Answer
See answers below
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 1+i \\
1-i & 1
\end{bmatrix}$
Using the Gauss-Jordan method we get:
$\begin{bmatrix}
1 & 1+i | 1 & 0\\
1-i & 1 | 0 & 1
\end{bmatrix}\approx^1\begin{bmatrix}
1 & 1+i | 1 & 0\\
0 & -1 | i-1 & 1
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & 1+i | 1 & 0\\
0 & 1 | 1-i & -1
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & 0 | -1 & 1+i\\
0 & 1 | 1-i & -1
\end{bmatrix}$
$1A_{12}(i-1)$
$2.M_2(-1)$
$3.A_{21}(-1-i)$
$\rightarrow A^{-1}=\begin{bmatrix}
-1& 1+i \\
1-i & -1
\end{bmatrix}$
Hence $A^{-1}$ exists. Check the answer by verifying that $AA^{-1} = I_n$
$AA^{-1}=\begin{bmatrix}
1 & 1+i \\
1-i & 1
\end{bmatrix}.\begin{bmatrix}
-1& 1+i \\
1-i & -1
\end{bmatrix}=\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}=I_2$
