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Finding the cofactors

  1. Nov 11, 2009 #1
    Find the cofactors of x,y,z, in the determinate:

    | 1 1 1 |
    | 2 3 4 |
    | x y z |

    I can make 3 equations from this by taking the determinate from each line
    z - 2y + x
    x - 2y + z
    z - 6y - x

    but how do I solve these? Have I done the right thing? I know they all equal each other

    Thanks
    Tom?
     
  2. jcsd
  3. Nov 11, 2009 #2

    lanedance

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    you can take the determinant starting at any column or row, as lonag as you;re careful with negatives

    i think the quetsion is asking you to take the determeinant using the bottom row to start off, then it will be something like

    det = +-(x(det(2x2)) -y(det(2x2))+ zdet(2x2)
     
  4. Nov 11, 2009 #3

    lanedance

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    the cofactors for x,y,z are the correspsonding 2x2 determinants
     
  5. Nov 11, 2009 #4

    lanedance

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    by the way how do you get these?
     
  6. Nov 12, 2009 #5

    HallsofIvy

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    First, those are not even equations. Second, why? The problem doesn't say anything about "solving", it only asks for cofactors. Do you know what cofactors are?
     
  7. Nov 12, 2009 #6
    + - +
    - + -
    + - +

    for a 3x3.

    The cofactor is taken from the matrix above and placed infront of the orignal matrix element.

    in a
    1 2 3
    4 5 6
    7 8 9

    the cofactors of row 2 are
    -4, +5, -6

    is that correct?

    EDIT:
    I got those equations by taking the determinants of each row. I know that they all equal each other, but I', not sure if that helps!?
    You're right HallsOfIvy - those arn't equations, they don't equal anything (except each other). they're just expressions! Silly me!
    Thanks
    Tom
     
    Last edited: Nov 12, 2009
  8. Nov 12, 2009 #7

    lanedance

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    have a look here
    http://en.wikipedia.org/wiki/Cofactor_(linear_algebra [Broken])

    the cofactor is specific to an element this is what i get

    so take the first element of the 2nd row (4) this is row i = 2, column j = 1

    the cofactor Cij of 4, is
    [tex] C_{12} = (-1)^{1+2}\left| \begin{matrix} 2&3\\ 8&9 \end{matrix} \right|
    = (-1)(2.9 - 3.8) = -(18 - 24) = 6
    [/tex]
     
    Last edited by a moderator: May 4, 2017
  9. Nov 12, 2009 #8
    ahh once again I've miss understood the workbook. Cofactors are minors with a sign attached, not just the sign!!!

    [tex]A_{31} = 1
    A_{32} = -2
    A_{33} = 1
    [/tex]

    Thank you very much!
     
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