# [SOLVED] Linear Equation

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• January 17th 2009, 02:13 AM
ronaldo_07
[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?
• January 17th 2009, 02:53 AM
mr fantastic
Quote:

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.
• January 17th 2009, 08:09 AM
1234567
you can easily solve linear equations like this 1 with Gaussian elimination
• January 17th 2009, 08:18 AM
TheMasterMind
Quote:

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.
• January 17th 2009, 08:31 AM
ronaldo_07
Quote:

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.
• January 17th 2009, 11:06 AM
running-gag
Quote:

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.
• January 17th 2009, 12:07 PM
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.
• January 17th 2009, 01:55 PM
mr fantastic
Quote:

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).
• January 17th 2009, 02:15 PM
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.
• January 17th 2009, 02:37 PM
mr fantastic
Quote:

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?
• January 17th 2009, 02:39 PM
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
• January 17th 2009, 02:50 PM
mr fantastic
Quote:

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.
• January 17th 2009, 03:03 PM
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?
• January 17th 2009, 03:06 PM
mr fantastic
Quote:

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.
• January 17th 2009, 04:32 PM
ronaldo_07
$\left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]
\left[\begin{array} {cc} x \\ y \end{array}\right]=
\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$

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

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

After this it gets really messy and I am unable to get a solution. (Worried)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last