Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2 \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
1 & -1 & 1 \\
0 & 2 & -1 \\
2 & 3 & 0 \\
\end{matrix} \right]$
Compute matrix in the form of:
$\left[ \left. A \right|I \right]$
The augment with identity matrix is:
$\left[ \left. A \right|I \right]=\left[ \begin{matrix}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 2 & -1 & 0 & 1 & 0 \\
2 & 3 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below:
$\begin{align}
& {{R}_{1}}\leftrightarrow {{R}_{3}}, \\
& {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{1}}, \\
& {{R}_{2}}\leftrightarrow {{R}_{3}}, \\
& {{R}_{3}}\to {{R}_{3}}+\frac{4}{5}\times {{R}_{2}} \\
\end{align}$
The resulting matrix is:
$\left[ \begin{matrix}
2 & 3 & 0 & 0 & 0 & 1 \\
0 & -\frac{5}{2} & 1 & 1 & 0 & -\frac{1}{2} \\
0 & 0 & -\frac{1}{5} & \frac{4}{5} & 1 & -\frac{2}{5} \\
\end{matrix} \right]$
Again, apply the row operations as below:
$\begin{align}
& {{R}_{3}}\to -5\times {{R}_{3}}, \\
& {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{3}}, \\
& {{R}_{2}}\to -\frac{2}{5}\times {{R}_{2}}, \\
& {{R}_{1}}\to {{R}_{1}}-3\times {{R}_{2}}, \\
& {{R}_{1}}\to \frac{1}{2}\times {{R}_{1}} \\
\end{align}$
The resulting matrix is:
$\begin{align}
& \left[ \left. A \right|I \right]=\left[ \begin{matrix}
1 & 0 & 0 & 3 & 3 & -1 \\
0 & 1 & 0 & -2 & -2 & 1 \\
0 & 0 & 1 & -4 & -5 & 2 \\
\end{matrix} \right] \\
& =\left[ \left. I \right|B \right]
\end{align}$
Where ${{A}^{-1}}=\left[ B \right]$
Therefore, the inverse of the matrix is:
${{A}^{-1}}=\left[ \begin{matrix}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2 \\
\end{matrix} \right]$
Now, check the result for
$ A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$
Here, $ A=\left[ \begin{matrix}
1 & -1 & 1 \\
0 & 2 & -1 \\
2 & 3 & 0 \\
\end{matrix} \right]$
So, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
1 & -1 & 1 \\
0 & 2 & -1 \\
2 & 3 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2 \\
\end{matrix} \right] \\
& A{{A}^{-1}}=\left[ \begin{matrix}
3+4+\left( -4 \right) & 3+2+\left( -5 \right) & \left( -1 \right)+\left( -1 \right)+2 \\
0+\left( -4 \right)+4 & 0+\left( -4 \right)+5 & 0+1+\left( -2 \right) \\
6+\left( -6 \right)+0 & 6+\left( -6 \right)+0 & \left( -2 \right)+3+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
And, $\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -1 & 1 \\
0 & 2 & -1 \\
2 & 3 & 0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3+0+\left( -2 \right) & 1+0+\left( -2 \right) & 3+\left( -3 \right)+0 \\
\left( -2 \right)+0+2 & \left( -1 \right)+0+2 & \left( -2 \right)+2+0 \\
\left( -4 \right)+0+4 & \left( -4 \right)+0+4 & \left( -4 \right)+5+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Thus, $ A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.