Exercise 4.5 (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths

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Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Ex 4.5 Question 1:
Examine the consistency of the system of equations.
$
\begin{aligned}
& x+2 y=2 \\
& 2 x+3 y=3
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& x+2 y=2 \\
& 2 x+3 y=3
\end{aligned}
$

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}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{l}
2 \\
3
\end{array}\right] \text {. }
$

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.

Ex 4.5 Question 2:
Examine the consistency of the system of equations.
$
\begin{aligned}
& 2 x-y=5 \\
& x+y=4
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& 2 x-y=5 \\
& x+y=4
\end{aligned}
$

The given system of equations 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] \text { and } B=\left[\begin{array}{l}
5 \\
4
\end{array}\right] \text {. }
$

Now,
$
|A|=2(1)-(-1)(1)=2+1=3 \neq 0
$
$\therefore A$ is non-sinqular.
Therefore, $A^{-1}$ exists.
Hence, the given system of equations is consistent.

Ex 4.5 Question 3:
Examine the consistency of the system of equations.
$
\begin{aligned}
& x+3 y=5 \\
& 2 x+6 y=8
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& x+3 y=5 \\
& 2 x+6 y=8
\end{aligned}
$

The given system of equations 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] \text { and } B=\left[\begin{array}{l}
5 \\
8
\end{array}\right] \text {. }
$

Now,
$
|A|=1(6)-3(2)=6-6=0
$
$\therefore A$ is a singular matrix.
Now,
$
\begin{aligned}
& (\operatorname{adj} A)=\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right] \\
& (\operatorname{adj} 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 O
\end{aligned}
$

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

Ex 4.5 Question 4:
Examine the consistency of the system of equations.
$
\begin{aligned}
& x+y+z=1 \\
& 2 x+3 y+2 z=2 \\
& a x+a y+2 a z=4
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& x+y+z=1 \\
& 2 x+3 y+2 z=2 \\
& a x+a y+2 a z=4
\end{aligned}
$

This system of equations can be written in the form $A X=B$, where
$
A=\left[\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2 a
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{l}
1 \\
2 \\
4
\end{array}\right] \text {. }
$

Now,
$
\begin{aligned}
|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
\end{aligned}
$
$\therefore A$ is non-singular.

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

Ex 4.5 Question 5:
Examine the consistency of the system of equations.
$
\begin{aligned}
& 3 x-y-2 z=2 \\
& 2 y-z=-1 \\
& 3 x-5 y=3
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& 3 x-y-2 z=2 \\
& 2 y-z=-1 \\
& 3 x-5 y=3
\end{aligned}
$

This system of equations can be written in the form of $A X=B$, 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] \text { and } B=\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right] \text {. }
$

Now,
$
|A|=3(0-5)-0+3(1+4)=-15+15=0
$
$\therefore A$ is a singular matrix.
Now,
$
\begin{aligned}
& (\text { adj } A)=\left[\begin{array}{lll}
-5 & 10 & 5 \\
-3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right] \\
& \therefore(\operatorname{adj} A) B=\left[\begin{array}{lll}
-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}{l}
-10-10+15 \\
-6-6+9 \\
-12-12+18
\end{array}\right]=\left[\begin{array}{l}
-5 \\
-3 \\
-6
\end{array}\right] \neq O
\end{aligned}
$

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

Ex 4.5 Question 6:
Examine the consistency of the system of equations.
$
\begin{aligned}
& 5 x-y+4 z=5 \\
& 2 x+3 y+5 z=2 \\
& 5 x-2 y+6 z=-1
\end{aligned}
$

Answer
The given system of equations is:
$
\begin{aligned}
& 5 x-y+4 z=5 \\
& 2 x+3 y+5 z=2 \\
& 5 x-2 y+6 z=-1
\end{aligned}
$

This system of equations can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{lll}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{c}
5 \\
2 \\
-1
\end{array}\right] \text {. }
$

Now,
$
\begin{aligned}
|A| & =5(18+10)+1(12-25)+4(-4-15) \\
& =5(28)+1(-13)+4(-19) \\
& =140-13-76 \\
& =51 \neq 0
\end{aligned}
$
$\therefore A$ is non-singular.
Therefore, $A^{-1}$ exists.
Hence, the given system of equations is consistent.

