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