Find the general solution of linear system

Question

$$\begin{cases}(2i)z+(i-1)w=-2\\(2+i)z+(1+i)w=1+4i\end{cases}$$

How to solve this linear system over $\mathbb{C}$?

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what I did :

let $$z=x+iy$$

$$w=x'+iy'$$

so we get :

$$(2i)(x+iy)+(i-1)(x'+y'i)=-2\implies(-2y+x'-y')+(2x+x'-y')i=-2$$ $$(2+i)(x+iy)+(1+i)(x'+y'i)=1+4i\implies(2x-y+x'-y')+(x+y+x'+y')i=-1+4i$$

so i need to solve this linear system:

$$\begin{cases}-2y+x'-y'=-2\\2x+x'-y'=0\\2x-y+x'-y'=-1\\x+y+x'+y'=4\end{cases}$$

and find the general solution for it ...

Is this a correct way to solve it , if not any other idea how to solve it ?

Answer 1

$$\Delta=(2i(1+i)-(2+i)(i-1)=1+i$$ $$\Delta_z=-2(1+i)-(1+4i)(i-1)=3+i$$ and $$\Delta_w=2i(1+4i)-(2+i)(-2)=-4+4i,$$ which gives $$w=\frac{\Delta_w}{\Delta}=4i$$ and $$z=\frac{\Delta_z}{\Delta}=2-i.$$

Answer 2

It is a very bad idea to express every complex number as $x+yi$ here. It follows from the first equation that $z+\frac{i-1}{2i}w=-\frac2{2i}$. Simplify this, express $z$ as a multiple of $w$ and replace the $z$ of the second equation by this value to solve the problem.

Answer 3

Hint:  eliminate $\,w\,$ between the equations:

$$ \begin{cases} \begin{align} (2i)z+(i-1)w &=-2 &\mid \;\cdot\, (1+i) \\(2+i)z+(1+i)w&=1+4i &\mid \;\cdot\, (i-1) \end{align} \end{cases} \quad \Bigg| \;-\; \Bigg| $$

$$ \implies \quad \big(2i(1+i) - (2+i)(i-1)\big)z = -2(1+i) - (i-1)(1+4i) \;\;\iff\;\; (1+i)z=3+i $$