# Homework Help: A matrix question

1. Sep 25, 2008

### Melawrghk

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

Solve: -8x-5y-9z=4

3. The attempt at a solution

I just need to know if I'm doing this correctly...

There is only one equation given, so I expressed x in terms of y and z and got:
x=-5/8y-9/8z-1/2
then I set y=s and z=t
x=-(5/8)s -(9/8)t -1/2

so I'd enter the answers as:
x = -(5/8)s-(9/8)t-1/2
y = s
z = t

Is that correct? I only have one attempt for the online assignment.

Thanks!

2. Sep 25, 2008

### buffordboy23

Try checking it yourself. You have three variables and one equation, so there must be multiple solutions as you suggested. You should be able to pick arbitrary values for "s" (y) and "t" (z) to determine your "x". Do they satisfy the given equation?

3. Sep 25, 2008

### Defennder

Yeah it looks ok.

4. Sep 25, 2008

### Melawrghk

Thanks! It was right. I have another problem though. This one I'm actually stuck on...

solve the system:
6A-2B+3C=4
1A+6B-6C=20
3A-1B-2C=9

6, -2, 3, 4
1, 6, -6, 20
3, -1, -2, 9

And through some simple things I got to:
1, -13, 11, -31
0, 19, -17, 51
0, 0, 5, -14

After which I don't know what to do... Help?

5. Sep 26, 2008

### Defennder

Well, you've reduced the matrix to a row-echelon form. Why not perform further row reduction to reduce it to its reduced-row echelon from? Remember that the leading entry of every pivot column of a RREF matrix is 1.

6. Sep 26, 2008

### HallsofIvy

Assuming that is correct (I haven't checked the calculations), those now correspond to x- 13y+ 11z= -31, 19y- 17z= 51, and 5z= -14.

Now you can "back substitute". Solve the last equation for z, substitute that into the second equation and solve for y, then substitute those two values for y and z into the first equation and solve for x.