# The cost of 4 kg onion , 3 kg wheat and 2 kg rice of 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^{-1}=\large\frac{1}{|A|}$$adj \;A • adj \;A=\begin{bmatrix}A_{11} & A_{21}& A_{31}\\A_{12} & A_{22}& A_{32}\\A_{13} & A_{23}& A_{33}\end{bmatrix} Step 1: Let the cost of onion is Rs.x Let the cost of wheat is Rs.y Let the cost of rice is Rs.z Thus the above given situation can be represented as 4x+3y+2z=60 2x+4y+6z=90 6x+2y+3z=70 A=\begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix} X=\begin{bmatrix}x\\y\\z\end{bmatrix} B=\begin{bmatrix}60\\90\\70\end{bmatrix} |A|=4(12-12)-3(6-36)+2(4-24) \quad=0+90-40 \quad=50\neq 0 Step 2: A_{11}=0 A_{12}=30 A_{13}=-20 A_{21}=-5 A_{22}=0 A_{23}=10 A_{31}=10 A_{32}=-20 A_{33}=10 adj \;A=\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix} \Rightarrow X=\large\frac{1}{50}$$\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$
Step 3:
$A^{-1}=\large\frac{1}{|A|}$$adj \;A=\large\frac{1}{50}$$\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\large\frac{1}{50}$$\begin{bmatrix}0-450+700\\1800+0-1400\\-1200+900+700\end{bmatrix} \quad\;\;=\large\frac{1}{50}$$\begin{bmatrix}250\\400\\400\end{bmatrix}$
$\quad\;\;=\begin{bmatrix}5\\8\\8\end{bmatrix}$
$x=5,y=8,z=8$
Step 4:
Hence the cost of onions is $Rs.5/kg.$
The cost of wheat is $Rs.8/kg.$
The cost of rice is $Rs.8/kg.$