# Question about how to solve a simple matrix (beginner linear algebra).

1. Jan 25, 2012

### skyturnred

1. The problem statement, all variables and given/known data

Solve the following (obviously using matrices):

a=[2 1 1 3 8; 3 -1 4 7 7; 0 1 -1 -1 2; 1 -1 2 3 1]

2. Relevant equations

3. The attempt at a solution

OK so my linear algebra prof is useless. So I am now on my own (well plus the infinite power of the internet that is!)

So on our assignment, we must solve the following system

a=[2 1 1 3 8; 3 -1 4 7 7; 0 1 -1 -1 2; 1 -1 2 3 1] (where each ; denotes a new line in the matrix, so in total there are 4 rows and 5 columns. Note: This is an augmented matrix.)

So, no matter WHAT I do in row reducing, I always seem to get two rows of complete zeroes on the bottom, which leaves me with only 2 leading variables, and 2 free variables. The only problem is, since I have 4 unknowns and 4 equations, shouldn't it be very possible for me to find what each variable is equal to?

Thanks!

PS: the system I CONTINUOUSLY get is the following:

a=[1 0 1 2 3; 0 1 -1 -1 2; 0 0 0 0 0; 0 0 0 0 0]

2. Jan 25, 2012

### Ray Vickson

The answer to your question " shouldn't it be very possible for me to find what each variable is equal to?" is NO. Your 4x4 matrix A =[[2 1 1 3],[3 -1 4 7],[0 1 -1 -1],[1 -1 2 3]] is singular, so either the system of equations Ax = b (b = [8 7 2 1]^T) has no solution, or else it has infinitely many. You are getting infinitely many. Just counting the number of equations and the number of variables is not enough; we also need a non-singular matrix on the left.

I think your solution is correct. When I put the system through Maple I get a solution of the form x1 = 2 - u - 2v, x2 = 2 + u + v, x3 = u, x4 = v for arbitrary values u, v.

RGV

3. Jan 25, 2012

### skyturnred

OK thanks, I appreciate the help.