# Linear Algebra Solving a System of LE

1. Jan 24, 2013

### Workout

Okay here's my system of equations:

x − 3y − 2z = 0
−x + 2y + z = 0
2x + 4y + 6z = 0

Solve the following systems using Gaussian elimination

I put it in a matrix and did Gaussian elimination.

But I can't find a unique solution and it doesn't end up working out. Is this true? Or am I just making a mistake?

2. Jan 24, 2013

### LCKurtz

How can we tell without seeing what you did? Why would you expect a unique solution? Show us your work.

3. Jan 24, 2013

### Staff: Mentor

As LCKurtz said, show us what you did.

Also, don't blow away the homework template - it's there for a reason.

4. Jan 24, 2013

### Workout

2R2 + R2 = | 1 -3 -2 0 |
-1 2 1 0
0 8 8 0

R1 + R2 = | 0 -3 -2 0 |
0 -1 -1 0
0 8 8 0

8R + R3 = | 1 -3 -2 0 |
0 -1 -1 0
0 0 0 0

So I was left with 2 equations.

x -3y -2z = 0
-y -z = 0

And I'm just stuck.

5. Jan 24, 2013

### Workout

Basically when I solve it I get z = 0
then y = -z
and x = -z

There's no unique solution I guess. I was specifically looking for a unique solution. I think that was my problem.

6. Jan 24, 2013

### Staff: Mentor

So far, so good, but you can do just a bit more to completely reduce your matrix. You should end up here:

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}$$

This is shorthand for the system
x + z = 0
y + z = 0

Geometrically, what you have is two planes in space that intersect in a line.

7. Jan 25, 2013

### HallsofIvy

It really makes no sense to "get z= 0" and then have a solution that involves z!

It should have been obvious from the start that, since we have "= 0" on the right side of each equation (this was a "homogeneous" system), x= y= z= 0 is a solution. So if there had been a unique solution, it would have been trivial.