left(-5-0\right)+1\left(3-0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=26+5+3=34\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ Now,\text{\hspace{0.17em}}\\ {A}_{11}=13,{A}_{12}=5,{A}_{13}=3\\ {A}_{21}=8,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{22}=-10,{A}_{23}=-6\\ {A}_{31}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{32}=3,{A}_{33}=-5\\ Adj\text{\hspace{0.17em}}A={\left[\begin{array}{ccc}13& 5& 3\\ 8& -10& -6\\ 1& 3& -5\end{array}\right]}^{‘}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\frac{1}{|A|}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \therefore X=\text{\hspace{0.17em}}{A}^{-1}B\\ \text{}=\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]\\ \text{}=\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}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}34\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}17\\ -51\end{array}\right]\text{\hspace{0.17em}}=\left[\begin{array}{l}1\\ \frac{1}{2}\\ -\frac{3}{2}\end{array}\right]\\ Hence,\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}y=\frac{1}{2}\text{and z}=-\frac{3}{2}.\end{array}$

Q.12 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}:\\ \mathrm{x}-\mathrm{y}+\mathrm{z}=4\\ \mathrm{ }2\mathrm{x}+\mathrm{y}-3\mathrm{z}=0\\ \mathrm{ }\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& \mathrm{ }1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right], \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},\\ |\mathrm{A}|=|\begin{array}{ccc}1& -1& \mathrm{ }1\\ 2& 1& -3\\ 1& 1& 1\end{array}|\\ =1\left(1+3\right)+1\left(2+3\right)+1\left(2-1\right)\\ =4+5+1=10\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=4,{\mathrm{A}}_{12}=-5,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=2, \mathrm{ }{\mathrm{A}}_{22}=0,{\mathrm{A}}_{23}=-2\\ {\mathrm{A}}_{31}=2, {\mathrm{A}}_{32}=5,{\mathrm{A}}_{33}=3\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ =\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\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} \mathrm{ }20\\ -10\\ \mathrm{ }10\end{array}\right]\\ =\left[\begin{array}{l}2\\ -1\\ 1\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=-1\mathrm{and}\mathrm{z}=1.\end{array}$

Q.13 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}:\\ \mathrm{ }2\mathrm{x}+3\mathrm{y}+3\mathrm{z}=5\\ \mathrm{ }\mathrm{x}-2\mathrm{y}+\mathrm{z}=-4\\ \mathrm{ }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{ }3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l} \mathrm{ }5\\ -4\\ \mathrm{ }3\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}2& 3& \mathrm{ }3\\ 1& -2& 1\\ 3& -1& -2\end{array}|=2\left(4+1\right)-3\left(-2-3\right)+3\left(-1+6\right)\\ =10+15+15=40\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=5, {\mathrm{A}}_{12}=5,{\mathrm{A}}_{13}=5\\ {\mathrm{A}}_{21}=3, \mathrm{ }{\mathrm{A}}_{22}=-13, {\mathrm{A}}_{23}=11\\ {\mathrm{A}}_{31}=9, {\mathrm{A}}_{32}=1, {\mathrm{A}}_{33}=-7\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}5& 5& 5\\ 3& -13& 11\\ 9& 1& -7\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\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} \mathrm{ }5\\ -4\\ \mathrm{ }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} \mathrm{ }40\\ \mathrm{ }80\\ -40\end{array}\right]\\ =\left[\begin{array}{l}1\\ 2\\ -1\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=1,\mathrm{ }\mathrm{y}=2\mathrm{and}\mathrm{z}=-1.\end{array}$

Q.14 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}:\\ \mathrm{ }\mathrm{x}-\mathrm{y}+2\mathrm{z}=7\\ 3\mathrm{x}+4\mathrm{y}-5\mathrm{z}=-5\\ \mathrm{ }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{ }2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l} \mathrm{ }7\\ -5\\ \mathrm{ }12\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}|\\ =1\left(12-5\right)+1\left(9+10\right)+2\left(-3-8\right)\\ =7+19-22=4\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=7, {\mathrm{A}}_{12}=-19,{\mathrm{A}}_{13}=-11\\ {\mathrm{A}}_{21}=1, \mathrm{ }{\mathrm{A}}_{22}=-1, {\mathrm{A}}_{23}=-1\\ {\mathrm{A}}_{31}=-3, {\mathrm{A}}_{32}=11, {\mathrm{A}}_{33}=7\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\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} \mathrm{ }7\\ -5\\ \mathrm{ }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{ }=\left[\begin{array}{l}2\\ 1\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=1\mathrm{and}\mathrm{z}=3.\end{array}$

Q.15

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

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}|\\ =2\left(-4+4\right)+3\left(-6+4\right)+5\left(3-2\right)\\ =0-6+5\\ =-1\ne 0\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=0, {\mathrm{A}}_{12}=2,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=-1, \mathrm{ }{\mathrm{A}}_{22}=-9, {\mathrm{A}}_{23}=-5\\ {\mathrm{A}}_{31}=2, {\mathrm{A}}_{32}=23, {\mathrm{A}}_{33}=13\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \mathrm{ }\therefore {\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{ }\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{ }{\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]\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\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{ }\mathrm{x}=1,\mathrm{ }\mathrm{y}=2\mathrm{and}\mathrm{z}=3.\end{array}$

Q.16 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],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]\text{ }\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|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\left(12-12\right)-3\left(6-36\right)+2\left(4-24\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0+90-40=50\ne 0\\ Now,\text{\hspace{0.17em}}\\ {A}_{11}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{12}=30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{13}=-20\\ {A}_{21}=-5,\text{\hspace{0.17em}}{A}_{22}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{23}=10\\ {A}_{31}=10,\text{\hspace{0.17em}}{A}_{32}=-20,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{33}=10\\ \therefore Adj\text{\hspace{0.17em}}A={\left[\begin{array}{ccc}0& 30& -20\\ -5& 0& 10\\ 10& -20& 10\end{array}\right]}^{‘}\\ \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}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ \therefore {A}^{-1}=\frac{1}{\left|A\right|}Adj\text{\hspace{0.17em}}A\\ =\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ Now,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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]\\ ⇒\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}5\\ 8\\ 8\end{array}\right]\\ ⇒x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=8\text{\hspace{0.17em}}and\text{\hspace{0.17em}}z=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}$

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### 1. How to check if the system of equations is consistent or inconsistent?

If one or more solutions exist in the system of equations, it is said to be consistent. If no solution exists in the system of equations; it is said to be inconsistent. For step-by-step authentic answers, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 are available at the Extramarks website prepared by experts. Extramarks benefits by interlinking concepts for better retention.

### 2. What are the properties of Determinants?

There are four properties of Determinants:

1) If the rows and columns of a determinant are interchanged, the value of the determinants remains unchanged.

2) The sign of the determinant changes, if any two rows of determinants are interchanged.

3) The value of the determinant is 0 if any two rows or columns of the determinant are equal or identical.

4) The value of the determinant originally obtained is multiplied by k, if each element of row and column is multiplied by a constant value k.

Learn more about the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 that are accessible at the Extramarks website prepared by our subject matter experts. There is a curriculum mapping and full syllabus coverage to help you score well.