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

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

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 (Ex 4.6) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 4 Determinants Exercise 4.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

### Important Topics

The topics covered in the NCERT Class 12 Maths Chapter 4 are given below.

 Sections Topics 4.1 Introduction 4.2 Determinant 4.3 Properties of Determinants 4.4 Area of a Triangle 4.5 Adjoint and Inverse of a Matrix 4.6 Applications of Determinants and Matrices
Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline) Book your Free Demo session
Get a flavour of LIVE classes here at Vedantu ## Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

1. Examine the consistency of the system of equations.

$x + 2y = 2$

$2x + 3y = 3$

Ans: The given system of equations is:

$x + 2y = 2$

$2x + 3y = 3$

The given system of equations is:

$\begin{array}{l} x+2y=2 \\ 2x+3y=3 \end{array}$

The given system of equations can be written in the form of $A X=B$, where $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], X=\left[\begin{array}{l}2 \\ 3\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

Now, $|A|=1(3)-2(2)=3-4=-1 \neq 0$

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

2. Examine the consistency of the system of equations.

$2x - y = 5$

$x + y = 4$

Ans: The given system of equations is:

$\begin{array}{l} 2 x-y=5 \\ x+y=4 \end{array}$

The given system of equation can be written in the form of $A X=B$, where $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 4\end{array}\right]$ $|A|=2(1)-(-1)(1)=2+1=3 \neq 0$

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

3. Examine the consistency of the system of equations.

$x + 3y = 5$

$2x + 6y = 8$

Ans: The given system of equations is:

$\begin{array}{l} x+3 y=5 \\ 2 x+6 y=8 \end{array}$

The given system of equation can be written in the form of $A X=B$,

where $A=\left[\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 8\end{array}\right]$

Now, $|A|=1(6)-3(2)=6-6=0$

$\therefore A$ is a singular matrix. $(\operatorname{ad} j A)=\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$

$(a d j A) B=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \end{array}\right]=\left[\begin{array}{l} 30-24 \\ -10+8 \end{array}\right]=\left[\begin{array}{c} 6 \\ -2 \end{array}\right] \neq 0$

Thus, the solution of the given system of equations does not exists. Hence, the given system of equations is inconsistent.

4. Examine the consistency

$x + y + z = 1$

$2x + 3y + 2z = 2$

$ax + ay + 2az = 4$

Ans: The given system of equations is:

$x + y + z = 1$

$2x + 3y + 2z = 2$

$ax + ay + 2az = 4$

The system of equation can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right]$

Now, $|A|=1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)$

$=4 a-2 a-a=4 a-3 a=a \neq 0$

$\therefore A$ is a non-singular matrix. Therefore, $A^{-1}$ exists. Hence, the given system of equation is consistent.

5. Examine the consistency of the system of equations.

$3x - y - 2z = 2$

$2y - z = - 1$

$3x - 5y = 3$

Ans: The given system of equation is:

$3x - y - 2z = 2$

$2y - z = - 1$

$3x - 5y = 3$

This system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$

Now, $|A|=3(-5)-0+3(1+4)=-15+15=0$

$\therefore A$ is a singular matrix.

Now $(\operatorname{adj} A)=\left[\begin{array}{ccc}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]$

$\therefore(a d j A) B=\left[\begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$

$=\left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right]=\left[\begin{array}{l} -5 \\ -3 \\ -6 \end{array}\right] \neq 0$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

6. Examine the consistency of the system of equations.

$5x - y + 4z = 5$

$2x + 3y + 5z = 2$

$5x - 2y + 6z = - 1$

Ans: The given system of equation is:

$5x - y + 4z = 5$

$2x + 3y + 5z = 2$

$5x - 2y + 6z = - 1$

The system of equation can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 3 & -2 & 6\end{array}\right], X\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 2 \\ -1\end{array}\right]$

Now, $|A|=5(18+10)+1(12-25)+4(-4-15)$

$\begin{array}{l} =5(28)+1(-13)+4(-19) \\ =140-13-76 \\ =51 \neq 0 \end{array}$

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists. Hence, the given system of equations is consistent.

