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?

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- Jan 17th 2009, 02:13 AMronaldo_07[SOLVED] Linear Equation
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? - Jan 17th 2009, 02:53 AMmr fantastic
- Jan 17th 2009, 08:09 AM1234567
you can easily solve linear equations like this 1 with Gaussian elimination

- Jan 17th 2009, 08:18 AMTheMasterMind
- Jan 17th 2009, 08:31 AMronaldo_07
- Jan 17th 2009, 11:06 AMrunning-gag
- Jan 17th 2009, 12:07 PMronaldo_07
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. - Jan 17th 2009, 01:55 PMmr fantastic
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). - Jan 17th 2009, 02:15 PMronaldo_07
Both equations are one It is a "system of linear equations" I think you use matrices to solve but im not sure how to when it has complex numbers.

- Jan 17th 2009, 02:37 PMmr fantastic
- Jan 17th 2009, 02:39 PMronaldo_07
Im not sure how to do it as if it was real as I dont know what to do when there is i values

- Jan 17th 2009, 02:50 PMmr fantastic
- Jan 17th 2009, 03:03 PMronaldo_07
$\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]$

Is this correct? - Jan 17th 2009, 03:06 PMmr fantastic
- Jan 17th 2009, 04:32 PMronaldo_07
$\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. (Worried)