Problem when row reducing matrix with complex entries I have to find a row-reduced matrix which is row-equivalent to
$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right]$$
On one hand
$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right] \\ \xrightarrow{R_1 \leftrightarrow R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ i & -(1+i) & 0 \\ 1 & 2i & -1\\ \end{array}\right]$$
$$\xrightarrow{R_2 \gets -iR_1 + R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 1 & 2i & -1\\ \end{array}\right] $$
$$\xrightarrow{R_3 \gets -R_1 + R_3} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ \frac{1}{(-1+i)}R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ R_1\gets 2R_2+R_1} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ R_3\gets -(2+2i)R_2+R_3} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 0 & -4\\ \end{array}\right] $$
and this last matrix is equivalent to the identity matrix $I_{3x3}$.
On the other hand I reduced it with other row operations and concluded the same as WOLFRAM ALPHA. Which one is wrong? Any help is appreciated.

Your mistake is in 4th step. When you are taking $(-1+i)$ common from $R_2$. Then element at 2nd row and 3rd column is wrong I think.