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

$5x + 2y = 4$

$7x + 3y = 5$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$

Now $|A|=15-14-1 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

$\begin{array}{l} A^{-1}=\dfrac{1}{|\mathrm{~A}|}(\operatorname{adj} A) \\ \therefore A^{-1}=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right] \\ \therefore X=A^{-1} B=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]\left[\begin{array}{l} 4 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 12-10 \\ -28+25 \end{array}\right]=\left[\begin{array}{c} 2 \\ -3 \end{array}\right] \end{array}$

Hence, $x=2$ and $y=-3$

8 Solve the system of linear equations, using the matrix method.

$2x - y = - 2$

$3x + 4y = 3$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}-2 \\ 3\end{array}\right]$

Now, $|A|=8+3=11 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

$\begin{array}{l} A^{-1}=|A|^{1}(\operatorname{adj} A)=\dfrac{1}{11}\left[\begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array}\right] \\ \therefore X=A^{-1} B=\dfrac{1}{11}\left[\begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} -2 \\ 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{c} -8+3 \\ 6+6 \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{l} -5 \\ 12 \end{array}\right]=\left[\begin{array}{c} -\dfrac{5}{11} \\ \dfrac{12}{11} \end{array}\right] \end{array}$

Hence, $x=\dfrac{-5}{11}$ and $y=\dfrac{12}{11}$.

9. Solve the system of linear equations, using the matrix method.

$4x - 3y = 3$

$3x - 5y = 7$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$

Now, $|A|=-20+9=-11 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now, $A^{-1}=\dfrac{1}{\mid A}(a d i-A)=-\dfrac{1}{11}\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right]=\dfrac{1}{11}\left[\begin{array}{cc}5 & -3 \\ 3 & -4\end{array}\right]$

$\therefore X=A^{-1} B=\dfrac{1}{11}\left[\begin{array}{ll} 5 & -3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{l} 3 \\ 7 \end{array}\right]$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{ll} 5 & -3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{l} 3 \\ 7 \end{array}\right]$

\begin{aligned} =& \dfrac{1}{11}\left[\begin{array}{c} 15-21 \\ 9-28 \end{array}\right] \\ &=\dfrac{1}{11}\left[\begin{array}{c} -6 \\ -19 \end{array}\right] \\ =\left[\begin{array}{r} -\dfrac{6}{11} \\ -\dfrac{19}{11} \end{array}\right] \end{aligned}

Hence, $x=\dfrac{-6}{11}$ and $y=\dfrac{-19}{11}$

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

$5x + 2y = 3$

$3x + 2y = 5$

Ans: The system of equation is

$\begin{array}{l} 5 x+2 y=3 \\ 3 x+2 y=5 \end{array}$

Writing the above equation as $\mathrm{AX}=\mathrm{B}$

$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$

Hence $A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right] \&\; B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$

Calculating |A|

\begin{aligned} |\mathrm{A}| &=\left|\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right| \\ &=5(2)-3(2)=10-6=4 \end{aligned}

Since $|\mathrm{A}| \neq 0$

The System of equation is consistent and has a unique solution

Now,

$\begin{array}{l} A X=B \\ X=A^{-1} B \end{array}$

Calculating $\mathrm{A}^{-1}$

$A^{-1}=\dfrac{1}{|A|} \operatorname{adj}(A)$

Interchange sign

$\operatorname{adj} A=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$

Now,

$\begin{array}{l} \mathrm{A}^{-1}=\dfrac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A} \\ \mathrm{A}^{-1}=\dfrac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \end{array}$

Thus,

$X=A^{-1} B$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 2(3)+(-2) 5 \\ -3(3)+5(5) \end{array}\right]$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 6-10 \\ -9+25 \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} -4 \\ 16 \end{array}\right]$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -1 \\ 4 \end{array}\right]$

Hence, $x=-1 \;\&\; y=4$

11. Solve the system of linear equations, using the matrix method.

$2x + y + z = 1$

$x - 2y - z = \dfrac{3}{2}$

$3y - 5z = 9$

Ans: The given system can be written as $A X=B$, where

$A=\left[\begin{array}{ccc} 2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 3 \\ 9 \end{array}\right]$

$\begin{array}{l}\left|\begin{array}{lll}2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5\end{array}\right| \\ = & 2(20+6)-1(-10-0)+1(6-0) \\ = & 52+10+6=68 \neq 0\end{array}$

Thus, $\mathrm{A}$ is non-singular, Therefore, its inverse exists.

