# Solve for variable in augmented matrix

1. Sep 25, 2011

### Gurvir

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

Given the augmented matrix for a system of equations below, find m such that the system will have infinitely many solutions.
3 3 -9 15
1 -1 3 -3
-4 -4 12 m

m = ?

2. Relevant equations
None

3. The attempt at a solution

Well I tried to reduce the matrix to row-echelon form but I'm not sure what to do with the 'm' at all. I am not familiar with variables in matrices. Here's how far I have gotten:

r1/3 and r3/-4

1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4

r2-r1 and r3-r1

1 1 -3 5
0 1 -3 4
0 0 0 m-5/-4

I don't know how to answer for m. I am pretty sure that infinite solutions means that the matrix needs as many variables as equations which I read somewhere else on the forum.

Last edited: Sep 25, 2011
2. Sep 25, 2011

### Staff: Mentor

No, it doesn't mean this. If the system represented by your augmented matrix has an infinite number of solutions, there will be fewer equations than variables.

What are you doing here?
Dividing a row by another row is not one of the three valid row operations. In the matrix above that, you should notice that row 1 and row 3 are almost the same, and will be the same for some value of m.

3. Sep 25, 2011

### Gurvir

So wait, you make one row equal to the one with the variable to solve for it? also, dividing is not one of the row operations but i believe multiplying is. where you can multiply 1/-4 which is the same as dividing

4. Sep 25, 2011

### Staff: Mentor

Yes, you can replace a row by a multiple of itself, but you can't divide one row by another, which is what your notation implied that you were doing.

5. Sep 25, 2011

### Gurvir

My bad, i didn't mean r2/r1...it was r2-r1 and r3-r1

Anyways, back to my question:
You make one row equal to the one with the variable to solve for it? Or is there some other way

6. Sep 25, 2011

### Staff: Mentor

What does m need to be so that the first and third rows are the same?
1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4

7. Sep 25, 2011

### Gurvir

Oh! m = -20?

Would this be the same idea?
Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.

5 -3 h
-20 15 5

I can't solve for it, you can never make the first to variables in the matrix equal
You can get:

r2/-5
5 -3 h
4 -3 1

or

r1 * -4
-20 12 -4h
-20 15 5

Last edited: Sep 25, 2011
8. Sep 25, 2011

### Staff: Mentor

You have a mistake in the work above.
In this augmented matrix...

5 -3 h
-20 15 5

it's possible to find a solution for any value of h.

9. Sep 25, 2011

### flyingpig

If you have trouble seeing the "big picture", write it out as an equation.

10. Sep 26, 2011

### Gurvir

Sorry I was just brain dead yesterday. You just divide r2 by -4 and then it gives you h. Thanks for your help guys.

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