# Solving Matrix Vector Equation

1. Oct 12, 2011

### Gurvir

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

2. Relevant equations
None

3. The attempt at a solution
Well I am pretty sure you just put it into a matrix and solve in row echelon form. So I tried it.

1 -1 -1 0
2 1 7 0
1 2 9 0

r2 = r2 - 2r1
r3 = r3 - r1

1 -1 -1 0
0 3 9 0
0 3 10 0

r3 = r3 - r2

1 -1 -1 0
0 3 9 0
0 0 1 0

r2 = r2 - 9r3

1 -1 -1 0
0 3 0 0
0 0 1 0

r2 = r2/3
r1 = r1 + r3

1 -1 0 0
0 1 0 0
0 0 1 0

r1 = r1 + r2

1 0 0 0
0 1 0 0
0 0 1 0

But that means a = 0, b = 0, and c = 0?
Edit: Oh, it asks if they are linearly independent, I am pretty sure they are not because it cannot equal zero, is there a easier way to figure this out or should I have noticed that because the answer was 0 from the beginning. But I am pretty sure you still have to show work. /edit

I do not have an answer for this equation solved so that is why I need to check. I don't even know if I am doing this correct at all, so if I am not, please advice me because I have a quiz tomorrow. Thanks in advance.

2. Oct 12, 2011

### Dick

If the only solution is a=b=c=0, then they ARE linearly independent. I didn't check your derivation in detail but I know that's the correct conclusion. You have shown they are linearly independent.

3. Oct 12, 2011

### Gurvir

So they are linearly independent if a = b = c = 0?

I thought it was not suppose to equal zero like the determinant or whatever it is called det A. When A doesn't equal 0, it is not or is it linearly independent? Or does this have nothing to do with this?

4. Oct 12, 2011

### Dick

If det(A) isn't zero, then they are linearly independent. If you can prove that a=b=c=0 then they are also linearly independent. Both work fine. You did the second.

5. Oct 12, 2011

### Gurvir

OK! Thank you I understand. I seriously love this forum. University profs go to fast, it helps having like a online tutor! Thanks again!

6. Oct 12, 2011

### Staff: Mentor

They are linearly independent if that is the only solution for the constants. For three vectors that are linearly dependent, that will be one solution, but there will be many more where not all of the constants are zero.