Correct option: A.$\frac{{ - 1}}{8}\,\left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$
a
(a) Let $A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&2\end{array}} \right]\,,{\rm{ }}\therefore \,|A|\, = \,\left| {\,\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&2\end{array}\,} \right| = 4 - 12 = - 8$
The matrix of cofactors of the elements of $A $ viz.
$\left[ {\begin{array}{*{20}{c}}{{c_{11}}}&{{c_{12}}}\\{{c_{21}}}&{{c_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - ( - 4)}\\{ - ( - 3)}&2\end{array}} \right] = \left[ {\,\begin{array}{*{20}{c}}2&4\\3&2\end{array}\,} \right]$
$\therefore $ $adjA = $ transpose of the matrix of cofactors of elements of $A = \left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{\Delta }(adj\,A) = \frac{1}{{ - 8}}\,\left[ {\begin{array}{*{20}{c}}
2&3 \\
4&2
\end{array}} \right]$