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Math Help - Cofactor of a Determinant

  1. #1
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    Cofactor of a Determinant

    A 4th order determinant (aij) is given by

    e(ijkl) a1i a2j a3k a4l

    and the summation convention over repeated indices. e(ijkl) is the general kronecker delta (for ex e(1234)=1, e(2134)=-1, etc)

    The a23 term in the expansion is

    e(i3kl) a1i a3k a4l,

    which is the cofactor of a23 by def. i,k,l is the column subscript and varies over 1,2,4 so that cofactor is the minor except for sign. How do you extract the sign from above formula? The answer is (-1)^(2+3), but how do you get it?
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Hartlw View Post
    A 4th order determinant (aij) is given by

    e(ijkl) a1i a2j a3k a4l

    and the summation convention over repeated indices. e(ijkl) is the general kronecker delta (for ex e(1234)=1, e(2134)=-1, etc)

    The a23 term in the expansion is

    e(i3kl) a1i a3k a4l,

    which is the cofactor of a23 by def. i,k,l is the column subscript and varies over 1,2,4 so that cofactor is the minor except for sign. How do you extract the sign from above formula? The answer is (-1)^(2+3), but how do you get it?
    that is just sum of the indexes (row and column) from the term by which you do expansion ...

     a_{11} \Rightarrow (-1) ^{(1+1)} ....

     a_{35} \Rightarrow (-1) ^{(3+5)} ....

     a_{ij} \Rightarrow (-1) ^{(i+j)} ....


    like

    \displaystyle D= \begin{vmatrix}<br />
 a_{11} & a_{12}   &  a_{13}  & . . .  & a_{1n}  \\ <br />
  a_{21} & a_{22}   &  a_{23}  & . . .  & a_{2n}  \\ <br />
. . . & . . . & . . .  & . . .   &. . .  \\ <br />
. . . & . . . & . . .  & . . .   &. . .  \\ <br />
a_{n1} & a_{n2}   &  a_{n3}  & . . .  & a_{nn}<br />
\end{vmatrix} = (-1)^{(1+1)} a_{11}A_{11} +(-1)^{(1+2)}a_{12}A_{12}+ .... + (-1)^{(1+n)}a_{1n}A_{1n} = \sum_{k=1} ^n (-1)^{1+k} a_{1k}A_{1k}

    where

     A_{11} = \begin{vmatrix}<br />
 a_{22}   &  a_{23}  & . . .  & a_{2n}  \\ <br />
 . . . & . . .  & . . .   &. . .  \\ <br />
 . . . & . . .  & . . .   &. . .  \\ <br />
a_{n2}   &  a_{n3}  & . . .  & a_{nn}<br />
\end{vmatrix}

    and so on
    Last edited by yeKciM; September 3rd 2010 at 08:25 AM.
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  3. #3
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    The sign can be extracted as follows:

    Let D=e(i,j,k,l) (a1i, a2j, a3k, a4l)

    Then define cofactor ARS of aRS by multiplicand of ARS
    Then A23 = e(i,3,k,l)(a1i, a3k, a4k) where the 3 is two places over in the permutation symbol.

    Properties of the permutation symbol:

    consider e(i,j,k...S..l) where S is R places to the right. Then
    e(i,j,k,...S..l) = (-1)^R (e(S,i,j,k...l)

    What is the sign of e(S,i,j,k,...l)? Consider (S,i,j,k)

    e(1,i,j,k) = +1 when i,j,k = 2,3,4
    e(2,i,j,k) = -1 when i,j,k = 1,3,4
    e(3,i,j,k) = +1 when i,j,k = 1,2,4
    e(4,i,j,k) = -1 when i,j,k = 1,2,3

    In general e(S,i,j,k..l) = (-1)^S e(i,j,k...l)

    Then ARS = (-1)^(R+S) DRS, where DRS is the determinant of what"s left after crossing out the Rth row and Sth column

    You need some familiarity with the permutation symbol, such as the first few pgs of "Mirsky, Intro to Linear Algebra)" which are a bear to follow. The principles are easy but following them in terms of general subscripted variables is a nightmare, for me anyhow. Mirsky simply defines the cofactor as (-1)^(R+S)DRS and then shows arsArs = D, sum over r,s.
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