A*x=B, linear algebra problem.

[colossus__]

Homework Statement

The problem is rather simple. A set of linear equations, in the form V=Z*I, is given to represent a circuit in the frequency domain. The values for the V and I vectors are given and i have the Z(impedance) matrix written in function of the Z1, Z2, Z3,...,Zn variables?

In a simpler way: How to solve the A*x=B equation, when the values for the x and B vectors are given, and A is written in function of A1, A2, A3,...,An variables.
In my case, the A matrix looks like this:
|A1 -A2 0 |
|A1 0 -A3|
|1 1 1 |

x vector:
-0.26
0.259 - i0.966
0.259 + i0.966

B vector:
150 + i0.342
-150 + i0.342
0

Homework Equations

As simple as stated before, the only equation is the A*x=B. My example with all the complex numbers is not the best to ask for help in this subject, but it's the one that has led me to it.

The Attempt at a Solution

So far i have tried algebraic manipulation of the A*x=B equation multiplying it by A^-1 in the attempt to somehow reduce the matrix into a vector or any of the vectors into a matrix.

I tought maybe eigenvectors and eigenvalues could be involved in the solution of this problem, but since i don't have any linear algebra books around and don't really remember how to use this stuff, I'm hoping you guys could help me with this one.

 

[Dick]

It's not that hard. Just multiply A by the vector x and equate the resulting vector to B. That gives you three linear equations in the three unknowns A1, A2, A3. Actually, I'll save you a little time. The (1,1,1) row of the matrix times x gives (-0.26)+2*0.259. That's not zero but the third component of B is zero. That makes the system inconsistent. Is there a typo in there somewhere?

 

[colossus__]

You pointed it right, there is something wrong with the system. I went back to the question and found that the data given is wrong. The sum of the currents in the node in question is not equal to zero, therefore we have this problem.

I found this question in a test that is now 2 years old, and I'm actually surprised that none of the students who took the test noticed this mistake.

We could try to redefine the x vector to:
-0.52
0.26 - i0.966
0.26 + i0.966

Now the sum is really equal to zero. I just don't know if this will mess up with the balance in the rest of the system.

 

[colossus__]

I tried using your suggestion to solve the problem, it ended up looking like this:
(in a step by step version)

<br /> \begin{bmatrix}<br /> A_1 &amp; -A_2 &amp; 0\\<br /> A_1 &amp; 0 &amp; -A_3\\<br /> 1 &amp; 1 &amp; 1<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> x_1\\<br /> x_2\\<br /> x_3<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> B_1\\<br /> B_2\\<br /> B_3<br /> \end{bmatrix}<br />

<br /> \begin{bmatrix}<br /> x_1A_1 &amp; -x_2A_2 &amp; 0\\<br /> x_1A_1 &amp; 0 &amp; -x_3A_3\\<br /> x_1 &amp; x_2 &amp; x_3<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> 1\\<br /> 1\\<br /> 1<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> B_1\\<br /> B_2\\<br /> B_3<br /> \end{bmatrix}

<br /> \begin{bmatrix}<br /> x_1 &amp; -x_2 &amp; 0\\<br /> x_1 &amp; 0 &amp; -x_3\\<br /> \frac{x_1}{A_1} &amp; \frac{x_2}{A_2} &amp; \frac{x_3}{A_3}<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> A_1\\<br /> A_2\\<br /> A_3<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> B_1\\<br /> B_2\\<br /> B_3<br /> \end{bmatrix}

Using the actual values,

<br /> \begin{bmatrix}<br /> -0.52 &amp; -0.26 + i0.966 &amp; 0\\<br /> -0.52 &amp; 0 &amp; -0.26 - i0.966\\<br /> \frac{-0.52}{A_1} &amp; \frac{0.26 - i0.966}{A_2} &amp; \frac{0.26 + i0.966}{A_3}<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> A_1\\<br /> A_2\\<br /> A_3<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 150 + i0.342\\<br /> -150 + i0.342\\<br /> 0<br /> \end{bmatrix}

And i still got no clue on how to solve it with those divisions by An that showed up now.

 

[Dick]

You are doing some strange matrix manipulations. If you really want to write this as a matrix equation the last row should really be [0,0,0] and B3 should be 0. There's no A dependency in the last equation. Which brings us to the next problem, now you only have two equations in three unknowns. There are many solutions.