# Finding a solution to (A)(B)=0

1. Oct 23, 2012

### Ribena

Hi all,

So I've been trying to solve a system of 'linear' equations and I understand that for a non trivial solution to exist you have to get the determinant to reduce to zero.

Given that you have an equation which takes the form

(f(x))*A=0, where f(x) is an arbitary function and A is a constant, the solution is either f(x)=0 or A=0.

Now say that for a non trivial solution, I'd require that f(x)=0, is there any methods in general in which I can obtain an expression for A? Let's assume it's a 4 by 4 system of equation.

2. Oct 23, 2012

### Erland

What are f(x) and A? Matrices?

3. Oct 23, 2012

### Ribena

Polynomials.

4. Oct 23, 2012

### Erland

How can it then be a 4x4 system of equations?

No, you must more clearly specify what you mean...

5. Oct 24, 2012

### Ribena

Ah, I get what you mean now. I'm not from a mathematical background so do bear with me is some of my definitions are not precise. So I basically have 4 equations with 4 unknowns to solve lets assume that they are w,x,y,z. By eliminating the variables, I've managed to bring them down a single equation that looks something like (2g+h)*(x)=0 where g and h are predefined constants.

So for a non trivial solution to exist, then (2g+h)=0. Will it be possible then to know the expression for x in terms of g and h (if one exists at all)?

Hopefully it's not too confusing

6. Oct 24, 2012

### Erland

But if 2g+h=0, then any value of x will do.

7. Oct 29, 2012

### Ribena

Exactly, so I was wondering if there was any way in general to infer the expression of x. I think the only way is to perhaps add a constraint that does not result in an infinite solution case, but if possible, I didn't want to change the original constraints.

Just wanted to seek some opinions to see if there actually is some method that I was unaware of.