Systems of Equations with Complex Numbers

[inshooter]

Mod note: This thread was moved from a technical math section, so doesn't include the homework template.
I know this has been asked before, but none of the other posts have helped me. I cannot for the life of me figure out how to solve a system of equations with complex numbers. Here is a very simple example (from and older post):

(1+i)x + (2+i)y =5
(3+2i)x + (4+i)y =10 WolframAlpha says x=2 + i, and y = 1 - 2i

One suggestion was to split it up into two separate systems, which I believe would give:

x + 2y = 5 and ix + iy = 0i
3x + 4y = 10 and 2ix + iy = 0i Is that correct?

Obviously for the second equation, x and y would = 0, which is clearly not the original answer WolframAlpha gives. (For the second equation it turns out x = 0, and y = 2.5, which is still not like the original system)

Please help! How do you do this? And I know you can substitute for this equation, but that becomes impractical for systems with 3 and 4 equations and unknowns.

Thank you

 

[Mark44]

I know this has been asked before, but none of the other posts have helped me. I cannot for the life of me figure out how to solve a system of equations with complex numbers. Here is a very simple example (from and older post):

(1+i)x + (2+i)y =5
(3+2i)x + (4+i)y =10 WolframAlpha says x=2 + i, and y = 1 - 2i

One suggestion was to split it up into two separate systems, which I believe would give:

x + 2y = 5 and ix + iy = 0i
3x + 4y = 10 and 2ix + iy = 0i Is that correct?

I don't believe so. Your approach would work if you knew that x and y were real, not complex.

Obviously for the second equation, x and y would = 0, which is clearly not the original answer WolframAlpha gives. (For the second equation it turns out x = 0, and y = 2.5, which is still not like the original system)

For the 2nd equation, are you referring to ix + iy = 0i? That's equivalent to x + y = 0, of which x = y = 0 is just one solution of an infinite number of them.

Please help! How do you do this? And I know you can substitute for this equation, but that becomes impractical for systems with 3 and 4 equations and unknowns.

The simplest approach to this problem is to solve for one variable in terms of the other in one of the equations, and then substitute for the variable you solved for in the other equation. That will give you one equation in one unknown.

 

[inshooter]

I don't believe so. Your approach would work if you knew that x and y were real, not complex.
For the 2nd equation, are you referring to ix + iy = 0i? That's equivalent to x + y = 0, of which x = y = 0 is just one solution of an infinite number of them.
The simplest approach to this problem is to solve for one variable in terms of the other in one of the equations, and then substitute for the variable you solved for in the other equation. That will give you one equation in one unknown.

Yeah I know, but I encounter systems with 3 equations and 3 unknowns, sometimes 4. Like I said, making the substitution wouldn't be practical for those. I'm just using a system with 2 equations as an easy example.

 

[Mark44]

Note: I realize that the system of equations shown here is not the system shown in the OP, but the work shown here needs to be addressed.

x + 2y = 5 and ix + iy = 0i
3x + 4y = 10 and 2ix + iy = 0i Is that correct?

Obviously for the second equation, x and y would = 0, which is clearly not the original answer WolframAlpha gives. (For the second equation it turns out x = 0, and y = 2.5, which is still not like the original system)

You're missing an important point about linear equations. Although (0, 0) is a solution of ix + iy = 0i, it is only one of many solutions. Likewise, (0, 2.5) is a solution of 3x + 4y = 10, but there are an infinite number of solutions, one for each point on this line. The goal of solving a system of linear equations is to find the point of intersection, a point that is on both lines.

 

[inshooter]

Ah yeah good point lol. So if it can be split into two separate systems, how would you do that?

 

[Mark44]

Yeah I know, but I encounter systems with 3 equations and 3 unknowns, sometimes 4. Like I said, making the substitution wouldn't be practical for those. I'm just using a system with 2 equations as an easy example.

For the example you showed, substitution is the quickest method, I believe.

For examples with three or more equations, you can write the system of equations as an augmented matrix, and then use Gaussian elimination to reduce the matrix to a simpler form that gives the solution.

 

[Mark44]

Ah yeah good point lol. So if it can be split into two separate systems, how would you do that?

I wouldn't do that for the example in your post, for the reason I already gave.

 

[inshooter]

What I'm ultimately getting at here is: How can I solve a system of equations with 3 unknowns and complex numbers with my TI-84 plus Silver Edition calculator. Regular 3x3 matrices with no complex numbers are easy enough to do on there. But I have no idea how I would do it with complex numbers. Doing it by hand on a 1 hour exam is not ideal too. From a post on this forum from like 2009, it said something that you can split up the real and complex numbers. I don't know how to do that.

 

[Mark44]

For the system in post #1, you could rewrite it as
(1 + i)(a + bi) + (2 + i)(c + di) = 5
(3 + 2i)(a + bi) + (4 + i)(c + di) = 10

where x = a + bi and y = c + di

Expand the left side and separate out the real parts and the imaginary parts. This will give you four equations in four unknowns. The real part of the first equation above has to equal 5 and the imaginary part has to equal 0. The real part of the second equation has to equal 10, and the imaginary part has to equal 0.

Doing problems like this by hand is good practice, since many classes won't let you use calculators on exams. I would also recommend doing some problems by hand, since you seem to be missing some very basic ideas about how to solve a system of equations.

 

[inshooter]

So that comes out to be (if I did it right):

a-b+2c-d=5
3a-2b+4c-d=10


2ai+3bi+ci+4di=0i
ai+bi+2di+ci=0i

What should I do after that? Maybe this can't be done with my calculator and the only way to do this is with Gaussian elimination?

 

[Mark44]

So that comes out to be (if I did it right):

a-b+2c-d=5
3a-2b+4c-d=10


2ai+3bi+ci+4di=0i
ai+bi+2di+ci=0i

What should I do after that? Maybe this can't be done with my calculator and the only way to do this is with Gaussian elimination?

These equations look OK, but you don't need the 'i' in your last two equations, since you can divide both sides of each equation by i. Here's your system:
a-b+2c-d=5
3a-2b+4c-d=10
2a+3b+c+4d=0
a+b+2d+c=0

I recommend that you do Gaussian elimination by hand. When you get a solution, check that your values for a, b, c, and d satisify the original pair of equations.

 

[inshooter]

Wow, thank you! Yes my hand calculation, my calculator, and wolframalpha all agree! Replacing x = a + bi and y = c + di and removing the "i"s are what helped. Thank you!

If it were a system with 3 original equations and unknowns, you'd replace:
x=a+bi
y=c+di
z=e+fi
...and so on

 

[Mark44]

Wow, thank you! Yes my hand calculation, my calculator, and wolframalpha all agree! Replacing x = a + bi and y = c + di and removing the "i"s are what helped. Thank you!

If it were a system with 3 original equations and unknowns, you'd replace:
x=a+bi
y=c+di
z=e+fi
...and so on

Yes, and you would get 6 equations in 6 unknowns.

It's good that you did the calculations so many ways. It's also a boost in confidence to know that your work by hand agrees with the results produced by your calculator and WA.

If you don't have access to a calculator or computer, you can still verify that your results are correct by plugging your solutions into the original equations. If your solution is correct, each equation should be identically true (i.e., like 4 = 4).

 

[SteamKing]

You can write your complex linear equations as a special system of real equations as described in the following article:

http://www.math.hkbu.edu.hk/icsm2006/allabstract/Bertaccini.pdf