NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

Question:2 Examine the consistency of the system of equations

$2x-y=5$

$x+y=4$

Answer:

We have given the system of equations:

$2x-y=5$

$x+y=4$

The given system of equations can be written in the form of matrix; $AX=B$

where $A=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]$,

$X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and

$B=\left[\begin{array}{c}5\\ 4\end{array}\right]$.

So, we want to check for the consistency of the equations;

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

Here A is non -singular therefore there exists $A^{-1}$.

Hence, the given system of equations is consistent.

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

$x+3y=5$

$2x+6y=8$

Answer:

We have given the system of equations:

$x+3y=5$

$2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX=B$

where $A=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$,

$X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and

$B=\left[\begin{array}{c}5\\ 8\end{array}\right]$.

So, we want to check for the consistency of the equations;

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

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]$

So, $(adjA)B=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]\left[\begin{array}{c}5\\ 8\end{array}\right]=\left[\begin{array}{c}30-24\\ -10+8\end{array}\right]=\left[\begin{array}{c}6\\ -2\end{array}\right]\ne 0$

As, $(adjA)B\ne 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

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

$x+y+z=1$

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

$ax+ay+2az=4$

Answer:

We have given the system of equations:

$x+y+z=1$

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

$ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX=B$

where $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ a& a& 2a\end{array}\right]$,

$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and

$B=\left[\begin{array}{c}1\\ 2\\ 4\end{array}\right]$.

So, we want to check for the consistency of the equations;

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

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

[If zero then it won't satisfy the third equation]

Here A is non- singular matrix therefore there exist $A^{-1}$.

Hence, the given system of equations is consistent.

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

$3x-y-2z=2$

$2y-z=-1$

$3x-5y=3$

Answer:

We have given the system of equations:

$3x-y-2z=2$

$2y-z=-1$

$3x-5y=3$

The given system of equations can be written in the form of matrix; $AX=B$

where $A=\left[\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right]$,

$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and

$B=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right]$.

So, we want to check for the consistency of the equations;

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

$=-15+3+12=0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

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

$\therefore (adjA)B=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right]=\left[\begin{array}{c}-10-10+15\\ -6-6+9\\ -12-12+18\end{array}\right]=\left[\begin{array}{c}-5\\ -3\\ -6\end{array}\right]\ne 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

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

$5x-y+4z=5$

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

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

Answer:

We have given the system of equations:

$5x-y+4z=5$

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

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

The given system of equations can be written in the form of the matrix; $AX=B$

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

So, we want to check for the consistency of the equations;

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

$=140-13-76=51\ne 0$

Here A is non- singular matrix therefore there exist $A^{-1}$.

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using matrix method.

$5x+2y=4$

$7x+3y=5$

Answer:

The given system of equations

$5x+2y=4$

$7x+3y=5$

can be written in the matrix form of AX =B, where

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

we have,

$|A|=15-14=1\ne 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{|A|}(adjA)=(adjA)=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]$

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

Hence the solutions of the given system of equations;

x = 2 and y =-3.

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

$2x-y=-2$

$3x+4y=3$

Answer:

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

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

we have,

$|A|=8+3=11\ne 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-8+3\\ 6+6\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-5\\ 12\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11}\\ \\ -\frac{12}{11}\end{array}\right]$

Hence the solutions of the given system of equations;

x = -5/11 and y = 12/11.

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

$4x-3y=3$

$3x-5y=7$

Answer:

The given system of equations

$4x-3y=3$

$3x-5y=7$

can be written in the matrix form of AX =B, where

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

we have,

$|A|=-20+9=-11\ne 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{-1}{11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}15-21\\ 9-28\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-6\\ -19\end{array}\right]=\left[\begin{array}{c}-\frac{6}{11}\\ \\ -\frac{19}{11}\end{array}\right]$

Hence the solutions of the given system of equations;

x = -6/11 and y = -19/11.

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

$5x+2y=3$

$3x+2y=5$

Answer:

The given system of equations

$5x+2y=3$

$3x+2y=5$

can be written in the matrix form of AX =B, where

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

we have,

$|A|=10-6=4\ne 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\left[\begin{array}{c}3\\ 5\end{array}\right]$

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

Hence the solutions of the given system of equations;

x = -1 and y = 4

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

$2x+y+z=1$

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

$3y-5z=9$

Answer:

The given system of equations

$2x+y+z=1$

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

$3y-5z=9$

can be written in the matrix form of AX =B, where

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

we have,

$|A|=2(10+3)-1(-5-0)+1(3-0)=26+5+3=34\ne 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

Now, we will find the cofactors;

$A_{11}=(-1{)}^{1+1}(10+3)=13$ $A_{12}=(-1{)}^{1+2}(-5-0)=5$

$A_{13}=(-1{)}^{1+3}(3-0)=3$ $A_{21}=(-1{)}^{2+1}(-5-3)=8$

$A_{22}=(-1{)}^{2+2}(-10-0)=-10$ $A_{23}=(-1{)}^{2+3}(6-0)=-6$

$A_{31}=(-1{)}^{3+1}(-1+2)=1$ $A_{32}=(-1{)}^{3+2}(-2-1)=3$

$A_{33}=(-1{)}^{3+3}(-4-1)=-5$

$(adjA)=\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\left[\begin{array}{c}1\\ \frac{3}{2}\\ 9\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{c}13+12+9\\ 5-15+27\\ 3-9-45\end{array}\right]=\frac{1}{34}\left[\begin{array}{c}34\\ 17\\ -51\end{array}\right]=\left[\begin{array}{c}1\\ \frac{1}{2}\\ -\frac{3}{2}\end{array}\right]$

Hence the solutions of the given system of equations;

x = 1, y = 1/2, and z = -3/2.

