Solve the system of linear equations in complex numbers:
$\displaystyle (2+i) x + (-3i) y = 3+i$
$\displaystyle 3i x + (3-2i) y = 3-3i$
How would I do this?
I am still unsure as to how to solve this.
If I put the real and imaginary numbers together I would get
$\displaystyle 2x=3$ $\displaystyle x-3y=1$
$\displaystyle 3y=3$ $\displaystyle 3x-2y=-3$
But I do not see how I can solve this with these solutions.
I understood them as two seperate problems and I did the first. Implicit in that is that x and y are both real (which is the usual convention).
If they are meant to be solved simultaneously then x and y are not real. You need to provide clarification: are they two seperate questions (x and y are real) or a single question (x and y are not real).
$\displaystyle \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]
\left[\begin{array} {cc} x \\ y \end{array}\right]=
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$
$\displaystyle \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1}$ $\displaystyle \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$=$\displaystyle \left[\begin{array} {cc} x \\ y \end{array}\right]$
$\displaystyle \frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]$
After this it gets really messy and I am unable to get a solution.