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).
