Answer
$|A|=12$ and $|A^{-1}|=\frac{1}{12}$.
Work Step by Step
Let $A$ be given by $$A=\left[ \begin {array}{ccc} -1&1&2\\ 2&4&8
\\ -1&1&0\end {array} \right]
,$$
then we have $A^{-1}=\left[ \begin {array}{ccc} -\frac{2}{3}&\frac{1}{6}&0\\ -\frac{2}{3}&\frac{1}{6}&1
\\ \frac{1}{2}&0&-\frac{1}{2}\end {array} \right]
$.
Now, $|A|=12$ and $|A^{-1}|=\frac{1}{12}$. Hence, we verify that
$$|A^{-1}=\frac{1}{|A|}.$$
