Solution: We will use elementary row operations on the coefficient matrix to solve the problem. Note that the coefficient matrix is \begin{align*} A = \begin{bmatrix} 1 - \iota & -\iota \\ 2 & 1 - \iota \\ \end{bmatrix}. \end{align*} Using elementary row operations, we have \begin{align*} \begin{bmatrix} 1 - \iota & -\iota \\ 2 & 1 - \iota \\ \end{bmatrix} & \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 2 & 1 - \iota \\ 1 - \iota & -\iota \\ \end{bmatrix} \\ & \xrightarrow{R_1 \rightarrow \frac{R_1}{2}} \begin{bmatrix} 1 & \frac{1}{2} - \frac{\iota}{2} \\ 1 - \iota & -\iota \\ \end{bmatrix} \\ & \xrightarrow{R_2 \rightarrow R_2 - (1 - \iota ) R_1} \begin{bmatrix} 1 & \frac{1}{2} - \frac{\iota}{2}\\ 0 & 0 \\ \end{bmatrix} . \end{align*} Thus, the given system is equivalent to the following equation \begin{align*} x_1 + x_2 \left( \frac{1}{2} - \frac{\iota}{2} \right) = 0 \implies x_1 = -x_2 \left( \frac{1}{2} - \frac{\iota}{2} \right). \end{align*} Therefore, the solution set is \[ \mathcal{S} = \left\{ x_1 = (\iota - 1) \alpha , x_2 = 2\alpha : \alpha \in \mathbb{C} \right\} . \]