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 the each item per kg by matrix method.

Let cost of 1 kg onion $=$ Rs. x,

cost of 1 kg wheat $=$ Rs. y and cost of 1 kg rice $=$ Rs. z

$\therefore$ According to question, wo have

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

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

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

This system can be written as AX$=$B, where

$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&3&2\\2&4&6\\6&2&3\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{60}\\{90}\\{70}\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&3&2\\2&4&6\\6&2&3\end{array}} \right|$

$= 4\left( {12 - 12} \right) - 3\left( {6 - 36} \right) + 2\left( {4 - 24} \right)90 - 40 = 50 \ne 0$

$\therefore$ A is a non-singular matrix and system has a unique solution.

Cofactors of elements of A are :
${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&6\\2&3\end{array}} \right| = 0,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&6\\6&3\end{array}} \right| = 30,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&4\\6&2\end{array}} \right| = - 20,$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\2&3\end{array}} \right| = - 5,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2\\6&3\end{array}} \right| = 0,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&3\\6&2\end{array}} \right| = 10,$,

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\4&6\end{array}} \right| = 10,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&2\\2&6\end{array}} \right| = - 20,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&3\\2&4\end{array}} \right| = 10$

$\therefore$ $adj\;A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{30}&{ - 20}\\{ - 5}&0&{10}\\{10}&{ - 20}&{10}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A)$

$= \cfrac{1}{{50}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]$

As,$AX = B \Rightarrow X = {A^{ - 1}}B$

$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \cfrac{1}{{50}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{60}\\{90}\\{70}\end{array}} \right]$

$= \cfrac{1}{5}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 5}&{10}\\{30}&0&{ - 20}\\{ - 20}&{10}&{10}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6\\9\\7\end{array}} \right]$

$= \cfrac{1}{5}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{0 - 45 + 70}\\{180 + 0 - 140}\\{ - 120 + 90 + 70}\end{array}} \right] = \cfrac{1}{5}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{25}\\{40}\\{40}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5\\8\\8\end{array}} \right]$

$\therefore$ $x = 5,y = 8,z = 8$

Hence, cost of 1 kg of onion$=$Rs. 5,

cost of 1 kg of wheat $=$ Rs. 8 and cost of 1 kg of rice $=$ Rs. 8