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Linear algebra - gaussian elimination

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data

    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



    3. 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/
     
  2. jcsd
  3. Sep 2, 2008 #2
    If you don't have time to show us steps, we don't have time to check your answers! :)))
     
  4. Sep 2, 2008 #3

    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?
     
  5. Sep 2, 2008 #4
    You made a mistake here, this should be -7.
     
  6. Sep 3, 2008 #5
    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 :-/
     
    Last edited: Sep 4, 2008
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