Answer
$$A^{-1}= \left[ \begin {array}{ccccccccc} 1&-\frac{3}{2}&-4&{\frac {13}{5}}
\\ 0&\frac{1}{2}&1&-\frac{4}{5}\\ 0&0&-\frac{1}{2}&\frac{1}{10}\\ 0&0&0&\frac{1}{5}\end {array}
\right]
.$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]=\left[ \begin {array}{ccccccccc} 1&3&-2&0&1&0&0&0
\\ 0&2&4&6&0&1&0&0\\ 0&0&-2&1&0&0
&1&0\\ 0&0&0&5&0&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{ccccccccc} 1&0&0&0&1&-\frac{3}{2}&-4&{\frac {13}{5}}
\\ 0&1&0&0&0&\frac{1}{2}&1&-\frac{4}{5}\\ 0&0&1&0
&0&0&-\frac{1}{2}&\frac{1}{10}\\ 0&0&0&1&0&0&0&\frac{1}{5}\end {array}
\right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}= \left[ \begin {array}{ccccccccc} 1&-\frac{3}{2}&-4&{\frac {13}{5}}
\\ 0&\frac{1}{2}&1&-\frac{4}{5}\\ 0&0&-\frac{1}{2}&\frac{1}{10}\\ 0&0&0&\frac{1}{5}\end {array}
\right]
.$$
