Linear algebra - gaussian elimination

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The_ArtofScience
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Homework Statement



Use Gauss-Jordan elimination to solve this sys of linear eqs

2w+3x-y+4z =0
3w-x+z =1
3w-4x+y-z=2



The Attempt at a Solution



I wrote 9 tablabeaus and ended up with three arbitrary solns. I just want to know if there anything wrong, here they are: 225a/121 = 39/605 550b/121 =-488/605, 148c/11=-126/55.

If there is a unique soln please let me know. Sorry I can't type everything out, don't have time. Gotta rush to/
 
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If you don't have time to show us steps, we don't have time to check your answers! :)))
 
rocomath said:
If you don't have time to show us steps, we don't have time to check your answers! :)))


2 3 -1 4 0 1/2R1=R1, R2-3R1=R2
3 -1 0 1 1
3 -4 1 -1 2

1, 3/2, -1/2, 2, 0 R3-3R1=R3
0, -11/2, 3/2, -5, 1
3, -4, 1, -1

1, 3/2, -1/2, 2, 0 2/5R3=R3, 2/-11R2
0, -11/2, 3/2, 5, 1
0, -17/2, 5/2, -5, 2

1, 3/2 -1/2, 2, 0 R3+ 17/5R2= R3
0, 1, -3/11, 10/11, 2/-11
0, -17/5, 1, -2, 4/5

1, 3/2, -1/2, 2, 0
0, 1, -3/11, 10/11, -2/11
0, 0,

...Ok I took a long look at my paper. I realize I made one mistake but I still can't figure out how to get back at 1 after that 0. The eq R3+ 17/5R2= R3 doesn't really work out. Its a bit messy to work back. Is there a better method?
 
1, 3/2, -1/2, 2, 0 2/5R3=R3, 2/-11R2
0, -11/2, 3/2, 5, 1
0, -17/2, 5/2, -5, 2

You made a mistake here, this should be -7.
 
I worked it out again and still got 3 different arbitrary solns. Can a sys of equations have varying or infinite number of arbitrary solns?

My method is as follows:

(1) 1/2R1
(2) -2/11R2
(3) R3+17/5R2=R3
(4) R1+ 55/44R3=R1
(5) R2 +15/4R3=R2
(6) 55/4R3
(7) R1- 2/15R3 =R1

Solving I get 2w=1/2 5x=31/11 15y=11 and no z. I just hope I never see this again in cal 1 :-/
 
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