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

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1. Sep 3, 2015

### Erithacus

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. Sep 3, 2015

### Orodruin

Staff Emeritus
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.

3. Sep 3, 2015

### Erithacus

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
4. Sep 3, 2015

### Orodruin

Staff Emeritus
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.

5. Sep 3, 2015

### Erithacus

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?

6. Sep 3, 2015

### Orodruin

Staff Emeritus
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.