# Systems of linear equations

1. Sep 26, 2008

### war485

1. The problem statement, all variables and given/known data
Taking first year matrix/linear algebra theory at college.

I got a system of linear equations:
x+2y+3z=0
-5y-2z=0
4z=0

2. The attempt at a solution
This question looks so funny and easy but I think there must be a trick answer.
I got: z = 0, y = 0 and x = 0 (used back substitution).
The real problem for me is interpreting/accepting this answer.
Since all the variables are 0, then the whole thing is going to be zero. Is there a special interpretation for this? Does this mean that there are infinite solutions for x y and z? If there are infinite solutions, does this imply that x y and z are "free variables"?

I'm not even sure if I should put this in the "precalculus" part of the forum.
Thanks

2. Sep 26, 2008

### gabbagabbahey

Your solution is correct. Your interpretation of the solution is not; why would x,y,z be considered as free variables, when you have just shown that they must equal zero. It seems like (0,0,0) is only one unique solution not infinite.

P.S. this thread probably does belong in the precalc forum.

3. Sep 27, 2008

### war485

thanks.

4. Sep 28, 2008

### Melawrghk

What you have is a matrix with a trivial solution (0, 0, 0) which you will get every time if the last column in the matrix is entirely zeros. The problem is, sometimes you will also get other solutions (called "non-trivial). That's not the case here because the number of variables you have equals the number of equations. But if you were to cross out the bottom equation altogether, you'd have 3 variables, but only 2 equations, which would result in having an additional parameter.