Answer
See below
Work Step by Step
$\begin{bmatrix}
1 & i&2|1 & 0&0\\
1+i &-i &2i |0 & 1 & 0\\
2&2i&5|0 & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & i&2|1 & 0&0\\
0 &-i &2i |-1-i & 1 & 0\\
2&2i&5|0 & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & i&2|1 & 0&0\\
0 &-i &2i |-1-i & 1 & 0\\
0 & 0& 1|-2& 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0|-i & 0&0\\
0 &-i &2i |-1-i & 1 & 0\\
0 & 0& 1|-2& 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0|-i & 0&0\\
0 &-i & 0 |-5-i & 1 & 2\\
0 & 0& 1|-2& 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0|-i & 0&0\\
0 &1 & 0 |1-5i & i & 2i\\
0 & 0& 1|-2& 0 & 1
\end{bmatrix}$
$1.A_{12}[-(1+i)]\\
2. A_{13}(-2)\\
3.A_{21}(1)\\
4.A_{32}(2)\\
5.M_{2}(i)$
Thus, the inverse of the matrix $A$ is:
$A^{-1}=\begin{bmatrix}
-i & 1 & 0\\1-5i & i& 2i\\
-2 &0 &1
\end{bmatrix}$
Using the standard matrix multiplication, we obtain:
$AA^{-1}=\begin{bmatrix}
1 & i&2\\1+i & -i &2i \\2 &2i & 5
\end{bmatrix}\begin{bmatrix}
-i & 1 & 0\\1-5i & i& 2i\\
-2 &0 &1
\end{bmatrix}\\
=\begin{bmatrix}
1& 0 & 0\\0& 1& 0\\
0 &0 &1
\end{bmatrix}\\
=I_3$
Hence, $A^{-1}$ is the inverse of the matrix $A$.
