# Gaussian reducation to solve

1. Oct 12, 2008

### kash-k

3x+2y-z=5
-2x+2y+5z=2
x-2y+3z=11

Now we are required to solve these linear equations via Gaussian method.

I put these in an echelon form and this how I did it:

Switch L1 with L3 to make life easier.

L1: x-2y+3z=11
L2:-2x+2y+5z=2
L3:3x+2y-z=5

Then I went to work to remove eliminate X from L2 and L3.

L2 + 2L1
L3 + (-3L1)

Which got me:

L2: 0 -2 11 23

L3: 0 8 -8 -28

Then I removed the Y from L3.

L3+(4L2)

However, I got a stupid number which didn't follow the law:

The bottom row had to be : 0 0 1 X

So if anyone can help me solve this I'd be much obliged. Thank you

2. Oct 12, 2008

### HallsofIvy

Staff Emeritus
What do you MEAN by "a stupid number". Following what you did I get, as the last line,
0 0 34 20.

Now divide that line (equation) by 34 to get 0 0 1 10/17.

Doesn't look stupid to me.

3. Oct 12, 2008

### kash-k

20 is wrong
way wrong

4. Oct 12, 2008

### HallsofIvy

Staff Emeritus
You are right. I just copied your error. L3- 3L1 gives 3-3(1)= 0, 2- 3(-2)= 8, -1-3(3)= -10, 5- 3(11)= -28. That last line should be
0 8 -10 -28.

Now L3+ 4L2 is 8+ 4(-2)= 0, -10+ 4(11)= 34, -28+ 4(24)= 68.[/quote]
Looks like 0 0 34 68

Now divide the last row by 34. By the way, which "law" are you referring to? No matter what numbers you get in the last two places, you can always divide by the "z" coefficient to get "0 0 1 X".

5. Oct 12, 2008

### kash-k

I guess at the end of the day it was a calculating mistake on my part.

And for the law I misunderstood. "The leading coefficient of each nonzero row is one"

Thank you for the help.

Regards

ps. so the last row is 0 0 1 2