Find $z_1$ and $z_2$ in the system of complex numbers

Question

\begin{cases} (-1+i)z_1+(2i-3)z_2=-2; \\ (2+2i)z_1+(3+5i)z_2=0. \end{cases}

I decided by the method of Cramer. Received such an answer: $$ z_1=-\frac{3 + 5i}{2} \;\;\;\;;\;\;\;\; z_2=1+i $$

Can you please tell me how to find $ z_1 $ and $ z_2 $ without Cramer's method?

Answer 1

Well, we have:

$$ \begin{cases} \left(-1+i\right)\cdot\text{z}_1+\left(2i-3\right)\cdot\text{z}_2=-2\\ \\ \left(2+2i\right)\cdot\text{z}_1+\left(3+5i\right)\cdot\text{z}_2=0 \end{cases}\tag1 $$

Now, we can write $\text{z}_1=\text{a}+\text{b}i$ and $\text{z}_2=\text{c}+\text{d}i$. We also know that the imaginary parts of the LHS of the system has to be 0. So we can write:

$$ \begin{cases} -\text{a}-\text{b}-3\text{c}-2\text{d}=-2\\ \\ \text{a}-\text{b}+2\text{c}-3\text{d}=0\\ \\ 2\text{a}-2\text{b}+3\text{c}-5\text{d}=0\\ \\ 2\text{a}+2\text{b}+5\text{c}+3\text{d}=0 \end{cases}\tag2 $$

Solving that system gives $\text{a}=-3$, $\text{b}=-5$, $\text{c}=2$ and $\text{d}=2$.

Answer 2

HINT: plug $$z_2=\left(-\frac{11}{34}+\frac{7 i}{34}\right)z_1$$ into your first equation and solve this equation for $$z_1$$ HINT: $$-\frac{2+2i}{3+5i}\cdot \frac{3-5i}{3+5i}=...$$

Answer 3

Substitution is another method you can use. Substitute value for $z_2$ from second equation in first equation. $$z_2=\frac{-(2+2i)}{(3+5i)}z_1$$

Answer 4

Let$$A=\begin{bmatrix}-1+i & -3+2 i \\ 2+2 i & 3+5 i\end{bmatrix}.$$Then your system is equivalent to$$A.\begin{bmatrix}z_1\\z_2\end{bmatrix}=\begin{bmatrix}-2\\0\end{bmatrix}.$$Therefore, you can solve it doing\begin{align}\begin{bmatrix}z_1\\z_2\end{bmatrix}&=A^{-1}.\begin{bmatrix}-2\\0\end{bmatrix}\\&=\begin{bmatrix}\frac{3}{2}+\frac{5 i}{2} & \frac{3}{2}-i \\ -1-i & -\frac{1}{2}+\frac{i}{2}\end{bmatrix}.\begin{bmatrix}-2\\0\end{bmatrix}\\&=\begin{bmatrix}-3-5i\\2+2i\end{bmatrix}.\end{align}