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Determinant formula with einstein notation proof

  1. Sep 4, 2010 #1
    Hello, Im supposed to prove that the determinant of a second order tensor (a matrix) is equal to the following:

    det[A] = [tex] \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{pi} A_{qj} A_{rk} [/tex]

    anyone have any idea how i would go about this? any method is welcome

    where the determinant of the matrix A is expressed below:

    det[A] =
    [tex] A_{11}(A_{23}A_{32}-A_{22}A_{33}) + A_{12}(A_{21}A_{33}-A_{23}A_{31}) + A_{13}(A_{22}A_{31}-A_{21}A_{32}) [/tex]
  2. jcsd
  3. Sep 5, 2010 #2


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    Science Advisor

    Assuming that last formula is your definition of the determinant, then the obvious way to do this is to write out the actual sum implied by the first formula and show that the two formulas are the same thing.
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