A (possibly) less cryptic way to calculate the cross product

Here is the latest entry in my "how to take simple, well-understood problems and convert them to horrible-looking exercises in matrix multiplication" file. (I'd say about 99% of my graduate work so far should also be stored in this rather unwieldy file...)

The cross product (which I think only exists in ordinary, three-dimensional space) of two vectors $\mathbf{a}$ and $\mathbf{b}$, written $\mathbf{a}\times\mathbf{b}$, is a way of combining $\mathbf{a}$ and $\mathbf{b}$ to produce a new vector, oriented perpendicularly to both $\mathbf{a}$ and $\mathbf{b}$. In symbols, this is
\[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin \left(\theta \right) \mathbf{\hat{n}}, \]
where $|\mathbf{a}| = \sqrt{{a_1}^2+{a_2}^2+{a_3}^2}$ is the length of $\mathbf{a}$, and $\mathbf{\hat{n}}$ is an as-yet-unknown unit vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$. The standard way of calculating the components of $\mathbf{a} \times \mathbf{b}$ is by constructing a very strange-looking determinant:
\[ \mathbf{a} \times \mathbf{b} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \times \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = \left| \begin{matrix} \mathbf{\hat{e}}_1 & \mathbf{\hat{e}}_2 & \mathbf{\hat{e}}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{matrix} \right| = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}. \]
I remember feeling puzzled when I first learned this as an undergraduate. It does give you the right answer, but seriously, what's going on with that determinant? Its first row is a set of unit vectors, and its second and third rows are populated by the elements of the vectors $\mathbf{a}$ and $\mathbf{b}$. There's probably some obvious explanation that I'm missing, but to me, that determinant has always been maddeningly cryptic.

Anyway, I was waiting for Maple to grind through some calculations for me this afternoon, so I thought I'd take a crack at writing the cross product in a somewhat less confusing way. Looking at the result of the above equation, it's clear that we're looking for the following three inner products:
\[ \mathbf{a} \times \mathbf{b} = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} a_2 & -a_3 \end{bmatrix} \begin{bmatrix} b_3 \\ b_2 \end{bmatrix} \\ \begin{bmatrix} -a_1 & a_3 \end{bmatrix} \begin{bmatrix} b_1 \\ b_3 \end{bmatrix} \\ \begin{bmatrix} a_1 & -a_2 \end{bmatrix} \begin{bmatrix} b_2 \\ b_1 \end{bmatrix} \end{bmatrix}. \]
It should be possible to form a matrix $\mathbf{A}$ from the components of $\mathbf{a}$, while leaving $\mathbf{b}$ untouched. (Or vice-versa.) Clearly, $\mathbf{A}$ will be a $3 \times 3$ matrix, and, since $\mathbf{a} \times \mathbf{b}$ is orthogonal to $\mathbf{a}$ and $\mathbf{b}$, it should have zeroes on its diagonal. After a bit of tinkering,
\[ \mathbf{A} = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}, \]
which allows us to compute the cross product by multiplying $\mathbf{b}$ on the left by the matrix $\mathbf{A}$:
\[ \mathbf{A} \mathbf{b} = \mathbf{a} \times \mathbf{b}. \]
Two immediately evident properties of $\mathbf{A}$ are that it is traceless (since its diagonal elements are all zero),
\[ \text{tr}\left(\mathbf{A}\right) = 0, \]
and it is skew-symmetric (equal to the negative of its transpose):
\[ \mathbf{A} = -\mathbf{A}^\mathrm{T}. \]
A quick calculation reveals that $\mathbf{A}$ only has two eigenvalues,
\[ \lambda_{\pm} = \pm i | \mathbf{a} |. \]

I don't know if this is otherwise useful, but it seems to me that it's quite a bit less confusing than the determinant method of calculating $\mathbf{a} \times \mathbf{b}$. One possible application might be to define the cross product in higher-dimensional spaces, using a larger matrix $\mathbf{A}$ with the same internal structure.