NCERT Solutions for Class 12 Maths Chapter 4 - Determinants Exercise 4.5

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 $\mathrm{kg}$.

Conclusion

Class 12 Maths Ex 4.5 Chapter 4 focuses on the concept of determinants and their properties. This exercise is essential for learning how to solve systems of linear equations using Cramer's Rule, which helps in simplifying complex problems. Students should concentrate on calculating determinants for 2x2 and 3x3 matrices and understanding cofactor expansions. Additionally, applying determinants to find the area of triangles provides practical applications that reinforce theoretical knowledge. Practicing these problems enhances problem-solving skills and prepares students effectively for exams. Mastering these concepts is crucial for further studies in mathematics and related fields, as emphasized by Vedantu.

Class 12 Maths Chapter 4: Exercises Breakdown

CBSE Class 12 Maths Chapter 4 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths