Answer
First method;
$$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{cc} {\frac {1}{2209}}&-{\frac {56}{2209}}
\\ {\frac {40}{2209}}&-{\frac {31}{2209}}
\end {array} \right]
.
$$
Second method;
$$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{cc} {\frac {1}{2209}}&-{\frac {56}{2209}}
\\ {\frac {40}{2209}}&-{\frac {31}{2209}}
\end {array} \right]
.$$
One can see that the two methods are equivalent.
Work Step by Step
First method; we find $A^2$ which is given by
$$A^{2}=\left[ \begin {array}{cc} -31&56\\ -40&1
\end {array} \right]
.$$
By calculating the inverse of $A^2$, we have
$$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{cc} {\frac {1}{2209}}&-{\frac {56}{2209}}
\\ {\frac {40}{2209}}&-{\frac {31}{2209}}
\end {array} \right]
.
$$
Second method; we find $A^{-1}$ as follows
$$A^{-1}= \left[ \begin {array}{cc} {\frac {6}{47}}&-{\frac {7}{47}}
\\ {\frac {5}{47}}&{\frac {2}{47}}\end {array}
\right]
.
$$
Now, we have
$$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{cc} {\frac {1}{2209}}&-{\frac {56}{2209}}
\\ {\frac {40}{2209}}&-{\frac {31}{2209}}
\end {array} \right]
.$$
One can see that the two methods are equivalent.