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

$x-y+z=4$

$2x+y-3z=0$

$x+y+z=2$

Answer:

The given system of equations

$x-y+z=4$

$2x+y-3z=0$

$x+y+z=2$

can be written in the matrix form of AX =B, where

$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ $andB=\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right].$

we have,

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

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

Now, we will find the cofactors;

$A_{11}=(-1{)}^{1+1}(1+3)=4$ $A_{12}=(-1{)}^{1+2}(2+3)=-5$

$A_{13}=(-1{)}^{1+3}(2-1)=1$ $A_{21}=(-1{)}^{2+1}(-1-1)=2$

$A_{22}=(-1{)}^{2+2}(1-1)=0$ $A_{23}=(-1{)}^{2+3}(1+1)=-2$

$A_{31}=(-1{)}^{3+1}(3-1)=2$ $A_{32}=(-1{)}^{3+2}(-3-2)=5$

$A_{33}=(-1{)}^{3+3}(1+2)=3$

$(adjA)=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

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

So, the solutions can be found by $X=A^{-1}B=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

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

Hence the solutions of the given system of equations;

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

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

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

$x-2y+z=-4$

$3x-y-2z=3$

Answer:

The given system of equations

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

$x-2y+z=-4$

$3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A=\left[\begin{array}{ccc}2& 3& 3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ $andB=\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right].$

we have,

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

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

Now, we will find the cofactors;

$A_{11}=(-1{)}^{1+1}(4+1)=5$ $A_{12}=(-1{)}^{1+2}(-2-3)=5$

$A_{13}=(-1{)}^{1+3}(-1+6)=5$ $A_{21}=(-1{)}^{2+1}(-6+3)=3$

$A_{22}=(-1{)}^{2+2}(-4-9)=-13$ $A_{23}=(-1{)}^{2+3}(-2-9)=11$

$A_{31}=(-1{)}^{3+1}(3+6)=9$ $A_{32}=(-1{)}^{3+2}(2-3)=1$

$A_{33}=(-1{)}^{3+3}(-4-3)=-7$

$(adjA)=\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\left[\begin{array}{c}5\\ -4\\ 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}25-12+27\\ 25+52+3\\ 25-44-21\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}40\\ 80\\ -40\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]$

Hence the solutions of the given system of equations;

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

Question:14 Solve system of linear equations, using matrix method.

$x-y+2z=7$

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

$2x-y+3z=12$

Answer:

The given system of equations

$x-y+2z=7$

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

$2x-y+3z=12$

can be written in the matrix form of AX =B, where

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

we have,

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

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

Now, we will find the cofactors;

$A_{11}=(-1{)}^{12-5}=7$ $A_{12}=(-1{)}^{1+2}(9+10)=-19$

$A_{13}=(-1{)}^{1+3}(-3-8)=-11$ $A_{21}=(-1{)}^{2+1}(-3+2)=1$

$A_{22}=(-1{)}^{2+2}(3-4)=-1$ $A_{23}=(-1{)}^{2+3}(-1+2)=-1$

$A_{31}=(-1{)}^{3+1}(5-8)=-3$ $A_{32}=(-1{)}^{3+2}(-5-6)=11$

$A_{33}=(-1{)}^{3+3}(4+3)=7$

$(adjA)=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36\\ -133+5+132\\ -77+5+84\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8\\ 4\\ 12\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]$

Hence the solutions of the given system of equations;

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

Question: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

$2x-3y+5z=11$

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

$x+y-2z=-3$

Answer:

The given system of equations

$2x-3y+5z=11$

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

$x+y-2z=-3$

can be written in the matrix form of AX =B, where

$A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ $andB=\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right].$

we have,

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

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1}=\frac{1}{|A|}(adjA)$

Now, we will find the cofactors;

$A_{11}=(-1{)}^{-4+4}=0$ $A_{12}=(-1{)}^{1+2}(-6+4)=2$

$A_{13}=(-1{)}^{1+3}(3-2)=1$ $A_{21}=(-1{)}^{2+1}(6-5)=-1$

$A_{22}=(-1{)}^{2+2}(-4-5)=-9$ $A_{23}=(-1{)}^{2+3}(2+3)=-5$

$A_{31}=(-1{)}^{3+1}(12-10)=2$ $A_{32}=(-1{)}^{3+2}(-8-15)=23$

$A_{33}=(-1{)}^{3+3}(4+9)=13$

$(adjA)=\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)=-1\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]$

So, the solutions can be found by $X=A^{-1}B=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{c}11\\ -5\\ -3\end{array}\right]$

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

Hence the solutions of the given system of equations;

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

Question:16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

So, let us assume the cost of onion, wheat, and rice be x, y and z respectively.

Then we have the equations for the given situation :

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

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

$6x+2y+3y=70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$.

We have;

$A=\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right],$ $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ $andB=\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right].$

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

Now, we will find the cofactors of A;

$A_{11}=(-1{)}^{1+1}(12-12)=0$ $A_{12}=(-1{)}^{1+2}(6-36)=30$

$A_{13}=(-1{)}^{1+3}(4-24)=-20$ $A_{21}=(-1{)}^{2+1}(9-4)=-5$

$A_{22}=(-1{)}^{2+2}(12-12)=0$ $A_{23}=(-1{)}^{2+3}(8-18)=10$

$A_{31}=(-1{)}^{3+1}(18-8)=10$ $A_{32}=(-1{)}^{3+2}(24-4)=-20$

$A_{33}=(-1{)}^{3+3}(16-6)=10$

Now we have adjA;

$adjA=\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$s

So, the solutions can be found by $X=A^{-1}B=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right]$