Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
3 & 2 & 6 \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
2 & 2 & -1 \\
0 & 3 & -1 \\
-1 & -2 & 1 \\
\end{matrix} \right]$.
Compute matrix in the form of:
$\left[ \left. A \right|I \right]$
The augment matrix with identity matrix is:
$\left[ \left. A \right|I \right]=\left[ \begin{matrix}
2 & 2 & -1 & 1 & 0 & 0 \\
0 & 3 & -1 & 0 & 1 & 0 \\
-1 & -2 & 1 & 0 & 0 & 1 \\
\end{matrix} \right]$
Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below:
$\begin{align}
& {{R}_{3}}\to {{R}_{3}}+\frac{1}{2}\times {{R}_{1}}, \\
& {{R}_{3}}\to {{R}_{3}}+\frac{1}{3}\times {{R}_{2}} \\
\end{align}$
The resulting matrix is:
$\left[ \begin{matrix}
2 & 2 & -1 & 1 & 0 & 0 \\
0 & 3 & -1 & 0 & 1 & 0 \\
0 & 0 & \frac{1}{6} & \frac{1}{2} & \frac{1}{3} & 1 \\
\end{matrix} \right]$
Again, apply the row operations as below:
$\begin{align}
& {{R}_{3}}\to 6\times {{R}_{3}}, \\
& {{R}_{2}}\to {{R}_{2}}+1\times {{R}_{3}}, \\
& {{R}_{1}}\to {{R}_{1}}+1\times {{R}_{2}}, \\
& {{R}_{2}}\to \frac{1}{3}\times {{R}_{2}}, \\
& {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}}, \\
& {{R}_{1}}\to \frac{1}{2}\times {{R}_{1}} \\
\end{align}$
The resulting matrix is:
$\begin{align}
& \left[ \left. A \right|I \right]=\left[ \begin{matrix}
1 & 0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 & 2 \\
0 & 0 & 1 & 3 & 2 & 6 \\
\end{matrix} \right] \\
& =\left[ \left. I \right|B \right]
\end{align}$
Where ${{A}^{-1}}=\left[ B \right]$
Therefore, the inverse of the matrix is:
${{A}^{-1}}=\left[ \begin{matrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
3 & 2 & 6 \\
\end{matrix} \right]$
Now, check the result for
$ A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$
Here, $ A=\left[ \begin{matrix}
2 & 2 & -1 \\
0 & 3 & -1 \\
-1 & -2 & 1 \\
\end{matrix} \right]$
So, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
2 & 2 & -1 \\
0 & 3 & -1 \\
-1 & -2 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
3 & 2 & 6 \\
\end{matrix} \right] \\
& A{{A}^{-1}}=\left[ \begin{matrix}
2+2+\left( -3 \right) & 0+2+\left( -2 \right) & 2+4+\left( -6 \right) \\
0+3+\left( -3 \right) & 0+3+\left( -2 \right) & 0+6+\left( -6 \right) \\
\left( -1 \right)+\left( -2 \right)+3 & 0+\left( -2 \right)+2 & \left( -1 \right)+\left( -4 \right)+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{matrix} \right] \\
& A{{A}^{-1}}={{I}_{3}}
\end{align}$
And, $\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
3 & 2 & 6 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 2 & -1 \\
0 & 3 & -1 \\
-1 & -2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2+0+1 & 2+0+\left( -2 \right) & \left( -1 \right)+0+1 \\
2+0+\left( -2 \right) & 2+3+\left( -4 \right) & \left( -1 \right)+\left( -1 \right)+2 \\
6+0+\left( -6 \right) & 6+6+\left( -12 \right) & \left( -3 \right)+\left( -2 \right)+6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& {{A}^{-1}}A={{I}_{3}}
\end{align}$
Thus, $ A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.