Best way to solve a system of complex equations?

In summary, solving systems of complex equations by hand can be done using various methods such as row reduction, Cramer's rule, or matrix inversion. However, it is important to note that the number of equations must be equal to or greater than the number of variables in order to get a unique solution. Complex numbers do not change the number of equations needed to solve a system.
  • #1
x86
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In my circuit analysis class I consistently need to solve system of complex equations, and I can't use MATLAB or anything for it. Suppose I have the following system:

(Va-Vs)/(-j15) + Va/33 + (Va-Vo)/(-j25)=0
(Vo-Va)/(-j25) + (Vo-Vs)/10 = 0

What is the best way to solve it by hand in a time convinient manner?
 
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  • #2
Can't you make into a set of linear equations and then use MATLAB to find the eigenvalues?

(1/33 - 1/j15 - 1/j25) Va + (1/j15) Vs + (1/25) Vo = 0
...

In this case, it seems you're missing a third equation.
 
  • #3
How do we usually solve systems of linear equations by hand? Row reduction, Cramer's rule, or matrix inversion. Pick whichever one you like best. I don't recommend matrix inversion except for 2x2 systems (since the inverse is "easy" to remember).

jason
 
  • #4
jedishrfu said:
In this case, it seems you're missing a third equation.

Probably one of the V or maybe of the differences is not an unknown, or else is an input in terms of which it is desired to express everything else.

Given that, if these are considered hard equations I am afraid there is some bad news... :bugeye:
 
  • #5
jasonRF said:
How do we usually solve systems of linear equations by hand? Row reduction, Cramer's rule, or matrix inversion. Pick whichever one you like best. I don't recommend matrix inversion except for 2x2 systems (since the inverse is "easy" to remember).

jason

I always found row reduction slower than just back-substitution. So essentially if I just do back substitution, I'll get an answer?

Also, i want to express Vo in terms of Vs, so I only have two equations.

If I want to solve for Vo, I'd just solve for Va, then back substitute?

jedishrfu said:
Can't you make into a set of linear equations and then use MATLAB to find the eigenvalues?

(1/33 - 1/j15 - 1/j25) Va + (1/j15) Vs + (1/25) Vo = 0
...

In this case, it seems you're missing a third equation.

Oh I wish, but on tests we have to do it by hand.

jasonRF said:
How do we usually solve systems of linear equations by hand? Row reduction, Cramer's rule, or matrix inversion. Pick whichever one you like best. I don't recommend matrix inversion except for 2x2 systems (since the inverse is "easy" to remember).

jason

I see. I guess complex numbers don't really change anything, even though you'd think it would mean you'd have 2 times more equations to solve
 
  • #6
jasonRF said:
How do we usually solve systems of linear equations by hand? Row reduction, Cramer's rule, or matrix inversion. Pick whichever one you like best. I don't recommend matrix inversion except for 2x2 systems (since the inverse is "easy" to remember).

x86 said:
I always found row reduction slower than just back-substitution. So essentially if I just do back substitution, I'll get an answer?

Also, i want to express Vo in terms of Vs, so I only have two equations.

If I want to solve for Vo, I'd just solve for Va, then back substitute?
If you're using matrices to solve for the variables, row reduction gets you a matrix with 1's on the diagonal. The 1 at the lower end of the diagonal gives you one of the variables, and you can then use back-substitution to find the values of the other variables. So back-substitution implies that you have already done row reduction, at least in the context of using matrices to solve the system. Otherwise I'm not sure what you're saying.

Also, and as pointed out already by jedishrfu, you have two equations and three variables, so it's not possible to get a unique solution. A system with fewer equations than variables is called under-determined.

jedishrfu said:
Can't you make into a set of linear equations and then use MATLAB to find the eigenvalues?

(1/33 - 1/j15 - 1/j25) Va + (1/j15) Vs + (1/25) Vo = 0
...

In this case, it seems you're missing a third equation.

x86 said:
Oh I wish, but on tests we have to do it by hand.

jasonRF said:
How do we usually solve systems of linear equations by hand? Row reduction, Cramer's rule, or matrix inversion. Pick whichever one you like best. I don't recommend matrix inversion except for 2x2 systems (since the inverse is "easy" to remember).

x86 said:
I see. I guess complex numbers don't really change anything, even though you'd think it would mean you'd have 2 times more equations to solve
No. Each complex solution counts as one solution.

Here's a simple example:
z + w = 2
z - w = 2i

Setting up an augmented matrix:
##\begin{bmatrix} 1 & 1 & | 2\\ 1 & -1 & | 2i \end{bmatrix}##
Using row reduction, we get
##\begin{bmatrix} 1 & 1 & | 2\\ 0 & 1 & | 1 - i \end{bmatrix}##
At this we can note that w = 1 - i, and then back substitute to get z, or we can continue to reduced row-echelon form (RREF) where each row starts with a 1:
##\begin{bmatrix} 1 & 0 & | 1 + i \\ 0 & 1 & | 1 - i \end{bmatrix}##
The single solution is (z, w) = (1 + i, 1 - i).
 
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1. What is the most efficient method for solving a system of complex equations?

The most efficient method for solving a system of complex equations is typically Gaussian elimination. This method involves reducing the system into simpler equations by eliminating variables until a solution can be found.

2. Is there a specific order in which the equations should be solved?

Yes, in general, it is best to solve the equations in the order in which they are given. However, if there is a particular variable that can be easily eliminated, it may be more efficient to solve for that variable first.

3. Are there any alternative methods for solving complex equations?

Yes, there are several other methods for solving complex equations such as Cramer's rule, elimination by substitution, and the method of determinants. However, Gaussian elimination is the most widely used and efficient method.

4. How do I know if my solution is correct?

To ensure that your solution is correct, you can substitute the values you found for each variable back into the original equations and see if they satisfy all of the equations. Additionally, you can use a calculator or software program to verify your solution.

5. Can I use complex numbers in a system of equations?

Yes, complex numbers can be used in a system of equations. In fact, some problems can only be solved by using complex numbers. However, the process of solving these equations may be more complex and require additional steps.

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