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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Determinant of 3x3 matrix equal to scalar triple product?

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