Answer
See below
Work Step by Step
$\begin{bmatrix}
2+i & i & 3-2i|0\\i & 1-i & 3i+4 |0 \\1-2i & i-1&1-3i|0
\end{bmatrix} \approx \begin{bmatrix}
i & 1-i & 3i+4 |0 \\2+i & i & 3-2i|0\\1-2i & i-1&1-3i|0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1-i & 3-4i |0 \\2+i & i & 3-2i|0\\1-2i & i-1&1-3i|0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1-i & 3-4i |0 \\0 & 4i+1 & 3i-7|0\\0& 3i+5 &20i-4|0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1-i & 3-4i |0 \\0 & 1 & \frac{31i+5}{17}|0\\0& 3i+5 &20i-4|0
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & \frac{25-32i}{17}|0 \\0 & 1 & \frac{31i+5}{17}|0\\0& 10i &0|0
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & \frac{25-32i}{17}|0 \\0 & 1 & \frac{31i+5}{17}|0\\0& 1 &0|0
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 0|0 \\0 & 1 & 0|0\\0& 0&1|0
\end{bmatrix}$
$1.P_{12}\\
2.M_1(-i)\\
3.A_{12}(-2-i),A_{13}(i-3)\\
4.M_2(\frac{1-4i}{17})\\
5.A_{21}(i+1),A_{23}(-3i-5)\\
6.M_3(-\frac{i}{10})\\
7.A_{31}(\frac{32i-25}{17}),A_{32}(\frac{x-31i-5}{17})$
We can notice that the only solution to this system is trivial solution.
