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Prove that the third invariant is equal to the determinant

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    This is all in summation notation.
    Given a 3x3 matrix [tex]A_{ij}[/tex], show that [tex]det[A]=1/6(A_{ii}A_{jj}A_{kk}+2A_{ij}A_{jk}A_{ki}-3A_{ij}A_{ji}A_{kk})[/tex]

    2. Relevant equations
    I've been told that we're supposed to begin with
    [tex]det[A]=1/6\epsilon_{ijk}\epsilon_{pqr}A_{ip}A_{jq}A_{kr}[/tex]

    3. The attempt at a solution
    I hate to say this, but I have no idea where to start. I would really appreciate if somebody could just give me the first step and push me in the right direction.
     
  2. jcsd
  3. Sep 9, 2009 #2

    Hurkyl

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    Definitions are often a good place to start.
     
  4. Sep 9, 2009 #3
    Never mind, I got it. You set [tex]A_{ij} = \delta_{ij}[/tex] and get the epsilons in terms of the deltas.
     
  5. Sep 9, 2009 #4

    Hurkyl

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    :confused: Why can you do that?
     
  6. Sep 9, 2009 #5
    Well, that's not exactly right. But you can work a little index magic and come up with
    [tex]\epsilon_{ijk}\epsilon_{pqr}=\left| \begin{matrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \end{matrix} \right|[/tex]
     
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