Therefore, the given system is consistent and has a unique solution given by $X=$ $A^{-1} B$

Cofactors of $A$ are

$\begin{array}{l} A_{11}=20+6=26 \\ A_{12}=-(-10+0)=10 \\ A_{13}=6+0=6 \\ A_{21}=-(-5-3)=8 \\ A_{22}=-10-0=-10 \\ A_{23}=-(6-0)=-6 \\ A_{31}=(-2+4)=2 \\ A_{32}=-(-4-2)=6 \\ A_{33}=-8-2=-10 \end{array}$

$\operatorname{adj}(A)=\left[\begin{array}{ccc}26 & 10 & 6 \\ 8 & -10 & -6 \\ 2 & 6 & -10\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{68}\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]$

Now, $X=A^{-1} B \Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$

$=\frac{1}{68}\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{68}\left[\begin{array}{c}26+24+18 \\ 10-30+54 \\ 6-18-90\end{array}\right]$

$=\frac{1}{68}\left[\begin{array}{c}68 \\ 34 \\ -102\end{array}\right]=\left[\begin{array}{c}1 \\ \frac{1}{2} \\ \frac{-3}{2}\end{array}\right]$

Hence, $x=1, y=\frac{1}{2}$ and $z=\frac{-3}{2}$

12. Solve a system of linear equations, using matrix method.

$x - y + z = 4$

$2x + y - 3z = 0$

$x + y + z = 2$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$

Now, $|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$

Thus $A$ is non-singular. Therefore, its inverse exists. Now, $A_{11}=4, A_{12}=-5, A_{13}=1$

$\begin{array}{l} A_{21}=2, A_{22}=0, A_{23}=-2 \\ A_{31}=2, A_{32}=5, A_{33}=3 \\ \therefore A^{-1}=\dfrac{1}{|A|}(a d j A)=\dfrac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right] \end{array}$

$\begin{array}{l} \therefore X=A^{-1} B=\dfrac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right]\left[\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]-\dfrac{1}{10}\left[\begin{array}{c} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{array}\right] \\ =\dfrac{1}{10}\left[\begin{array}{c} 20 \\ -10 \\ 10 \end{array}\right] \end{array}$

$=\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right]$

Hence, $x=2,\; y=-1,\;\text{&}\; z=1$

13. Solve the system of linear equations, using the matrix method. $2x + 3y + 3z = 5$

$x - 2y + z = - 4$

$3x - y - 2z = 3$

Ans: The given system of equation can be written in the form of $A X=B$ where

$\begin{array}{c} |A|=2(4+1)-3(2-3)+3(-1+6) \\ \quad=2(5)-3(-5)+3(5) \\ =10+15+15=40 \neq 0 \end{array}$

Thus, $A$ is non-singular. Therefore, its inverse exists. Now.

$\begin{array}{l} A_{11}=5, A_{2}=5, A_{13}=5 \\ A_{21}=3, A_{22}=-13, A_{23}-11 \\ A_{34}=9, A_{12}=1, A_{35}=-7 \\ \therefore A^{-1}=\dfrac{1}{|A|}(a d j A)=\dfrac{1}{40}\left[\begin{array}{ccc} 5 & 3 & 9 \\5 & -13 & 1 \\ 5 & 11 & -7 \end{array}\right] \end{array}$

$\begin{array}{l} \therefore X=A^{-1} B=\dfrac{1}{40}\left[\begin{array}{ccc} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array}\right]\left[\begin{array}{c} 5 \\ -4 \\ 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} y \\ z \end{array}\right]=\dfrac{1}{40}\left[\begin{array}{c} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{array}\right] \end{array}$

$=\dfrac{1}{40}\left[\begin{array}{c} 40 \\ 80 \\ -40 \end{array}\right]$

$=\left[\begin{array}{c} 1 \\ 2 \\ -1 \end{array}\right]$

Hence, $x=1, y=2$ and $z=-1$

14. Solve the system of linear equations, using the matrix method.

$x - y + 2z = 7$

$3x + 4y - 5z = - 5$

$2x - y + 3z = 12$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$

Now,

$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists. Now, $A_{11}=7, A_{12}=-19, A_{3}=11$

$\begin{array}{l} A_{21}=1, A_{22}=-1, A_{23}=-1 \\ A_{31}=-3, A_{12}=11, A_{35}=7 \end{array}$

$\therefore A^{-1}=\left.\left.\right|_{A}\right|^{1}(\operatorname{adj} A)=\dfrac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]$

$\therefore X=A^{-1} B=\dfrac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right]$

$\Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{array}\right]$

$=\dfrac{1}{4}\left[\begin{array}{c} 8 \\ 4 \\ 12 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right]$

Hence, $x=2, y=1$ and $z=3$.

