Suppose I have a complex system of equations in 3 unknowns, like this one:
$$
\pmatrix{
40 & -20 & 0\\
-20 & 20-20j & 30+10j\\
4 & -5 & 1
}
\pmatrix{
x_1+j x_2\\
y_1+j y_2\\
z_1+j z_2\\
}=
\pmatrix{
10\\
0\\
0\\
}
$$
Here are solutions:
I want to split the matrix with complex numbers to two matrices with real numbers. However the imaginary matrix would only have one equation:
$$-20y_2+10z_2=0$$
What am I doing wrong?
Or is it possible to split such a system to two systems in the first place? (two "real" systems?)
Best Answer
Also $X$, $Y$ and $Z$ should be split into real and imaginary parts.
The system becomes $$ \begin{cases} 40(x_1+jx_2)-20(y_1+jy_2)=10\\ -20(x_1+jx_2)+(20-20j)(y_1+jy_2)+(30+10j)(z_1+jz_2)=0\\ 4(x_1+jx_2)-5(y_1+jy_2)+(z_1+jz_2)=0 \end{cases} $$ so you get $$ \begin{cases} 40x_1-20y_1=10\\ 40x_1-20y_2=0\\ -20x_1+20y_1+20y_2+30z_1-10z_2=0\\ -20x_2-20y_1+20y_2+10z_1+30z_2=0\\ 4x_1-5y_1+z_1=0\\ 4x_2-5y_2+z_2=0 \end{cases} $$ which are six equations in six unknowns.
However, there's no need to do this, because complex numbers are as well behaved as the real numbers, when solving linear systems.