Ex 4.5 Question 7:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& 5 x+2 y=4 \\
& 7 x+3 y=5
\end{aligned}
$

Answer
The given system of equations can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{l}
4 \\
5
\end{array}\right] \text {. }
$

Now, $|A|=15-14=1 \neq 0$.
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
& A^{-1}=\frac{1}{|A|}(\text { adj } A) \\
& \therefore A^{-1}=\left[\begin{array}{rr}
3 & -2 \\
-7 & 5
\end{array}\right] \\
& \therefore X=A^{-1} B=\left[\begin{array}{rr}
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{aligned}
$

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

Question 8:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& 2 x-y=-2 \\
& 3 x+4 y=3
\end{aligned}
$

Answer
The given system of equations can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{cc}
2 & -1 \\
3 & 4
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{c}
-2 \\
3
\end{array}\right] \text {. }
$

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

Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
& A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{11}\left[\begin{array}{rr}
4 & 1 \\
-3 & 2
\end{array}\right] \\
& \therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{rr}
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]=\frac{1}{11}\left[\begin{array}{l}
-8+3 \\
6+6
\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}
-5 \\
12
\end{array}\right]=\left[\begin{array}{c}
-\frac{5}{11} \\
\frac{12}{11}
\end{array}\right]
\end{aligned}
$

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

Ex 4.5 Question 9:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& 4 x-3 y=3 \\
& 3 x-5 y=7
\end{aligned}
$

Answer
The given system of equations can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{ll}
4 & -3 \\
3 & -5
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{l}
3 \\
7
\end{array}\right] \text {. }
$

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

Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
& A^{-1}=\frac{1}{|A|}(\text { adj } A)=-\frac{1}{11}\left[\begin{array}{rr}
-5 & 3 \\
-3 & 4
\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}
5 & -3 \\
3 & -4
\end{array}\right] \\
& \therefore X=A^{-1} B=\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}{ll}
5 & -3 \\
3 & -4
\end{array}\right]\left[\begin{array}{l}
3 \\
7
\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}
15-21 \\
9-28
\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}
-6 \\
-19
\end{array}\right]=\left[\begin{array}{l}
-\frac{6}{11} \\
-\frac{19}{11}
\end{array}\right]
\end{aligned}
$

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

Ex 4.5 Question 10:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& 5 x+2 y=3 \\
& 3 x+2 y=5
\end{aligned}
$

Answer

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] \text {. }
$

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

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

Ex 4.5 Question 11:
Solve system of linear equations, using matrix method.
$
\begin{array}{r}
2 x+y+z=1 \\
x-2 y-z=\frac{3}{2} \\
3 y-5 z=9
\end{array}
$

Answer
The qiven system of equations can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{ccc}
2 & 1 & 1 \\
1 & -2 & -1 \\
0 & 3 & -5
\end{array}\right], X=\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{c}
1 \\
\frac{3}{2} \\
9
\end{array}\right] \text {. }
$

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

Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
\text { Now, } 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 \\
\therefore A^{-1}=\frac{1}{|A|}(\text { adj } A) & =\frac{1}{34}\left[\begin{array}{ccc}
13 & 8 & 1 \\
5 & -10 & 3 \\
3 & -6 & -5
\end{array}\right]
\end{aligned}
$
$
\therefore X=A^{-1} B=\frac{1}{34}\left[\begin{array}{ccc}
13 & 8 & 1 \\
5 & -10 & 3 \\
3 & -6 & -5
\end{array}\right]\left[\begin{array}{l}
1 \\
3 \\
\frac{3}{2} \\
9
\end{array}\right]
$
$
\begin{aligned}
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] & =\frac{1}{34}\left[\begin{array}{l}
13+12+9 \\
5-15+27 \\
3-9-45
\end{array}\right] \\
& =\frac{1}{34}\left[\begin{array}{l}
34 \\
17 \\
-51
\end{array}\right]=\left[\begin{array}{c}
1 \\
\frac{1}{2} \\
-\frac{3}{2}
\end{array}\right]
\end{aligned}
$

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

Ex 4.5 Question 12:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& x-y+z=4 \\
& 2 x+y-3 z=0 \\
& x+y+z=2
\end{aligned}
$

Answer
The given system of equations can be written in the form of $A X=B$, 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] \text { and } B=\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right] \text {. }
$

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.
$
\begin{aligned}
\text { 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 \\
\therefore A^{-1}=\frac{1}{|A|}(\text { adj } A) & =\frac{1}{10}\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]
\end{aligned}
$

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

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

Ex 4.5 Question 13:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& 2 x+3 y+3 z=5 \\
& x-2 y+z=-4 \\
& 3 x-y-2 z=3
\end{aligned}
$

Answer
The given system of equations can be written in the form $A X=B$, where
$
A=\left[\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{l}
5 \\
-4 \\
3
\end{array}\right] \text {. }
$

Now,
$
|A|=2(4+1)-3(-2-3)+3(-1+6)=2(5)-3(-5)+3(5)=10+15+15=40 \neq 0
$

Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
& 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
\end{aligned}
$
$
\begin{aligned}
& \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{40}\left[\begin{array}{ccc}
5 & 3 & 9 \\
5 & -13 & 1 \\
5 & 11 & -7
\end{array}\right] \\
& \therefore X=A^{-1} B=\frac{1}{40}\left[\begin{array}{ccc}
5 & 3 & 9 \\
5 & -13 & 1 \\
5 & 11 & -7
\end{array}\right]\left[\begin{array}{l}
5 \\
-4 \\
3
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{40}\left[\begin{array}{l}
25-12+27 \\
25+52+3 \\
25-44-21
\end{array}\right]
\end{aligned}
$

