Rajesh wants to purchase some fruits from fruit market. 4 kilograms (kgs) apples, 3 kgs grapes and…

Let cost per kg of apple, grapes and orange be ₹$x$, ₹$y$ and ₹$z$ respectively. Then according to the given condition, we have $4x + 3y + 2z = 600$, $2x + 4y + 6z = 900$, $6x + 2y + 3z = 700$. Which can be written in matrix form as $\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix} = \begin{bmatrix}600\\900\\700\end{bmatrix}$, $AX = B$

$|A| = 50 \neq 0$, $\therefore$ the system is consistent and has a unique solution given by $X = A^{-1}B$. Co-factor of $A = \begin{bmatrix}0&30&-20\\-5&0&10\\10&-20&10\end{bmatrix} \Rightarrow \text{adj}(A) = \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$ $A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$ $X = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}600\\900\\700\end{bmatrix} = \frac{1}{50}\begin{bmatrix}2500\\4000\\4000\end{bmatrix} = \begin{bmatrix}50\\80\\80\end{bmatrix}$

$x = 50$, $y = 80$ and $z = 80$. Hence, cost of 1 kg apple $= ₹\,50$, cost of 1 kg grapes $= ₹\,80$ and cost of 1 kg orange $= ₹\,80$.