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.
Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:
$4 x+3 y+2 z=60$
$2 x+4 y+6 z=90$
$6 x+2 y+3 z=70$
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, $\quad A_{11}=0, A_{12}=30, A_{13}=-20$
$A_{21}=-5, A_{22}=0, A_{23}=10$
$A_{31}=10, A_{32}=-20, A_{33}=10$
$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{50}\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]$
Now
$X=A^{-1} B$
$\Rightarrow X=\frac{1}{50}\left[\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right]\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}0-450+700 \\ 1800+0-1400 \\ -1200+900+700\end{array}\right]$
$=\frac{1}{50}\left[\begin{array}{l}250 \\ 400 \\ 400\end{array}\right]$
$=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$
$\therefore x=5, y=8$, and $z=8$
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.
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