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Homework Help: Gauss-jordan 3 equation matrix help

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    the original problem is in matrix form:

    1 1 2 8
    -1 -2 -3 1
    3 -7 4 10

    2. Relevant equations



    3. The attempt at a solution

    i got all the way to :

    1 1 0 -4
    0 1 0 1
    0 0 1 2

    however, the top row is supposed to = 3 but when i subtract row 2 from row 1 i get -5....i can't find where i messed up in the steps i did it. anyone get the same thing as me or know what i did wrong?
     
  2. jcsd
  3. Sep 14, 2012 #2

    Mark44

    Staff: Mentor

    I don't know what you did to get the matrix above, but it's incorrect. This says that y = 1 and z = 2. From these you can work back to get x.

    In any case, these values are incorrect. I've worked through the row reduction, and have a solution, which I have checked, so I know it's right.

    Can you show what work you did to get the above? From that, we can tell you where you went astray.
     
  4. Sep 15, 2012 #3
    that's weird because the book answer says that x = 3, y = 1, and z = 2, which is why i thought i only got the top row part wrong. the operations i used were:

    row 2 * -1 minus row 1

    row 3 minus 3 * row 1

    10 * row 2 + row 3

    row 3 / -52

    5 * row 3 plus row 2

    2 * row 3 minus row 1

    row 2 minus row 1
     
  5. Sep 15, 2012 #4

    Mark44

    Staff: Mentor

    That is not a solution to the problem you posted.
    The augmented matrix you posted is
    1 1 2 8
    -1 -2 -3 1
    3 -7 4 10

    The book solution works in the 1st and 3rd equations, but it doesn't work in the 2nd equation.

    -1(3) -2(1) -3(2) = -3 -2 -6 = -11 ≠ 1

    There are a couple of possibilities. It might be that you mistyped the original matrix. If the 2nd row was actually -1 -2 -3 11, then the book's answer is correct.


    Otherwise, the book's answer is wrong.
     
  6. Sep 15, 2012 #5

    You are correct, i accidentally put a -3 in the second row when it should have been a 3. My apologies to everyone. Also, is it a valid row operation to divide a row by itself?
     
    Last edited: Sep 15, 2012
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