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Gaussian Elimination, is this method OK?

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A=
    <1, -3, -1, 1>
    <2, -5, 0, 1>
    <-3, 5, -6, 3>


    What I did was Row2 = 2*Row1 - Row2 which renders Row2 as: <0, -1, -2, 1>

    However in the solutions, Row2 was given as: <0, 1, 2, -1>, which appears to be R2 = -2R1 + R2

    I'm guessing it makes no real difference, we were asked to calculate the basis of the kernel(A), to which I got <-8,-3,2,1>[itex]^{T}[/itex] and the solutions gave <8,3,-2,-1>[itex]^{T}[/itex], probably because of the difference in row-reducing the second row.

    Am I incorrect?
     
  2. jcsd
  3. Oct 11, 2011 #2

    Mark44

    Staff: Mentor

    No, your work is fine.
    I prefer the method used in in the solutions for this problem, though, with R2 being replaced by -2R1 + R2.

    The basis vector you got is a multiple of the one shown in the solutions. Both vectors span exactly the same space (a one-dimension subspace of R4).
     
  4. Oct 11, 2011 #3
    Ah yes, that makes sense, it can be gotten by just multiplying by -1. Thanks for pointing that outt.
     
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