Solving a complex equation $Ax=b$ with $A,b\in\mathbb C$ using linear algebra

Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have $$ (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \\ = [3x_r - 4x_c] + [4x_r + 3x_c]i $$ Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have $$ 3x_r - 4x_c = b_r\\ 4x_r + 3x_c = b_c $$ which is precisely the system of equations that you've come up with.

Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.

$$\begin{bmatrix}3&-4\\4&3\end{bmatrix}\begin{bmatrix}x_r&-x_c\\x_c&x_r\end{bmatrix}=\begin{bmatrix}b_r&-b_c\\b_c&b_r\end{bmatrix}$$

It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.

---

As an example where the simplified version will fail we can take a look at permutations. A permutation among three different elements can be represented with binary matrices:

$$P_1 = \left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right], P_2 = \left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]$$ Now let us say that if we apply $P_1$ to some permutation and it becomes $P_2$, then what was the original permutation?

This we can express as: $$P_1X = P_2$$

Here we need $X$ to be a 3x3 matrix, because there is no other way to represent it.

The answer is: $$X=P_1 ^{-1}P_2$$