Answer
First method: $$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401
\\ {\frac {61}{4}}&4&-28\\ -{
\frac {1317}{4}}&-86&605\end {array} \right]
.
$$
Second method; $$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401
\\ {\frac {61}{4}}&4&-28\\ -{
\frac {1317}{4}}&-86&605\end {array} \right]
.$$
One can see that the two methods are equivalent
Work Step by Step
First method; we find $A^2$ which is given by
$$A^{2}=\left[ \begin {array}{ccc} 48&4&32\\ -29&48&-17
\\ 22&9&15\end {array} \right]
.$$
By calculating the inverse of $A^2$, we have
$$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401
\\ {\frac {61}{4}}&4&-28\\ -{
\frac {1317}{4}}&-86&605\end {array} \right]
.
$$
Second method; we find $A^{-1}$ as follows
$$A^{-1}= \left[ \begin {array}{ccc} -\frac{15}{2}&-2&14\\ -\frac{1}{2}&0&1
\\ \frac{23}{2}&3&-21\end {array} \right]
.
$$
Now, we have
$$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401
\\ {\frac {61}{4}}&4&-28\\ -{
\frac {1317}{4}}&-86&605\end {array} \right]
.$$
One can see that the two methods are equivalent.