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