EXERCISE 4.6 ( Determinants )

Question 1: Examine the consistency of the system of equations:(Class 12 ncert solution math exercise 4.6)

$\begin{aligned} &x+2 y=2 \\ &2 x+3 y=3 \end{aligned}$

Solution: $x+2 y=2$

The given system of equations is: $2 x+3 y=3$

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

$A=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 2 \\ 3 \end{array}\right]$

Hence,

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

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Thus, the given system of equations is consistent.

Question 2: Examine the consistency of the system of equations:

$2 x-y=5$

$x+y=4$

Solution: $2 x-y=5$

The given system of equations is: $x+y=4$

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

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

Hence,

$|A| &=2(1)-1(-1)$
=2+1
$=3 \neq 0$

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Question 3: Examine the consistency of the system of equations:

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

Solution: $x+3 y=5$

The given system of equations is: $2 x+6 y=8$

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

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

Hence,

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

So, $A$ is a singular matrix.

Now,

$(\operatorname{adj} A)=\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$

Therefore,

$(\operatorname{adj} A) B =\begin{bmatrix} 6 & -5 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix}$

$=\begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix}$

$=\begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq 0$

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

Hence, the system of equations is inconsistent.

Question 4: Examine the consistency of the system of equations:

$x+y+z=1$
$2 x+3 y+2 z=2$
$a x+a y+2 a z=4$

Solution: $x+y+z=1$
$2 x+3 y+2 z=2$

The given system of equations is: $a x+a y+2 a z=4$

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

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

Hence,

$|A| =1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)$
$=4 a-2 a-a$
$=4 a-3 a$
$=a \neq 0$

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Thus, the given system of equations is consistent.

Question 5:Examine the consistency of the system of equations:

$3 x-y-2 z=2$
$2 y-z=-1$
$3 x-5 y=3$

Solution:The given system of equations is:

$3 x-y-2 z=2$
$2 y-z=-1$
$3 x-5 y=3$

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

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

Hence,

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

So, $A$ is a singular matrix.

Now,

$(\operatorname{adj} A)=\begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}$

Therefore,

$(\text { adji }) B =\begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$

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

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

Hence, the system of equations is inconsistent.

Question 6:Examine the consistency of the system of equations:

$5 x-y+4 z=5$

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

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

Solution:The given system of equations is:

$5 x-y+4 z=5$
$2 x+3 y+5 z=2$
$5 x-2 y+6 z=-1$

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

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

Hence,

$|A| =5(18+10)+1(12-25)+4(-4-15)$
$=5(28)+1(-13)+4(-19)$
$=140-13-76$
$=51 \neq 0$

So, $A$ is nonsingular. Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Question 7: Solve system of linear equations, using matrix method.
$5 x+2 y=4$
$7 x+3 y=5$

Solution: $5 x+2 y=4$

The given system of equations is: $7 x+3 y=5$

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

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

Hence,

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

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

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

$=\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$

Then,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]\left[\begin{array}{l} 4 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 12-10 \\ -28+25 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 2 \\ -3 \end{array}\right]$

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

Question 8: Solve system of linear equations, using matrix method.

$2 x-y=-2$

$3 x+4 y=3$

Solution: $2 x-y=-2$

The given system of equations is: $3 x+4 y=3$

The given system of equations can be written in the form of

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

Hence,

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

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

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

$=\frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}\left[\begin{array}{c} -2 \\ 3 \end{array}\right]$

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} -8+3 \\ 6+6 \end{array}\right]$

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{11}\left[\begin{array}{l} -5 \\ 12 \end{array}\right]$

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} \frac{-5}{11} \\ \frac{12}{11} \end{array}\right]$

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

Question 9: Solve system of linear equations, using matrix method.

$4 x-3 y=3$

$3 x-5 y=7$

Solution: $4 x-3 y=3$

The given system of equations is: $3 x-5 y=7$

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

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

Hence,

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

So, $A$ is nonsingular. Therefore, $A^{-1}$ exists.

Now,

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

$=-\frac{1}{11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}$

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]$

$=\frac{1}{11}\begin{bmatrix}5 & -3 \\3 & -4\end{bmatrix}\left[\begin{array}{l}3 \\7\end{array}\right]$

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

$\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}15-21 \\9-28\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-6 \\-19\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{c}\frac{-6}{11} \\\frac{-19}{11}\end{array}\right]$

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

Question 10: Solve system of linear equations, using matrix method.

