Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1 \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5 \\
\end{matrix} \right]$.
Compute matrix in the form of:
$\left[ \left. A \right|I \right]$
The augment matrix with identity matrix is:
$\left[ \left. A \right|I \right]=\left[ \begin{matrix}
3 & 2 & 6 & 1 & 0 & 0 \\
1 & 1 & 2 & 0 & 1 & 0 \\
2 & 2 & 5 & 0 & 0 & 1 \\
\end{matrix} \right]$
Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below:
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}-\frac{1}{3}\times {{R}_{1}}, \\
& {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{1}}, \\
& {{R}_{2}}\leftrightarrow {{R}_{_{3}}}, \\
& {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{2}} \\
\end{align}$
The resulting matrix is:
$\left[ \begin{matrix}
3 & 2 & 6 & 1 & 0 & 0 \\
0 & \frac{2}{3} & 1 & -\frac{2}{3} & 0 & 1 \\
0 & 0 & -\frac{1}{2} & 0 & 1 & -\frac{1}{2} \\
\end{matrix} \right]$
Again, apply the row operations as below:
$\begin{align}
& {{R}_{3}}\to -2\times {{R}_{3}}, \\
& {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{3}}, \\
& {{R}_{1}}\to {{R}_{1}}-6\times {{R}_{3}}, \\
& {{R}_{2}}\to \frac{3}{2}\times {{R}_{2}}, \\
& {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}}, \\
& {{R}_{1}}\to \frac{1}{3}\times {{R}_{1}} \\
\end{align}$
The resulting matrix is:
$\begin{align}
& \left[ \left. A \right|I \right]=\left[ \begin{matrix}
1 & 0 & 0 & 1 & 2 & -2 \\
0 & 1 & 0 & -1 & 3 & 0 \\
0 & 0 & 1 & 0 & -2 & 1 \\
\end{matrix} \right] \\
& =\left[ \left. I \right|B \right]
\end{align}$
Where ${{A}^{-1}}=\left[ B \right]$
So, The inverse of matrix is:
${{A}^{-1}}=\left[ \begin{matrix}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1 \\
\end{matrix} \right]$
Now, check the result for
$ A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$
So, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1 \\
\end{matrix} \right] \\
& A{{A}^{-1}}=\left[ \begin{matrix}
3+\left( -2 \right)+0 & 6+6+\left( -12 \right) & \left( -6 \right)+0+6 \\
1+\left( -1 \right)+0 & 2+3+\left( -4 \right) & \left( -2 \right)+0+2 \\
2+\left( -2 \right)+0 & 4+6+\left( -10 \right) & \left( -4 \right)+0+5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
And, $\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3+2+\left( -4 \right) & 2+2+\left( -4 \right) & 6+4+\left( -10 \right) \\
\left( -3 \right)+3+0 & \left( -2 \right)+3+0 & \left( -6 \right)+6+0 \\
0+\left( -2 \right)+2 & 0+\left( -2 \right)+2 & 0+\left( -4 \right)+5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Thus, $ A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.