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Homework Help: Which are the pivot columns? - Linear Algebra

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Given matrix

    A=

    1 1 -2 1 3
    2 -1 2 2 6
    3 2 -4 -3 -9

    x=

    x1
    x2
    x3
    x4
    x5

    b =

    1
    2
    3

    1. Solve the system by elimination (use the augmented matrix) until pivots are found (no backward elimination here).

    2. Show that pivots are respectively 1 and -3, and indicate clearly which are the pivot columns and free columns.


    2. Relevant equations

    Matrix A and vector b



    3. The attempt at a solution

    I perform elimination on augmented matrix to solve part 1.

    1 1 -2 1 3 | 1
    0 -3 6 0 0 | 0
    0 -1 2 -6 -18 | 0


    1 1 -2 1 3 | 1
    0 -3 6 0 0 | 0
    0 0 0 -6 -18 | 0

    What I don't understand is part 2.

    As far as I know, for an entry to become a pivot it must be the first non-zero entry on the row and there must be no non-zero entries under it.

    1 1 -2 1 3 | 1
    0 -3 6 0 0 | 0
    0 0 0 -6 -18 | 0

    If that rule applies I would find three pivots instead of two. (1,-3, -6) I know the right answer for this exercise is that there are only two pivots (1, -3), resulting col1 & col2 to be the pivot columns and the rest are free columns.

    Can someone explain why -6 is not a pivot in this case?
     
  2. jcsd
  3. Oct 18, 2011 #2

    Mark44

    Staff: Mentor

    I don't see an error on your part. Are you sure you have copied the matrix correctly?
     
  4. Oct 18, 2011 #3
    as far as I could tell, there should be 3 pivots in your reduced matrix.

    I suspect that you wrote the matrix down wrong, because if you make the last row
    -3 3 -4 -3 -9 instead of 3 3 -4 -3 -9, then there should be only two pivots.
     
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