Answer
The statement does make sense.
Work Step by Step
Consider the above statement such that the Gauss-Jordan elimination method can be used to find the multiplicative inverse of the $3\times 3$ matrix if the determinant of the given matrix is nonzero.
Example:
Consider the following matrix:
$A=\left[ \begin{matrix}
5 & 0 & 2 \\
2 & 2 & 1 \\
-3 & 1 & -1 \\
\end{matrix} \right]$
Calculate the determinant of the above matrix as below:
$\begin{align}
& \left| A \right|=5\left( -2-1 \right)+0\left( -2-3 \right)+2\left( 2+6 \right) \\
& =-15+16 \\
& =1 \\
& \ne 0
\end{align}$
Therefore, the inverse of the matrix exists.
Apply Gauss--Jordan elimination method:
The augmented matrix $\left[ A|{{I}_{3}} \right]$ is
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
5 & 0 & 2 & 1 & 0 & 0 \\
2 & 2 & 1 & 0 & 1 & 0 \\
-3 & 1 & -1 & 0 & 0 & 1 \\
\end{matrix} \right]$
Use the elementary row operations on $\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]$ to get a matrix of the form $\left[ \begin{matrix}
{{I}_{3}} & B \\
\end{matrix} \right]$. The matrix B is inverse of A, that is, ${{A}^{-1}}=B$.
${{R}_{1}}\to \frac{1}{5}{{R}_{1}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\
2 & 2 & 1 & 0 & 1 & 0 \\
-3 & 1 & -1 & 0 & 0 & 1 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( 3 \right){{R}_{1}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\
0 & 2 & \frac{1}{5} & -\frac{2}{5} & 1 & 0 \\
0 & 1 & -\frac{1}{5} & \frac{3}{5} & 0 & 1 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{1}{2}{{R}_{2}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\
0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\
0 & 1 & -\frac{1}{5} & \frac{3}{5} & 0 & 1 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{2}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\
0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{10} & \frac{4}{5} & -\frac{1}{2} & 1 \\
\end{matrix} \right]$
${{R}_{3}}\to 10{{R}_{3}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\
0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\
0 & 0 & 1 & 8 & -5 & 10 \\
\end{matrix} \right]$
${{R}_{1}}\to {{R}_{1}}+\left( -\frac{2}{5} \right){{R}_{3}},{{R}_{2}}\to {{R}_{2}}+\left( -\frac{1}{10} \right){{R}_{3}}$ gives
$\left[ \begin{matrix}
A & {{I}_{3}} \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & 0 & -3 & 2 & -4 \\
0 & 1 & 0 & -1 & 1 & -1 \\
0 & 0 & 1 & 8 & -5 & 10 \\
\end{matrix} \right]$
That is
$\left[ \begin{matrix}
{{I}_{3}} & B \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & 0 & -3 & 2 & -4 \\
0 & 1 & 0 & -1 & 1 & -1 \\
0 & 0 & 1 & 8 & -5 & 10 \\
\end{matrix} \right]$
That is
${{A}^{-1}}=B=\left[ \begin{matrix}
-3 & 2 & -4 \\
-1 & 1 & -1 \\
8 & -5 & 10 \\
\end{matrix} \right]$
$\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
5 & 0 & 2 \\
2 & 2 & 1 \\
-3 & 1 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
-3 & 2 & -4 \\
-1 & 1 & -1 \\
8 & -5 & 10 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
$\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
-3 & 2 & -4 \\
-1 & 1 & -1 \\
8 & -5 & 10 \\
\end{matrix} \right]\left[ \begin{matrix}
5 & 0 & 2 \\
2 & 2 & 1 \\
-3 & 1 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Hence
${{A}^{-1}}=\left[ \begin{matrix}
-3 & 2 & -4 \\
-1 & 1 & -1 \\
8 & -5 & 10 \\
\end{matrix} \right],A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{3}}$