$2x - y = - 2,3x + 4y = 3.$

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

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}\\3&4\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right]$

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}\\3\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 1}\\3&4\end{array}} \right| = 8 + 3 = 11 \ne 0$

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

Cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}(4) = 4,{A_{12}} = {( - 1)^{1 + 2}}(3) = - 3,$

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

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 3}\\1&2\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&1\\{ - 3}&2\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&1\\{ - 3}&2\end{array}} \right]$

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

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}X\\Y\end{array}} \right]$

$= \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&1\\{ - 3}&2\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 2}\\3\end{array}} \right] = \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 8 + 3}\\{6 + 6}\end{array}} \right]$

$= \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}\\{12}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{{ - 5}}{{11}}}\\{\cfrac{{12}}{{11}}}\end{array}} \right]$

$\therefore$ $x = - \cfrac{5}{{11}},y = \cfrac{{12}}{{11}}$