$5 x+2 y=3$

$3 x+2 y=5$

Solution: $5 x+2 y=3$

The given system of equations is: $3 x+2 y=5$

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

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

Hence,

$|A| =10-6$
$=4 \neq 0$

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

$A^{-1} =\frac{1}{|A|}(\text { adjA })$

$=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]$

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$

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

$\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{4}\left[\begin{array}{l} -4 \\ 16 \end{array}\right]$

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

Hence, $x=-1$ and $y=4$

Question 11: Solve system of linear equations, using matrix method.

$2 x+y+z=1$
$x-2 y-z=\frac{3}{2}$
$3 y-5 z=9$

Solution:The given system of equations is:
$2 x+y+z=1$
$x-2 y-z=\frac{3}{2}$
$3 y-5 z=9$

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

$A=\begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}$

Hence,

$|A| =2(10+3)-1(-5-3)+0$
$=2(13)-1(-8)$
$=26+8$
$=34 \neq 0$

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

$A_{11}=13, A_{12}=5, A_{13}=3$
$A_{21}=8, A_{22}=-10, A_{23}=-6$
$A_{31}=1, A_{32}=3, A_{33}=-5$

Hence,

Therefore,

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

$=\frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$

$\Rightarrow X=A^{-1} B$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 13+12+9 \\ 5-15+27 \\ 3-9-45 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

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

Question 12: Solve system of linear equations, using matrix method.

$x-y+z=4$

$2 x+y-3 z=0$

$x+y+z=2$

Solution:The given system of equations is:
$x-y+z=4$
$2 x+y-3 z=0$
$x+y+z=2$

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

$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

Hence,

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

So, $A$ is nonsingular.

Therefore, $A^{-1}$ exists.

Now,

$A_{11}=4, A_{12}=-5, A_{13}=1$
$A_{21}=2, A_{22}=0, A_{23}=-2$
$A_{31}=2, A_{32}=5, A_{33}=3$

Hence,

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

$=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix}{l} 4 \\ 0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$

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

Question 13: Solve system of linear equations, using matrix method.

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

$x-2 y+z=-4$

$3 x-y-2 z=3$

Solution:The given system of equations is:

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

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

$A=\begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$

Hence,

$|A| =2(4+1)-3(-2-3)+3(-1+6)$
$=10+15+15$
$=40 \neq 0$

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

$A_{11}=5, A_{12}=5, A_{13}=5$
$A_{21}=3, A_{22}=-13, A_{23}=11$
$A_{31}=9, A_{32}=1, A_{33}=-7$

Hence,

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

$=\frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{40}\begin{bmatrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{bmatrix}$ .

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{40}\begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$

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

Question 14:Solve system of linear equations, using matrix method.
$x-y+2 z=7$
$3 x+4 y-5 z=-5$
$2 x-y+3 z=12$

Solution:The given system of equations is:
$x-y+2 z=7$
$3 x+4 y-5 z=-5$
$2 x-y+3 z=12$

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

$A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$

Hence,

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

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists

$A_{11}=7, A_{12}=-19, A_{13}=-11$
$A_{21}=1, A_{22}=-1, A_{23}=-1$
$A_{31}=-3, A_{32}=11, A_{33}=7$

Hence,

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

$=\frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$

Therefore,

$\Rightarrow X=A^{-1} B$
$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$

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

Question 15: $A=\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4 \\1 & 1 & -2\end{bmatrix} \text {, find } A^{-1} \text {. Using } A^{-1} \text { solve the system of equations }$
$2 x-3 y+5 z=11$
$3 x+2 y-4 z=-5$
$x+y-2 z=-3$

Solution: It is given that $A=\begin{bmatrix}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{bmatrix}$

Therefore,

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

Now,

$A_{11}=0, A_{12}=2, A_{13}=1$
$A_{21}=-1, A_{22}=-9, A_{23}=-5$
$A_{31}=2, A_{32}=23, A_{33}=13$

Hence,

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

$=-\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}=\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

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

$A=\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

The solution of the system of equations is given by $X=A^{-1} B$ .

Therefore,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

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

Question 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.

Solution: Let the cost of onions, wheat, and rice per $\mathrm{kg}$ in $₹$ be $x, y$ and $z$ respectively.

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

$4 x+3 y+2 z=60$
$2 x+4 y+6 z=90$
$6 x+2 y+3 z=70$

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

$A=\begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \text { and } B=\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$

Therefore,

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

So, $A$ is non-singular.

Therefore, $A^{-1}$ exists.

Now,

$A_{11}=0, A_{12}=30, A_{13}=-20$
$A_{21}=-5, A_{22}=0, A_{23}=10$
$A_{31}=10, A_{32}=-20, A_{33}=10$

Therefore,

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

$=\frac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$

Hence,

$\Rightarrow X=A^{-1} B$

$\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$