# Simple gauss-jordan elimination

1. Sep 8, 2008

### The_ArtofScience

1. The problem statement, all variables and given/known data
I'm having trouble with these very strange (and deceptive-looking) systems of linear equations that I would like to know more about. Here they are:

(1) A+C=0
B-2C=4
-A+B+C=0

(2) 3x+y= 1
x-3y=17

3. The attempt at a solution

(1) Using algebra the solutions to the first are C=-1, B=2, A=1. When I did the Gauss method it failed. I've tried thinking of ways to make it work but it just won't come to me.

(2) Algebra gives x=2 and y=-5

This is how my Gauss method turned out:

a) 1/3 R1
b) R1 -1/9 R2 =R1 giving 8/9, 0, -14/9
c) 9/8R1 giving 1, 0, -7/4
d) -1/3R2 giving -1/3, 1, -17/3
e) R2 +1/3R1

but I get -7/4 and -25/4 (???)

2. Sep 8, 2008

### Defennder

1. Gaussian elimination should work just fine. You need to post your work here so that we'll know your mistakes (though it can be tedious to type out all your steps).

Here is your mistake. The 2nd row, 2nd entry is -3. You should multiply by 1/9 and not -1/9 to reduce the above row to zero.

3. Sep 8, 2008

### The_ArtofScience

Thanks Defender for tracing back to my steps. I got the soln

I couldn't get the first one to work. So here's the work

(1) R3+R1 = R3 giving 0, 1, 2
(2) 1/2 R3
(3) R1 -R3 =R1 getting 1, -1/2, 0

this is where I got stuck, can't figure out what to do because I keep going back and forth over and over again...

4. Sep 8, 2008

### Defennder

You need to understand how to perform the algorithm in a fashion which terminates meaningfully, because it seems you're performing row reductions without having any goal in mind.

(1.) Looks fine to me. What should do next is to reduce the 3rd row using the second row. Remember that you want to add a multiple of the 2nd row to the third row to make sure that the 2nd entry of the third row becomes zero (remember that the purpose of Gaussian-jordan elimination is to express every column as a pivot column if possible) so that the second row would be the only row which has a non-zero 2nd entry.

Once this is done, use row-reduction again to make sure that the third row has only one entry and this entry is at the last column. Observe that the end result is a matrix with 1's arranged in a staircase pattern. That is the purpose of Gauss Jordan elimination. Swap rows if necessary to achieve that pattern to get the end result.

5. Sep 8, 2008

### The_ArtofScience

Ok the answer seems very clear to me. Your help is amazing!

6. Sep 8, 2008

### Defennder

No prob, one last thing I should add:
It may not always be possible to reduce all the matrices you encounter into one in which every column is a pivot column. In this case, you should note that there are an infinite number of solutions to the system of equations, which you can express explicitly by reading them off the reduced-row echlon form and assigning each variable an arbitrary parameter like s or t.