Answer
To find the multiplicative inverse of an invertible matrix A, we perform row operations on $\left[ \begin{matrix}
A & {{I}_{n}} \\
\end{matrix} \right]$ to obtain a matrix of the form $\left[ \begin{matrix}
{{I}_{n}} & B \\
\end{matrix} \right]$, where $B={{A}^{-1}}$.
Work Step by Step
Since matrix A is invertible, that is, the matrix is non-singular, this means that the determinant of matrix A is nonzero. Now, to find the inverse of $n\times n$ matrix A, we will find matrix B, whose order is also $n\times n$, such that, $AB={{I}_{n}}$ and if such matrix B exists, this implies $BA={{I}_{n}}$. So, B is called the multiplicative inverse of A, that is, $B={{A}^{-1}}$.
Now, write an augmented matrix $\left[ \begin{matrix}
A & {{I}_{n}} \\
\end{matrix} \right]$ and use elementary row operations to make the left-hand side of the matrix as an identity matrix of the form $\left[ \begin{matrix}
{{I}_{n}} & B \\
\end{matrix} \right]$ and the corresponding right-hand side of the matrix is the inverse of matrix of A, that is:
$B={{A}^{-1}}$
Example:
We take an example to explain the whole concept. Consider the $2\times 2$ matrix,
$A=\left[ \begin{matrix}
1 & 2 \\
3 & 4 \\
\end{matrix} \right]$
Since $\det \left( A \right)=-2$, matrix A is non-singular and hence, ${{A}^{-1}}$ exists.
Write the augmented matrix as follows:
$\left[ \begin{matrix}
1 & 2 & 1 & 0 \\
3 & 4 & 0 & 1 \\
\end{matrix} \right]$
Perform the elementary row operations as follows:
${{R}_{2}}\to {{R}_{2}}-3{{R}_{1}}$ gives:
$\left[ \begin{matrix}
1 & 2 & 1 & 0 \\
0 & -2 & -3 & 1 \\
\end{matrix} \right]$
${{R}_{2}}\to -\frac{1}{2}{{R}_{2}}$ gives:
$\left[ \begin{matrix}
1 & 2 & 1 & 0 \\
0 & 1 & \frac{3}{2} & -\frac{1}{2} \\
\end{matrix} \right]$
${{R}_{1}}\to {{R}_{1}}-2{{R}_{2}}$ gives:
$\left[ \begin{matrix}
1 & 0 & -2 & -1 \\
0 & 1 & \frac{3}{2} & -\frac{1}{2} \\
\end{matrix} \right]$
So, $B={{A}^{-1}}=\left[ \begin{matrix}
-2 & -1 \\
\frac{3}{2} & -\frac{1}{2} \\
\end{matrix} \right]$
