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Best way to solve a system of complex equations?

  1. Nov 8, 2015 #1
    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?
  2. jcsd
  3. Nov 8, 2015 #2


    Staff: Mentor

    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.
  4. Nov 8, 2015 #3


    User Avatar
    Science Advisor
    Gold Member

    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).

  5. Nov 9, 2015 #4


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    Homework Helper
    Gold Member

    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:
  6. Nov 9, 2015 #5
    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?

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

    I see. I guess complex numbers dont really change anything, even though you'd think it would mean you'd have 2 times more equations to solve
  7. Nov 9, 2015 #6


    Staff: Mentor

    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.

    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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