Expert: Alon Mandes - 8/23/2008

Question
Hi!! I'm trying to solve this system of equations using the Gaussian elimination method. can you help me?

2x-3y+2z=2
x+4y-z=9
-3x+y-5z=5

following instructions for Gaussian, I came up with

2    -3    2    2
1     4   -1    9
-3    1   -5    5

I then switched the first and second row:

1    4    -1    9
2    -3    2    2
-3   1    -5    5

then multiplied the first row by (-2) to get a zero in first position of the second row, and then multiplied the first row by 3 to get a zero in the first position third row

1    4    -1    9
0   -11    4   -16
0    13   -8    32

and now I'm stuck. does this look right to you, or did I do something wrong?

Thanks!!

Answer
Ok, let's call the 1st row R1, the 2nd R2 & the 3rd R3.
Now we have the matrix :  
1    4    -1    9
0   -11    4   -16
0    13   -8    32
We need to make another zero in R3. In order to do that we perform :
13R2+11R3-->R3. That mean we will switch the row R3 by  13R2+11R3.
Let's do it : 13R2 = 0 -143 52 -206 , 11R3 = 0 143 -88 352
13R2+11R3 = 0 0 -36 146. Hence, our Matrix is :
1   4   -1     9
0  -11   4    -16
0   0   -36   146
The solution will be :
z=146/(-36)=1.27
-11y+4*146/(-36)=-16 --> y=1.91
x+4*1.91-1.27=9 -- > x=-6.4

Alon.