Show two systems of linear equations are equivalent.

Question

The two systems are

$$\left\{\begin{align*} &2x_1+(-1+i)x_2+x_4=0\\ &3x_2-3ix_3+5x_4=0\; \end{align*}\right.$$

and

$$\left\{\begin{align*} &\Big(1+\frac i2\Big)x_1+8x_2-ix_3-x_4=0\\ &\frac 23x_1-\frac12x_2+x_3+7x_4=0\;. \end{align*}\right.$$

However I cannot find a method of solving them. So far, in class, we have only learned row reduction, nothing else. Of course, the methods of substitution and elimination are assumed knowledge.

What would an effective way of solving each of these be? Every time I end up with an equation for one of the variables, I'm unable to derive an equation for another variable that doesn't result in me getting an equality like $0=0$ after substitution.

Is it possible to combine both systems?

I've been working on this question for hours and I feel that the answer is so simple yet there is something fundamentally important that I am missing.

Any help is greatly appreciated.

Answer 1

Notice that from $\left\{\begin{align*} &2x_1+(-1+i)x_2+x_4=0\\ &3x_2-3ix_3+5x_4=0\; \end{align*}\right.$, you can write $x_1$ as a function of $x_2$ and $x_4$ and the same goes for $x_3$.

You get $\left\{\begin{align*} &x_1=-\frac{1}{2}((-1+i)x_2+x_4)\\ &x_3=\frac{1}{3i}(3x_2+5x_4)\; \end{align*}\right.$

So your set of solutions is $\{(x_1,x_2,x_3,x_4):x_1=-\frac{1}{2}((-1+i)x_2+x_4) \wedge x_3=\frac{1}{3i}(3x_2+5x_4)\wedge x_2, x_4\in \mathbb{C}\}$=$\{(-\frac{1}{2}((-1+i)x_2+x_4), x_2, \frac{1}{3i}(3x_2+5x_4), x_4): x_2, x_4\in \mathbb{C}\}$.