$
\begin{aligned}
& =\frac{1}{40}\left[\begin{array}{l}
40 \\
80 \\
-40
\end{array}\right] \\
& =\left[\begin{array}{c}
1 \\
2 \\
-1
\end{array}\right]
\end{aligned}
$

Hence, $x=1, v=2$ and $z=-1$.
Ex 4.5 Question 14:
Solve system of linear equations, using matrix method.
$
\begin{aligned}
& x-y+2 z=7 \\
& 3 x+4 y-5 z=-5 \\
& 2 x-y+3 z=12
\end{aligned}
$

Answer
The given system of equations can be written in the form of $A X=B$, 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] \text { and } B=\left[\begin{array}{c}
7 \\
-5 \\
12
\end{array}\right] \text {. }
$

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,
$
\begin{aligned}
& 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
\end{aligned}
$

$
\begin{aligned}
& \therefore A^{-1}=\frac{1}{|A|}(\text { adj } A)=\frac{1}{4}\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right] \\
& \therefore X=A^{-1} B=\frac{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]=\frac{1}{4}\left[\begin{array}{c}
49-5-36 \\
-133+5+132 \\
-77+5+84
\end{array}\right] \\
& =\frac{1}{4}\left[\begin{array}{l}
8 \\
4 \\
12
\end{array}\right]=\left[\begin{array}{l}
2 \\
1 \\
3
\end{array}\right]
\end{aligned}
$

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

Ex 4.5 Question 15:
$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations
$
\begin{aligned}
2 x-3 y+5 z & =11 \\
3 x+2 y-4 z & =-5 \\
x+y-2 z & =-3
\end{aligned}
$

Answer
$
\begin{aligned}
& A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right] \\
& \therefore|A|=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0
\end{aligned}
$

Now,
$
\begin{aligned}
& \text { 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 \\
& \therefore A^{-1}=\frac{1}{|A|}(\text { 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{aligned}
$

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] \text { and } B=\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right] \text {. }
$

The solution of the system of equations is given by $X=A^{-1} B$.
$
\begin{aligned}
& X=A^{-1} B \\
& \Rightarrow\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}{c}
11 \\
-5 \\
-3
\end{array}\right] \quad[\text { 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] \\
&
\end{aligned}
$

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

Ex 4.5 Question 16:
The cost of $4 \mathrm{~kg}$ onion, $3 \mathrm{~kg}$ wheat and $2 \mathrm{~kg}$ rice is Rs 60 . The cost of $2 \mathrm{~kg}$ onion, $4 \mathrm{~kg}$ wheat and $6 \mathrm{~kg}$ rice is Rs 90 . The cost of $6 \mathrm{~kg}$ onion $2 \mathrm{~kg}$ wheat and $3 \mathrm{~kg}$ rice is Rs 70 .
Find cost of each item per $\mathrm{kg}$ by matrix method.
Answer
Let the cost of onions, wheat, and rice per $\mathrm{kg}$ be Rs $x$, Rs $y$, and Rs $z$ respectively.
Then, the given situation can be represented by a system of equations as:
$
\begin{aligned}
& 4 x+3 y+2 z=60 \\
& 2 x+4 y+6 z=90 \\
& 6 x+2 y+3 z=70
\end{aligned}
$

This system of equations can be written in the form of $A X=B$, where
$
\begin{aligned}
& A=\left[\begin{array}{ccc}
4 & 3 & 2 \\
2 & 4 & 6 \\
6 & 2 & 3
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { 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
\end{aligned}
$

Now,
$
\begin{aligned}
& 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
\end{aligned}
$

$
\begin{aligned}
& \therefore \operatorname{adj} A=\left[\begin{array}{ccc}
0 & -5 & 10 \\
30 & 0 & -20 \\
-20 & 10 & 10
\end{array}\right] \\
& \therefore A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{50}\left[\begin{array}{ccc}
0 & -5 & 10 \\
30 & 0 & -20 \\
-20 & 10 & 10
\end{array}\right]
\end{aligned}
$

Now,
$
\begin{aligned}
& X=A^{-1} B \\
& \Rightarrow X=\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]
\end{aligned}
$

$
\begin{aligned}
& \begin{aligned}
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\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}
5 \\
8 \\
8
\end{array}\right]
\end{aligned} \\
& \therefore x=5, y=8 \text {, and } z=8 .
\end{aligned}
$

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