Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix}
\frac{7}{2} & -2 & -2 \\
-\frac{5}{2} & 1 & 2 \\
-1 & 0 & 1 \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
2 & 4 & -4 \\
1 & 3 & -4 \\
2 & 4 & -3 \\
\end{matrix} \right]$.
Compute matrix in the form of:
$\left[ \left. A \right|I \right]$
The augment with identity matrix:
$\left[ \left. A \right|I \right]=\left[ \begin{matrix}
2 & 4 & -4 & 1 & 0 & 0 \\
1 & 3 & -4 & 0 & 1 & 0 \\
2 & 4 & -3 & 0 & 0 & 1 \\
\end{matrix} \right]$
Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below:
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}-\frac{1}{2}\times {{R}_{1}}, \\
& {{R}_{3}}\to {{R}_{3}}-1\times {{R}_{1}} \\
\end{align}$
The resulting matrix is:
$\left[ \begin{matrix}
2 & 4 & -4 & 1 & 0 & 0 \\
0 & 1 & -2 & -\frac{1}{2} & 1 & 0 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{matrix} \right]$
Again, apply the row operations as below:
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}+2\times {{R}_{3}}, \\
& {{R}_{1}}\to {{R}_{1}}+4\times {{R}_{3}}, \\
& {{R}_{1}}\to {{R}_{1}}-4\times {{R}_{2}}, \\
& {{R}_{1}}\to \frac{1}{2}\times {{R}_{1}} \\
\end{align}$
The resulting matrix is:
$\begin{align}
& \left[ \left. A \right|I \right]=\left[ \begin{matrix}
1 & 0 & 0 & \frac{7}{2} & -2 & -2 \\
0 & 1 & 0 & -\frac{5}{2} & 1 & 2 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \left. I \right|B \right]
\end{align}$
Where ${{A}^{-1}}=\left[ B \right]$
Therefore, the inverse of the matrix is;
${{A}^{-1}}=\left[ \begin{matrix}
\frac{7}{2} & -2 & -2 \\
-\frac{5}{2} & 1 & 2 \\
-1 & 0 & 1 \\
\end{matrix} \right]$
Now, check the result for
$ A{{A}^{-1}}={{I}_{3}}$
And, ${{A}^{-1}}A={{I}_{3}}$
Here, $ A=\left[ \begin{matrix}
2 & 4 & -4 \\
1 & 3 & -4 \\
2 & 4 & -3 \\
\end{matrix} \right]$
So, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
2 & 4 & -4 \\
1 & 3 & -4 \\
2 & 4 & -3 \\
\end{matrix} \right]\left[ \begin{matrix}
\frac{7}{2} & -2 & -2 \\
-\frac{5}{2} & 1 & 2 \\
-1 & 0 & 1 \\
\end{matrix} \right] \\
& A{{A}^{-1}}=\left[ \begin{matrix}
2\times \frac{7}{2}+4\times \left( -\frac{5}{2} \right)+\left( -4 \right)\times \left( -1 \right) & \left( -4 \right)+4+0 & \left( -4 \right)+8+\left( -4 \right) \\
1\times \frac{7}{2}+3\times \left( -\frac{5}{2} \right)+\left( -4 \right)\times \left( -1 \right) & \left( -2 \right)+3+0 & \left( -2 \right)+6+\left( -4 \right) \\
2\times \frac{7}{2}+4\times \left( -\frac{5}{2} \right)+\left( -3 \right)\times \left( -1 \right) & \left( -4 \right)+4+0 & \left( -4 \right)+8+\left( -3 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
And, $\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
\frac{7}{2} & -2 & -2 \\
-\frac{5}{2} & 1 & 2 \\
-1 & 0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 4 & -4 \\
1 & 3 & -4 \\
2 & 4 & -3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{7}{2}\times 2+\left( -2 \right)+\left( -4 \right) & \frac{7}{2}\times 4+\left( -6 \right)+\left( -8 \right) & \frac{7}{2}\times \left( -4 \right)+\left( -2 \right)\times \left( -4 \right)+\left( -2 \right)\times \left( -3 \right) \\
\left( -\frac{5}{2} \right)\times 2+1+4 & \left( -\frac{5}{2} \right)\times 4+3+8 & \left( -\frac{5}{2} \right)\times \left( -4 \right)+\left( -4 \right)+\left( -6 \right) \\
\left( -6 \right)+0+2 & \left( -4 \right)+0+4 & \left( -1 \right)\times \left( -4 \right)+0\times +\left( -3 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Thus, $ A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.
