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Prove Determinant Using the Triple Scalar Product

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to prove [itex]det A = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{kr} [/itex] using the triple scalar product.

    2. Relevant equations
    [itex] \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{ kr} [/itex]
    [itex] (\vec u \times \vec v) \cdot \vec w = u_i v_j w_k \epsilon_{ijk} [/itex]

    3. The attempt at a solution
    I have written out and understand that [itex]det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} [/itex] but I don't understand where the triple scalar product comes into play.

    I see the similarity between the shorthand notations of the triple scalar product and the determinant but don't see how I can relate the two. I figure I must be missing some geometric/mathematical interpretation that can help me.

    I've used Einstein summation notation in a couple previous problems but this is really the first involved problem I have to do and am a bit lost as to what to do.
     
  2. jcsd
  3. Aug 30, 2015 #2
    Both the scalar triple product and the determinant (in 3 dimensions) have the geometrical interpretation of volume. ##(\vec u \times \vec v) \cdot \vec w## is the volume of the prism formed by vectors u, v, and w. They are probably expecting you to make the identification

    $$ \det A \quad = (\vec{A}_1 \times \vec{A}_2) \cdot \vec{A}_3 ,$$

    where, for example, ##\vec{A}_1## is the first row in matrix A. The determinant of A is the volume spanned by its three row vectors.

    That gets you up to the point where you are, namely ##\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k}##. To get the rest of it, note that

    $$\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} = -\epsilon_{ijk} A_{2i} A_{1j} A_{3k} = \epsilon_{213}\epsilon_{ijk} A_{2i} A_{1j} A_{3k}$$

    for example.
     
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