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Determinant of Matrices

  1. Dec 2, 2007 #1
    [SOLVED] Determinant of Matrices

    Find the det. of the matrix:

    1 3 -2 2
    3 0 -1 4
    1 -3 4 2
    2 3 -3 3

    Using gaussian elimination to get it into upper triangular form (which I think is easiest) is where im struggling. I get as far as:

    1 3 -2 2
    0 -9 5 -2
    0 -6 6 0
    0 -3 0 -1

    Now im trying to get everythingbelow the diagonal all zeros, next i was going to use row3 and add on 2 lots of row1 to get a 0 for the -6, but it also removes one of my zero values and replaces it as a 2.

    I guess the method I am using is wrong to get it into upper traingular form, can anyone tell me where I am going wrong? Perhaps I should be using column operations instead.
     
  2. jcsd
  3. Dec 2, 2007 #2
    Actually perhaps I'm being stupid and doing this completely wrong, I think I can use cofactors instead, I don't know where I got the other method from, I guess it was for something else to do with matrices :)
     
  4. Dec 2, 2007 #3
    -12 is my answer if that looks correct.
     
  5. Dec 2, 2007 #4

    CompuChip

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    That's not going to work (as you already found out), instead add multiples of the second row to the third and fourth one (e.g. add the second to the first -2/3 times)

    By the way, the answer of -12 is not correct.
     
  6. Dec 2, 2007 #5
    awww, was the method of cofactors bad? :P cos I only did it with 3x3 matirces in practise.
     
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6

    CompuChip

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    Just looked up "cofactor" on Wikipedia -- obviously I know the method but I never heard that name :smile:
    That method will also work for any matrix (though the larger the matrices get, the more work it will be), so you probably just made a calculation error.
    Can you post a calculation?
     
  8. Dec 2, 2007 #7
    actually I just got the answer using my original method as -12 again, are you sure it's not correct?
     
  9. Dec 2, 2007 #8
    my end matrix for the original method is:

    1 3 -2 2
    0 -9 5 -2
    0 0 8/3 4/3
    0 0 0 0.5

    and 1*-9*(8/3)*0.5 is -12 :P
     
  10. Dec 2, 2007 #9

    cristo

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    Maple tells me the answer's zero. We cant really help you unless you show every row operation.
     
  11. Dec 2, 2007 #10
    Well in my original method my row operations were:

    R2 - 3R1 and R3 - R1 and R4 - 2R1

    to give

    1 3 -2 2
    0 -9 5 -2
    0 -6 6 0
    0 -3 0 -1

    then, R3 - (2/3)R2 and R4 - (1/3)R2

    to give

    1 3 -2 2
    0 -9 5 -2
    0 0 (8/3) (4/3)
    0 0 (-5/3) (-1/3)

    then finally R4 + (5/8)R3

    to give

    1 3 -2 2
    0 -9 5 -2
    0 0 8/3 4/3
    0 0 0 0.5
    :)
     
  12. Dec 3, 2007 #11
    It all seems ok to me, considering how I got the same answer for both methods, unless the methods were wrong.
     
  13. Dec 3, 2007 #12

    cristo

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    The last row is wrong; that is, check "R4-2R1."
     
  14. Dec 3, 2007 #13
    Ah dammit.

    Still weird I ended up with the same answer for both :P
     
  15. Dec 3, 2007 #14
    Nono, w8 w8, the start matrix i wrote down here wasn't correct :P it should have been:

    1 3 -2 2
    3 0 -1 4
    1 -3 4 2
    2 3 -4 3

    =P I hope I'm correct now, sorry lol, guess i was typing out the numbers too quick
     
  16. Dec 3, 2007 #15

    CompuChip

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    That has determinant -12 indeed, so you were correct after all :smile:
    It's often in the little steps
     
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