- #1
kash-k
- 17
- 0
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
-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