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Determinant of 3x3 matrix equal to scalar triple product?

  1. Sep 3, 2015 #1
    The determinant of a 3x3 matrix can be interpreted as the volume of a parallellepiped made up by the column vectors (well, could also be the row vectors but here I am using the columns), which is also the scalar triple product.
    I want to show that:
    ##det A \overset{!}{=} a_1 \cdot (a_2 \times a_3 ) ##
    with ## A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} ##

    I started to develop det A like this:
    ## det A = a_{11} (a_{22}a_{33}-a_{23}a_{32})-a_{12} (a_{21}a_{33}-a_{23}a_{31})+a_{13} (a_{21}a_{32}-a_{22}a_{31}) ##

    With ## e_i ## as unit vectors, this should equal:
    ## det A = a_1e_1\cdot (a_2 \times a_3)e_1-a_2e_1\cdot (a_1 \times a_3)e_1+a_3e_1\cdot (a_1 \times a_2)e_1 = a_1\cdot (a_2 \times a_3)-a_2\cdot (a_1 \times a_3)+a_3\cdot (a_1 \times a_2) ##

    With the rules for cross products I get:
    ## det A = a_1\cdot (a_2 \times a_3)+a_1\cdot (a_2 \times a_3)+a_1\cdot (a_2 \times a_3) = 3 \cdot a_1 (a_2 \times a_3) ##

    Shouldn't I get only ## a_1 (a_2 \times a_3) ##? Not times three. Please, can someone say what I am doing wrong? Or is it right? In that case, how should I interpret it?
     
  2. jcsd
  3. Sep 3, 2015 #2

    Orodruin

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    Your notation here is not very clear and likely the reason you are not getting the correct result, what do you mean by ##a_1 e_1 \cdot (a_2\times a_3)e_1##? The vector structure here makes no sense.
     
  4. Sep 3, 2015 #3
    Thank you for answering!
    My thought was, that to get ##a_{11}## I could do the multiplication
    ## a_1 \cdot e_1 = \begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = a_{11} ##

    And for the first row in the cross product I do the same thing:
    ## (a_2\times a_3) \cdot e_1= \begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix} \times \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix}\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = a_{22}a_{33}-a_{23}a_{32}##

    Or doesn't it work like that?
     
    Last edited: Sep 3, 2015
  5. Sep 3, 2015 #4

    Orodruin

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    It does, but you must be careful because you are mixing different types of multiplications. The quantity you call ##a_1e_1 \cdot (a_2 \times a_3)e_1## should really be ##(a_1\cdot e_1)((a_2\times a_3)\cdot e_1)## and is not a triple vector product. It is the product of two numbers.
     
  6. Sep 3, 2015 #5
    Okay, I see. Can I get around that or do I have to use a completely different approach to show what I want to show? Could I use the Levi-Civita symbol and do it all in components?
     
  7. Sep 3, 2015 #6

    Orodruin

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    You can write ##\vec a_i = a_{i1} \vec e_1 + a_{i2}\vec e_2 + a_{i3}\vec e_3## and just perform the triple product. Of course, this is equivalent to doing it in components using the Levi-Civita symbol.
     
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