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Homework Help: Determinant + indicial 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
     
  2. jcsd
  3. Sep 4, 2010 #2

    hunt_mat

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    One thing which I would do is look at the 3X3 case amd see if you can see where the formula comes from. You also need to tell us the definition of the determinant that you're using.
     
  4. Sep 4, 2010 #3
    well by det[A] i mean the following:

    det[[tex]A_{ij}[/tex]] = the determinant of a 3 by 3 matrix with the first row being [tex]A_{11}, A_{12}, A_{13} [/tex] and the second row would be [tex] A_{21}, A_{22}, A_{23} [/tex] and lastly the third row is [tex] A_{31}, A_{32}, A_{33} [/tex]
     
  5. Sep 4, 2010 #4

    hunt_mat

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    How would you write that as a sum?
     
  6. Sep 4, 2010 #5
    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]
     
  7. Sep 5, 2010 #6

    hunt_mat

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    I am assuming that you're aware of the notation, how can you write the above sum as a sum which includes [tex]\varepsilon_{ijk}[/tex]. A good start would be to write out one of the sums, say for example:
    [tex]
    \varepsilon_{ijk}A_{1i}A_{1j}A_{3k}
    [/tex]
    and see how this compares to your sum.
     
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