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Cofactor (4x4 matrix)

  1. Oct 19, 2009 #1
    For the matrix A given below, find C44 , where Cij is the i,j cofactor of A.

    A=

    | -5 +4 +5 -3 |
    | +2 -1 +5 +4 |
    | -5 -1 +2 -3 |
    | +4 -2 +4 -4 |

    So this is how I answer this question:

    = (4,4) cofactor of A

    =
    | -5 +4 +5 |
    | +2 -1 +5 | * (+)
    | -5 -1 +2 |

    =
    | -5 +4 +5 |
    | +2 -1 +5 | * (+)
    | +5 +1 -2 |
    R3*(-1)

    =
    | +0 +5 +3 |
    | +2 -1 +5 |
    | +5 +1 -2 |
    Add R1 and R3

    =
    | -1 +5 | *(+)(+)
    | +1 -2 |
    (1,1) cofactor of A

    =(+)(+)[(-1*-2)-(1*5)]

    =(+)(+)(2-5)

    =-3

    Please help me on this, is C44 = -3?
     
  2. jcsd
  3. Oct 19, 2009 #2

    tiny-tim

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    Hi shiri! :smile:
    No, that's the cofactor of the +0, and you get the determinant by multiplying +0 times its cofactor (and then adding the same for +5 and +3).

    If you're determined to save effort by getting down to a 2x2 determinant, you need another 0.

    I'd have started differently, and used one of the original -1s to get rid of the other -1 and the 4. :wink:
     
  4. Oct 19, 2009 #3
    Hi

    … =
    | +0 +5 +3 |
    | +2 -1 +5 |*(+)
    | +5 +1 -2 |
    Add R1 and R3

    =
    | +0 +5 +3 |
    | +2 -1 +5 |*(+)
    | +0 +7 -29 |
    2R3 - 5R2

    =
    | +5 +3 | *(+)(-)
    | +7 -29 |
    (2,1) cofactor of A

    =(+)(-)[(+5*-29)-(+7*+3)]

    =(+)(-)[(-145)-(21)]

    C44 = 166

    So after reading your message, is this a correct answer for C44?
     
  5. Oct 19, 2009 #4

    tiny-tim

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    hmm … you've made two mistakes, and still got the right answer! :rolleyes: :biggrin:

    i] you can add as many copies as you like of one row to another row,

    but you can't add copies of a row to itself … if you do that, you multiply the determinant (in this case by 2, when you added R3 - 5R2 to the original R3)

    ii] you should have multiplied the cofactor, 166, by the 2 that it's the cofactor of.

    Try again. :smile:
     
  6. Oct 19, 2009 #5
    So the C44 = 332?
     
  7. Oct 19, 2009 #6

    tiny-tim

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    No!!

    I said 166 was right …

    you made two mistakes, one doubled the answer, and the other halved it again.

    Do you understand why your 2R3 - 5R2 was wrong?​
     
  8. Oct 19, 2009 #7
    Strange. When I typed 166 in the online assignment. It came out wrong. Do you think the answer was suppose to be -166, not 166?
     
  9. Oct 20, 2009 #8

    tiny-tim

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    (just got up :zzz: …)
    oops! yes, I didn't notice that you made a third mistake (I thought that the two matrices were the same :rolleyes:) …
    when you multiply a row by minus one, you multiply the whole determinant by minus one!!

    As I said before, you can only add multiples of one row to another row! :smile:

    Now try the easier way I suggested before: use one of the original -1s to get rid of the other -1 and the 4. :wink:
     
  10. Oct 20, 2009 #9
    So it becomes like this:

    | -5 +4 +5 |
    | +2 -1 +5 | * (+)
    | -5 -1 +2 |

    =
    | -5 +5 | * (+) (+)
    | -5 +2 |

    =(+)(+)[(-5*+2)-(-5*+5)]

    =(+)(+)[-10-(-25)]

    =(+)(+)[-10+25]

    =(+)(+)[15]

    So is that means, C44 = 15?
     
  11. Oct 20, 2009 #10

    tiny-tim

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    That's bizarre! :redface:

    All you've done is calculate the cofactor of (an arbitrary) -1.

    Add multiples of the 2nd row to the 1st and 3rd rows so as to turn the -1 and the 4 into 0s …

    then!!! you can use the cofactor of the remaining -1 ! :smile:
     
  12. Oct 20, 2009 #11
    OK. I hope I know what I'm doing.

    | -5 +4 +5 |
    | +2 -1 +5 | * (+)
    | -5 -1 +2 |

    =
    | -5 +4 +5 |
    | +2 -1 +5 | * (+)
    | +7 0 +3 |
    R3(-1) + R2

    =
    | +3 +0 +25 |
    | +2 -1 +5 | * (+)
    | +7 +0 +3 |
    R1 + 4R2

    =
    | +3 +25 | *(+)(+)
    | +7 +3 |

    =(+)(+)[(+3*+3)-(+7*+25)

    =(+)(+)[+9-175]

    =-166

    C44 = -166

    Am I doing this right, tiny-tim?
     
  13. Oct 20, 2009 #12

    tiny-tim

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    Yes!! :smile:

    Do you see why this works? …

    the determinant is the sum of each element in one row (or one column) multiplied by its cofactor …

    so you made a row that was all 0s except for one element …

    so the determinant is that element multiplied by its cofactor!

    (and you chose -1 because keeping the numbers small, and preferably ±1, makes the arithmetic a lot easier than eg 2 and 5 :wink:)
     
  14. Oct 20, 2009 #13
    Great. Thank You! You're very helpful.

    Oh, one more thing. Can you take a look at this question [https://www.physicsforums.com/showthread.php?t=346994 Matrix Determinant]

    I just want to know whether I answer this question correctly. Thanks
     
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