Answer
$$A^{-1}=\left[ \begin {array}{cccccc} -\frac{3}{2}&\frac{3}{2}&1\\ \frac{9}{2}&-\frac{7}{2}&-3\\ -1&1&1\end {array} \right].$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&1&2&1&0&0\\ 3&1&0&0
&1&0\\ -2&0&3&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&-\frac{3}{2}&\frac{3}{2}&1\\ 0&
1&0&\frac{9}{2}&-\frac{7}{2}&-3\\ 0&0&1&-1&1&1\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}=\left[ \begin {array}{cccccc} -\frac{3}{2}&\frac{3}{2}&1\\ \frac{9}{2}&-\frac{7}{2}&-3\\ -1&1&1\end {array} \right].$$
