Compute $A^{-2}$ two different ways and show that the results are equal.

$A=\left[\begin{array}{cc}0& -2\\ -1& 3\end{array}\right]$

$A^{-1}$

$\left[\begin{array}{cc}-\frac{\frac{2}{3}}{?}& -1\\ -\frac{1}{2}& 0\end{array}\right]=>$

$(A^{-1}{)}^2$