# 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.

Toolbox:
• A matrix is said to be singular if |A|= 0.
• A matrix is said to be invertible if |A|$\neq 0$.
• If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
• Using this we can solve the system of equation which has unique solution.
Let the cost of 1kg onion be=x.
Let the cost of 1kg wheat be=y.
Let the cost of 1kg rice be=z.
Now we can frame equations according to the given information as
4x+3y+2z=60------(1)
2x+4y+6z=90-----(2)
6x+2y+3z=70-------(3)
This is of the form AX=B,then $X=A^{-1}B.$
$\begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}$
Where $A=\begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix}X=\begin{bmatrix}x\\y\\z\end{bmatrix}B=\begin{bmatrix}60\\90\\70\end{bmatrix}$
Let us first see if A is a singular or a non-singular matrix:
To verify this,we will have to find the determinant value,which can be got by expanding along $R_1$
$|A|=4(4\times 3 -6\times 2)-(-3)(2\times 3-6\times 6)+2(2\times 2-6\times 4)$
$\;\;\;=4(12-12)-3(6-36)+2(4-24)$
$\;\;\;=0-3\times (-30)+2(-20)=90-40=50\neq 0$
Hence inverse exists.
$A^{-1}=\frac{1}{|A|}(adj \;A)$
Now let us find the (adj A)
To find the (adj A) let us find the minors and cofactors of the respective elements.
$M_{11}=\begin{vmatrix}4 & 6\\2 & 3\end{vmatrix}=12-12=0$
$M_{12}=\begin{vmatrix}2 & 6\\6 & 3\end{vmatrix}=6-36=- 30$
$M_{13}=\begin{vmatrix}2 & 4\\6 & 2\end{vmatrix}=4-24=- 20$
$M_{21}=\begin{vmatrix}3 & 2\\2 & 3\end{vmatrix}=9-4=5$
$M_{22}=\begin{vmatrix}4 & 2\\6 & 3\end{vmatrix}=12-12=0$
$M_{23}=\begin{vmatrix}4 & 3\\6& 2\end{vmatrix}=8-18= - 10$
$M_{31}=\begin{vmatrix}3 & 2\\4 & 6\end{vmatrix}=18-8=10$
$M_{32}=\begin{vmatrix}4 & 2\\2 & 6\end{vmatrix}=24-4=20$
$M_{33}=\begin{vmatrix}4 & 3\\2 & 4\end{vmatrix}=16-6=10$
$A_{11}=(-1)^{1+1}.0=0$
$A_{12}=(-1)^{1+2}.-30=30$
$A_{13}=(-1)^{1+3}.-20= - 20$
$A_{21}=(-1)^{2+1}.5= - 5$
$A_{22}=(-1)^{2+2}.0=0$
$A_{23}=(-1)^{2+3}.-10=10$
$A_{31}=(-1)^{3+1}.10=10$
$A_{32}=(-1)^{3+2}.(20)= - 20$
$A_{33}=(-1)^{3+3}.10=10$
Therefore adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}$
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$.We know |A|=50
$A^{-1}=\frac{1}{50}\begin{bmatrix}0 & -5 & 10\\6 & 0 & -20\\-20 & 10 & 10\end{bmatrix}$
To solve the equation,$X=A^{-1}B.$
Therefore $\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{50}\begin{bmatrix}0 & -5 & 10\\6 & 0 & -20\\-20 & 10 & 10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$
Therefore $\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{50}\begin{bmatrix}0-450+700\\1800+0-1400\\-1200+900+700\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix}$
Therefore $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}$
Hence x=5,y=8,z=8.
Hence the cost of onion is Rs 5/kg.
The cost of wheat is Rs 8/kg.
The cost of rice is Rs 8/kg