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Tensor analysis

  1. Sep 4, 2012 #1
    ε1. The problem statement, all variables and given/known data
    show that the determinant of a matrix A can be calculated as followings:

    det[A]= 1/6 (A_ii A_jj A_kk + 2 A_ij A_jk A_ki - 3 A_ij A_ji A_kk


    2. Relevant equations



    3. The attempt at a solution

    use ε_pqr det[A]= ε_ijk A_ip A_jq A_kr

    ε_pqr ε_pqr det[A]= ε_ijk ε_pqr A_ip A_jq A_kr

    use ε_pqr ε_pqr = 6

    det[A]= 1/6 ( ε_ijk ε_pqr A_ip A_jq A_kr)

    and ε_ijk ε_pqr = det | δ_ip δ_iq δ_ir |
    | δ_jp δ_jq δ_jr |
    | δ_kp δ_kq δ_kr |

    from here i dont know what to do
    and δ - delta and ε - epsilon
     
  2. jcsd
  3. Sep 4, 2012 #2

    vela

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    So far so good. Just expand the determinant out now.
    $$\begin{vmatrix}
    \delta_{ip} & \delta_{iq} & \delta_{ir} \\
    \delta_{jp} & \delta_{jq} & \delta_{jr} \\
    \delta_{kp} & \delta_{kq} & \delta_{kr}
    \end{vmatrix} = \delta_{ip}\delta_{jq}\delta_{kp} + \cdots$$ Then plug the resulting expression into
    $$\det(A) = \frac{1}{6}(\varepsilon_{ijk}\varepsilon_{pqr} A_{ip}A_{jq}A_{kr})$$ The Kronecker deltas will allow you to do some of the summations.
     
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