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Linear Algebra - System of 2 Equations with 3 Variables-possible?

  1. Sep 7, 2007 #1
    Linear Algebra - System of 2 Equations with 3 Variables--possible?

    1. The problem statement, all variables and given/known data
    Solve: x1-3x2+4x3=-4
    3x1-7x2+7x3=-8
    -4x1+6x2-x3=7


    3. The attempt at a solution
    I was able to make it to:
    1 -3 4 -4
    0 -10 25 -11
    0 0 0 0
    So the third row goes away, and I am left with:
    1 -3 4 -4
    0 -10 25 -11
    I am pretty sure that cannot be solved, or am I overlooking something? Thank you in advance!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 7, 2007 #2
    It looks like you have a free variable. So you can solve for the other two in terms of it.
     
  4. Sep 7, 2007 #3

    NateTG

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    The solution isn't unique, you can try plugging in a value for [itex]x_1[/itex] and solving the rest.
    [tex]x_1=1,2,3,4,5,6....\pi, e,...[/tex]
     
  5. Sep 8, 2007 #4
    whenever you have more unknowns than equations, you get infinitely many solutions and one or more variables become free varaibles
     
  6. Sep 10, 2007 #5

    NateTG

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    Sometimes there can still be zero solutions:
    [tex]x+y+z=0[/tex]
    [tex]x+y+z=1[/tex]
     
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