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/Author (John Kerl)
/Title  (Geometry-topology notes)
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\beginex
Similarly, for $3 \times 3$ matrices:
\begin{eqnarray*}
	\det(\vecu,\vecv,\vecw) &=& \det\left(
		\sum_{i=1}^3 u_i \vece_i,
		\sum_{j=1}^3 v_j \vece_j,
		\sum_{k=1}^3 w_k \vece_k
		\right). \\
	&=& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3
		\det\left(u_i \vece_i, v_j \vece_j, w_k \vece_k \right) \\
	&=& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3
		u_i v_j w_k \det\left(\vece_i, \vece_j, \vece_k \right).
\end{eqnarray*}
It remains to find the values of $\det(\vece_i, \vece_j, \vece_k)$.
There are 27 of them, but only 6 with non-zero values.  They are
\begin{eqnarray*}
\det(\vece_1, \vece_2, \vece_3)=\det\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}=1,&&
\det(\vece_1, \vece_3, \vece_2)=\det\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0 \end{pmatrix}=-1,\\
\det(\vece_2, \vece_3, \vece_1)=\det\begin{pmatrix}0&1&0\\ 0&0&1\\ 1&0&0 \end{pmatrix}=1,&&
\det(\vece_3, \vece_2, \vece_1)=\det\begin{pmatrix}0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix}=-1,\\
\det(\vece_3, \vece_1, \vece_2)=\det\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix}=1,&&
\det(\vece_2, \vece_1, \vece_3)=\det\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}=-1.\\
\end{eqnarray*}
The determinant tensor on rows of $3 \times 3$ matrices may be thought
of as
$$ \det(\vecu, \vecv, \vecw) = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 u_i v_j w_k d_{ijk}$$
where $d_{ijk}$ are the entries of the following $3 \times 3 \times 3$ array
where $i$ indexes the face (front, middle, back), $j$ indexes rows, and $k$
indexes columns:
$$
\begin{matrix}
\textrm{front} &
\textrm{middle} &
\textrm{back} \\
\left(
\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 &-1 & 0 \\
\end{matrix}\right)
\right.
&
\left.
\left(\begin{matrix}
0 & 0 &-1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{matrix}\right)
\right.
&
\left.
\left(\begin{matrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix}\right)
\right)
\end{matrix}
.$$
\mendex


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