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Problem: Determine the solutions in $\mathbb{C}^3$ of the following system over $\mathbb{C}$: \begin{align*} \begin{cases} 2x+iy-(1+i)z &=1 \\ x-2y+ iz &= 0 \\ -ix +y -(2-i)z &= 1 \end{cases} \end{align*}

Attempt at solution: I'm not sure what's the best method to handle such problems. I just used the same technique as I would use for real numbers. From the problem, we see that the augmented matrix representing this system is: \begin{align*} A= \left(\begin{array}{ccc|c} 2 & i & -(1+i) & 1 \\ 1 & -2 & i & 0 \\ -i & 1 & -(2-i) & 1 \end{array}\right) \end{align*} I first interchanged the first two rows to get a leading 1 in the top left corner. Then I applied the operations: $R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow - \frac{1}{i} R_3, R_3 \rightarrow R_3 - R_1$. This gives: \begin{align*} A \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 2 & i & -(1+i) & 1 \\ -i & 1 & -(2-i) & 1 \end{array}\right) \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & i+4 & -1-3i & 1 \\ -i & 1 & -(2-i) & 1 \end{array}\right) \\ \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & i+4 & -1-3i & 1 \\ i & -\frac{1}{i} & \frac{2-i}{i} & -\frac{1}{i} \end{array}\right) \\ \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & i+4 & -1-3i & 1 \\ 0 & -\frac{1}{i} +2 & \frac{2}{i}-2 & -\frac{1}{i} \end{array}\right) \end{align*} Now I want a leading 1 at the position $a_{22}$. If I thus apply $R_2 \rightarrow \frac{1}{i+4} R_2$ I get: \begin{align*} \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & 1 & \frac{-1+3i}{i+4} & \frac{1}{i+4} \\ 0 & -\frac{1}{i} +2 & \frac{2}{i}-2 & -\frac{1}{i} \end{array}\right) \end{align*}

If I divide the third row now by $-\frac{1}{i} +2$ to get a leading 1 at the position $a_{23}$ I get some complicating fractions. So I'm not sure if this is the right way to go. Is there no easier way to handle such problems?

Any help would be greatly appreciated.

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You are going about it in the right way, but one thing you should do is to always keep denominators real. Thus for example $-\frac{1}{i} = i$ and $\frac{1}{4+i} = \frac{4-i}{(4-i)(4+i)} = \frac{4-i}{17}$.

So your next steps would be \begin{align*} \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & 1 & \frac{-1+3i}{i+4} & \frac{1}{i+4} \\ 0 & -\frac{1}{i} +2 & \frac{2}{i}-2 & -\frac{1}{i} \end{array}\right) = \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & 1 & \frac{-1+13i}{17} & \frac{4-i}{17} \\ 0 & 2+i & -2-2i & i \\ \end{array}\right) \\ \end{align*} \begin{align*} \sim \left(\begin{array}{ccc|c} 1 & 0 & \frac{-2+30i}{17} & \frac{8-2i}{17} \\ 0 & 1 & \frac{-1+13i}{17} &\frac{4-i}{17} \\ 0 & 2+i & -2-2i & i \end{array}\right) \sim \left(\begin{array}{ccc|c} 1 & 0 & \frac{-2+30i}{17} & \frac{8-2i}{17} \\ 0 & 1 & \frac{-1+13i}{17} & \frac{4-i}{17} \\ 0 & 0 & \frac{-19-59i}{17} & \frac{-9+15i}{17} \end{array}\right) \\ \end{align*} \begin{align*} \sim \left(\begin{array}{ccc|c} 1 & 0 & \frac{-2+30i}{17} & \frac{8-2i}{17} \\ 0 & 1 & \frac{-1+13i}{17} & \frac{4-i}{17} \\ 0 & 0 & 1 & \frac{814+816i}{3842} \end{array}\right) \\ \end{align*}

I have to admit the next (and last) step is going to look messy, but there you have it.

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