# Thread: [SOLVED] Linear Equation

1. ## [SOLVED] Linear Equation

Solve the system of linear equations in complex numbers:

$(2+i) x + (-3i) y = 3+i$

$3i x + (3-2i) y = 3-3i$

How would I do this?

2. Originally Posted by ronaldo_07
Solve the system of linear equations in complex numbers:

$(2+i) x + (-3i) y = 3+i$

$3i x + (3-2i) y = 3-3i$

How would I do this?
Equate the real and imaginary parts on each side:

$(2+i) x + (-3i) y = 3+i$: Solve 2x = 3 and x - 3y = 1.

3. you can easily solve linear equations like this 1 with Gaussian elimination

4. Originally Posted by 1234567
you can easily solve linear equations like this 1 with Gaussian elimination
If it's so easy why don't you solve it yourself? Mr.F provided a very simple approach to the problem and you should be able to easily solve from what he has provided.

5. Originally Posted by mr fantastic
Equate the real and imaginary parts on each side:

$(2+i) x + (-3i) y = 3+i$: Solve 2x = 3 and x - 3y = 1.
I don't think this question is solved individually or am I wrong? I believe the question wants to find x and y for which supports both equations.

6. Originally Posted by ronaldo_07
I don't think this question is solved individually or am I wrong? I believe the question wants to find x and y for which supports both equations.
Mr Fantastic supposed that x and y are real, which is usually the case for unknowns called x and y.

If x and y are complex then you can use substitution to calculate them. But you need of course both equations.

7. I am still unsure as to how to solve this.

If I put the real and imaginary numbers together I would get

$2x=3$ $x-3y=1$
$3y=3$ $3x-2y=-3$

But I do not see how I can solve this with these solutions.

8. Originally Posted by ronaldo_07
I am still unsure as to how to solve this.

If I put the real and imaginary numbers together I would get

$2x=3$ $x-3y=1$
$3y=3$ $3x-2y=-3$

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).

9. 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.

10. Originally Posted by ronaldo_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.
Do it in the same way as you would if both equations were real. Where are you stuck using that approach?

11. Im not sure how to do it as if it was real as I dont know what to do when there is i values

12. Originally Posted by ronaldo_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
Get the inverse of the 2x2 coefficient matrix in the usual way. If you understand the arithmetic of complex numbers there is nothing difficult here. Have you actually tried working through the process? Post your working of this and state where you get stuck.

13. $\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]
\left[\begin{array} {cc} x \\ y \end{array}\right]=
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$

Is this correct?

14. Originally Posted by ronaldo_07
$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]
\left[\begin{array} {cc} x \\ y \end{array}\right]=
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$

Is this correct?
Yes. Now calculate the inverse matrix. Then left multiply both sides by the inverse matrix. Then read off the answer for x and y.

15. $\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]
\left[\begin{array} {cc} x \\ y \end{array}\right]=
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$

$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1}$ $\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$= $\left[\begin{array} {cc} x \\ y \end{array}\right]$

$\frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]$

After this it gets really messy and I am unable to get a solution.

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