Answer
See answers below
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 2 \\
1 & 3
\end{bmatrix}$
Using the Gauss-Jordan method we get:
$\begin{bmatrix}
1 & 2|1 &0 \\
1 & 3| 0 &1
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & 2|1 &0 \\
0 & 1| -1 &1
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & 2|3 &-2 \\
0 & 1| -1 &1
\end{bmatrix} $
1. $A_{12}(-1)$
2. $A_{21}(-2)$
$\rightarrow A^{-1}=\begin{bmatrix}
3 & -2 \\
-1 & 1
\end{bmatrix}$
Hence $A^{-1}$ exists. Check the answer by verifying that $AA^{-1} = I_n$
$AA^{-1}=\begin{bmatrix}
1 & 2 \\
1 & 3
\end{bmatrix}.\begin{bmatrix}
3 & -2 \\
-1 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}=I_2$
