# Basic Matrix Question

1. Jan 22, 2010

### maherelharake

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

Determine, without performing any calculations, whether a linear system with the given augmented matrix has a unique solution, infinitely many solutions, or no solution. Justify your answer.

| 3 -2 0 1 | 1 |
| 1 2 -3 1 | -1|
| 2 4 -6 2 | 0 |

*Not sure how to input matrices into a computer, so sorry for the way I did it.

2. Relevant equations

3. I am not entirely sure how to begin this problem. Any suggestions will be helpful to get me started, and if I still need help I will ask. Please help. Thanks.

2. Jan 22, 2010

### tnutty

The key thing is on row3.

3. Jan 22, 2010

### maherelharake

Thanks for the response! I noticed before I posted my question that the final row had a constant equal to 0. I tried to mentally rearrange the matrix into row echelon form, but I didn't really get much further. I see that it has 4 variables. and 2 nonzero rows. Does that mean that there are 4-2=2 free variables. This is the part that I am getting confused on mostly. Thanks again.

4. Jan 22, 2010

### tnutty

Oh wait, there are 3 equation, and 4 variables. Do you think we have enough information
to solve this matrix?

5. Jan 22, 2010

### maherelharake

I thought about that as well, but I didn't think that choice would be an option according to the directions. :(
Is it possible there is some sort of cancellation after rearranging them mentally? I tried but couldn't get it to work, but it's possible I made a careless error.

6. Jan 22, 2010

### tnutty

Oh, wait see this :
The original matrix :
[3 -2 0 1 1]
[1 2 -3 1 -1]
[2 4 -6 2 0]

realize that the last row is a multiple of row2. Lets scale the last row3 so we have :

[3 -2 0 1 1]
[1 2 -3 1 -1]
[1 2 -3 1 0]

Notice something there?

7. Jan 22, 2010

### maherelharake

Does that mean there is no solution since those two equations are the same, and they can't be equal to two different values simultaneously?

8. Jan 22, 2010

### maherelharake

The left part of the equation is the same, but the right is different. Hence, no solution?

9. Jan 22, 2010

### tnutty

well try to solve it out and see

10. Jan 22, 2010

### maherelharake

I will give it a shot, but the problem doesn't want me to solve it out. I will try it anyways and repost when I give it my best attempt.

11. Jan 22, 2010

### tnutty

Its ok, the problem is trying to make you think hard, but I think once you thought about
the problem a little, you should then confirm it by solving the matrix.

12. Jan 22, 2010

### maherelharake

When I try to solve it I come up with
[1 -2/3 0 1/3 1/3]
[0 0 0 0 -1]
[0 0 0 0 1]

that means that x1=x2=x3=x4=0, however using these values, it is not possible to attain both 1 and -1 simultaneously. Thus the answer is no solution.

Did I make a mistake along the way?

13. Jan 22, 2010

### Dick

Yes. [1 2 -3 1 -1] and [1 2 -3 1 0] and translating them into equations, they can't be true simultaneously. Why are we dragging this out?

14. Jan 22, 2010

### tnutty

In the last row you have this : [0 0 0 0 1] which means this in equation form,
0X1 + 0X2 + 0X3 + 0X4 = 1, which means 0 = 1. This is impossible mathematically, right?
Because a 0 cannot never equal 1, which means that the system is inconsistent, or no solution. So Yes you are correct as to your final conclusion.

15. Jan 22, 2010

### maherelharake

Ok guys thanks a lot. I really appreciate it. I'm new to all of this stuff and I am still getting my feet under me. Thanks again.

16. Jan 23, 2010

### HallsofIvy

This is the crucial point, and why you don't need to do any calculation. All numbers in the third row are two times the numbers in the second row except the last. That tells us there is NO solution. If all numbers in the third row had been a multiple of the second row (and first and second rows were independent) there would have been an infinite number of solutions.

17. Jan 23, 2010

### maherelharake

Thanks a lot. That clear explanation really helped out even more.