# 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
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## 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.

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