. $5x + 2y = 3,3x + 2y = 5.$

The given system of equations can be written in the form $AX = B$,

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&2\\3&2\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right]$

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\5\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&2\\3&2\end{array}} \right| = 10 - 6 = 4 \ne 0$

$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution.

Cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}(2) = 2,{A_{12}} = {( - 1)^{1 + 2}}(3) = - 3,$

${A_{21}} = {( - 1)^{2 + 1}}(2) = - 2,{A_{22}} = {( - 1)^{2 + 2}}(5) = 5$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}\\{ - 2}&5\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\{ - 3}&5\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\{ - 3}&5\end{array}} \right]$

Solution of given system is given by$X = {A^{ - 1}}B.$

$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right] = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\{ - 3}&5\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\5\end{array}} \right];\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right] = \cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{6 - 10}\\{ - 9 + 25}\end{array}} \right] =$

$\cfrac{1}{4}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 4}\\{16}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}\\4\end{array}} \right]$

Hence, $x = - 1,y = 4$