15. If $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$, find $A^{-1}$ Using $A^{-1}$ solve the system of equations $2 x-3 y+5 z=11$

$3 x+2 y-4 z=-5$ $x+y-2 z=-3$

Ans: $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]$

$\therefore A \mid=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0$

Now, $A_{11}=0, A_{2}=2, A_{3}=1$

$\begin{array}{l} A_{31}=-1, A_{22}=-9, A_{23}=-5 \\ A_{31}=2, A_{32}=23, A_{33}=13 \\ \therefore A^{-1}=\dfrac{1}{|A|}(\operatorname{adj} A)=-\left[\begin{array}{lll} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] \end{array}$

Now, the given system of equations can be written in the form of $A X=B$,

where $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]$

The solution of the system of equations is given by $X=A^{-1} B\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]$ Using (1)

$=\left[\begin{array}{c} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$

Hence $x=1,\; y=2$, and $z=3$

16. The cost of $4{\text{Kg}}$ onion, $3\;{\text{kg}}$ wheat and $2\;{\text{kg}}$ rice is ${\text{Rs}}60$. The cost of $2\;{\text{kg}}$ onion, $4\;{\text{kg}}$ wheat and 6Kg rice is Rs 90. The cost of $6\;{\text{kg}}$ onion $2\;{\text{kg}}$ wheat and $3\;{\text{kg}}$ rice is Rs 70 .

Find cost of each item per kg by matrix method

Ans: Let the cost of onions, wheat and rice per ${\text{kg}}$ be Rs. X and Rs. Z respectively.

Then, the given situation can be represented by a system of equations as:

$4x + 3y + 2z = 60$

$2x + 4y + 6z = 90$

$6x + 2y + 3z - 70$

This system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right], X\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$

$|A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 \neq 0$

Now,

$\begin{array}{l} A_{11}=0, A_{2}=30, A_{13}=-20 \\ A_{21}=-5, A_{22}=0, A_{23}=10 \\ A_{31}=10, A_{32}=-20, A_{33}=10 \\ \therefore \operatorname{adj} A=\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] \\ \therefore A^{-1}=\left.A\right|^{1} \operatorname{adj} A=\dfrac{1}{50}\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] \end{array}$

Now,

$\begin{array}{l} X=A^{-1} B \\ \Rightarrow X=\dfrac{1}{50}\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right]\left[\begin{array}{l} 60 \\ 90 \\ 70 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\dfrac{1}{50}\left[\begin{array}{c} 0-450+700 \\ 1800+0-1400 \\ -1200+900+700 \end{array}\right] \\ =\dfrac{1}{50}\left[\begin{array}{l} 250 \\ 400 \\ 400 \end{array}\right] \end{array}$

$=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$

$\therefore x=5, y=8$, and $z=8$

Hence, the cost of onions is $5 R s$ per $\mathrm{kg}$, the cost of wheat is $8 \mathrm{Rs}$ per $\mathrm{kg}$, and the cost of rice is $8 \mathrm{Rs}$ per $\mathrm{kg}$.

### Important Points

• Determinant is defined as the numerical value of the square matrix. If A is a square matrix i.e A = [aij] of order n, then the determinant of this matrix is denoted by det A or |A|.

• The adjoint of a square matrix ‘A’ is defined as the transpose of the matrix obtained by co-factors of each element of a determinant corresponding to that given matrix. It is denoted by adj(A).

• Hence the adjoint of a matrix A = [aij] n×n is a matrix [Aji] n×n, where Aji is a cofactor of element aji.

• The properties of an Adjoint Matrix are given below.

NCERT Solution Class 12 Maths of Chapter 4 All Exercises

NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6

Opting for the NCERT solutions for Ex 4.6 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.6 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 4 Exercise 4.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 4 Exercise 4.6, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 4 Exercise 4.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. SHARE TWEET SHARE SUBSCRIBE