Answer
$|A|=74$ and $|A^{-1}|= \frac{1}{74}$
Work Step by Step
Let $A$ be given by $$A=\left[ \begin {array}{cccc} 10&2\\ -2&7
\end {array} \right]
,$$
then we have $A^{-1}=\frac{1}{74}\left[ \begin {array}{cccc} 7&-2\\ 2&10
\end {array} \right]$.
Now, $|A|=74$ and $|A^{-1}|=\frac{1}{74^2}74=\frac{1}{74}$. Hence, we verify that
$$|A^{-1}=\frac{1}{|A|}.$$
