Question

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 1

Let the cost of 1 kg onion , 1 kg wheat and 1 kg of rice be $Rsx,Rsy,Rsz$ respectively.

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

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

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

These equations can be written as

$AX=B$

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

Here,

$|A|=4(12-12)-3(6-36)+2(4-24)$

$\Rightarrow |A|=90-40=50$

Since, $|A|\ne 0$

Hence, the system of equations is consistent and has a unique solution given by $X==A^{-1}B$

$A^{-1}=\frac{adjA}{|A|}$ and $adjA=C^T$

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}4& 6\\ 2& 3\end{array}\right|$

$\Rightarrow C_{11}=12-12=0$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 6\\ 6& 3\end{array}\right|$

$\Rightarrow C_{12}=-(6-36)=30$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& 4\\ 6& 2\end{array}\right|$

$\Rightarrow C_{13}=4-24=-20$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}3& 2\\ 2& 3\end{array}\right|$

$\Rightarrow C_{21}=-(9-4)=-5$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}4& 2\\ 6& 3\end{array}\right|$

$\Rightarrow C_{22}=12-12=0$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}4& 3\\ 6& 2\end{array}\right|$

$\Rightarrow C_{23}=-(8-18)=10$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}3& 2\\ 4& 6\end{array}\right|$

$\Rightarrow C_{31}=18-8=10$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}4& 2\\ 2& 6\end{array}\right|$

$\Rightarrow C_{32}=-(24-4)=-20$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}4& 3\\ 2& 4\end{array}\right|$

$\Rightarrow C_{33}=16-6=10$

Hence, the co-factor matrix is $C=\left[\begin{array}{ccc}0& 30& -20\\ -5& 0& 10\\ 10& -20& 10\end{array}\right]$

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

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

Solution is given by

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}0-450+700\\ 1800+0-1400\\ -1200+900+700\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}250\\ 400\\ 400\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ 8\\ 8\end{array}\right]$

Hence, $x=5,y=8,z=8$

So, the cost of $1$ kg onion is Rs $5$ , $1$ kg wheat is Rs $8$ , $1$ kg rice is Rs $8$.