Q.The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find the cost of each item per kg by matrix method.

Part 1: set up $AX=B$ with the cost matrix, find $A^{-1}$ via the adjoint, then compute $X=A^{-1}B$. Part 2 (OR): substitute $u=1/x,v=1/y,w=1/z$ to linearise the system, then solve by elimination.

Part 1: Matrix method for costs.

Let the cost per kg of onion, wheat, rice be $x,y,z$ respectively. From the data:

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

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

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

In matrix form $AX=B$ where $A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}$, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, $B=\begin{bmatrix}60\\90\\70\end{bmatrix}$.

Determinant:

$|A| = 4(4\cdot3-6\cdot2) - 3(2\cdot3-6\cdot6) + 2(2\cdot2-4\cdot6) = 4(0) - 3(-30) + 2(-20) = 0+90-40=50$

Since $|A|\ne0$, $A$ is invertible.

Cofactors:

$C_{11}=0,\ C_{12}=30,\ C_{13}=-20$

$C_{21}=-5,\ C_{22}=0,\ C_{23}=10$

$C_{31}=10,\ C_{32}=-20,\ C_{33}=10$

$\text{Adj}(A) = \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}, \qquad A^{-1}=\frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$

Solve $X=A^{-1}B$:

$x = \frac{0(60)+(-5)(90)+10(70)}{50} = \frac{0-450+700}{50} = \frac{250}{50} = 5$

$y = \frac{30(60)+0(90)+(-20)(70)}{50} = \frac{1800+0-1400}{50} = \frac{400}{50} = 8$

$z = \frac{-20(60)+10(90)+10(70)}{50} = \frac{-1200+900+700}{50} = \frac{400}{50} = 8$

So onion = ₹5/kg, wheat = ₹8/kg, rice = ₹8/kg. (Check: $4(5)+3(8)+2(8)=20+24+16=60$ ✓, similarly the other two equations check out.)

OR: Solve the system with $u=1/x,v=1/y,w=1/z$.

$2u+3v+10w=4 \quad(1)$

$4u-6v+5w=1 \quad(2)$ …