$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\4&3\end{array}} \right]$

Let A$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\4&3\end{array}} \right]$

Then, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\4&3\end{array}} \right| = 6 + 8 = 14 \ne 0.$

So, A is a non-singular matrix and therefore, it is invertible.

Let ${A_{ij}}$ be cofactor of ${a_{ij}}$in A.

Then, the cofactors of elements of A are given by

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

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

% $\therefore$ $adj$ $A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\2&2\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\{ - 4}&2\end{array}} \right]$
Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}adj\;A = \cfrac{1}{{14}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\{ - 4}&2\end{array}} \right]$