Question

Use the Gauss-Jordan method to solve each of the given systems of equations:
x+2y+z=5
2x+y-3z=-2
3x+y+4z=-5

Answer 1

The Gauss-Jordan algorithm = putting equations into augmented matrix form and then reducing to row echelon form.
So, your matrix is

1 2 1 | 5 R1
2 1 -3 | -2 R2
3 1 4 | -5 R3

Then take it down to reduced row echelon form.

Answer 2

how do I solve for x, y, z though?

Answer 3

So the augmented matrix is the matrix created from the coefficients of x, y, and z. The first column represents the coefficients of x, the second column is y, and the third is z. The goal of the Gauss-Jordan method is to try to get your matrix to look like this:

1 0 0 | A
0 1 0 | B
0 0 1 | C

So then you get 1x = A, 1y = B, 1z = C. If you're doing this problem, you should know how to reduce a matrix. So given the matrix I wrote out in my first reply, reduce it to the matrix here, which will give you what x, y, and z equal.

Answer 4

When reducing this equation I get:
5x+7y=13
17x+2y=-23
-1x-7y=-25

I'm not sure what to do next to solve for x, y, and z since all of my solutions still have 2 variables in them and I need to have at least one with only one variable to plug into the other equations.

Answer 5

Attached is what I have done, I think we do it differently in my class.

Answer 6

What you have isn't Gauss-Jordan elimination, it's regular equation substitution. Have you learned matrices yet?

Answer 7

no

Answer 8

If you haven't done matrices, it'll be difficult to explain Gauss-Jordan.

Answer 9

Well that's the way we been taught so far. I've attached an example problem the instructor did in class. That one was easy to solve because one of the equations ends with one variable so it can be plugged into the others. The problem I am struggling with doesn't end up the same way.

Answer 10

Aaah. Gotcha. That's not Gauss-Jordan but I can explain this now :)

Answer 11

So what you had so far was good then. Take the equation with two variables and pick a variable, let's say x.

Rewrite x in terms of y: x = ay + b

Substitute that into your equation with only x and y variables and solve for y. Plug y into the equation to solve for x and then plug x and y into any equation with three variables to solve for z.