Now, ${A_{11}} = 4$, ${A_{12}}=-3$, ${A_{21}} = - ( - 2) = 2$, ${A_{22}} = 1$
$\therefore $ $adj\,(A) = \left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$
$\therefore $ ${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$.
Trick : Check from the options $A{A^{ - 1}} = I\,$
==> $A{A^{ - 1}}$=$\left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{\frac{4}{{10}}}&{\frac{2}{{10}}}\\{\frac{{ - 3}}{{10}}}&{\frac{1}{{10}}}\end{array}} \right]$= $\left[ {\begin{array}{*{20}{c}}{\frac{{10}}{{10}}}&0\\0&{\frac{{10}}{{10}}}\end{array}} \right]$
= $A{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$.
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If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
$\text{I}$
$0$
$2\text{I}$
$\frac{1}{2}\text{I}$