Cross product matrix I accidentally found that curl can be represented as a matrix, and conformed it on wiki:
$$[\nabla ]_{\times}=\begin{bmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{bmatrix}$$ 
then something quite strange I am not sure
since for any cross product vector:
$A\times{(B}\times{C})$
is not always equal to 
$(A\times{B})\times{C}$
but matrix can do this: 
$
(MN)Q=M(NQ)
$
So at the end of the day, is curl or cross product actually matrix? 
 A: You are not dealing with two cross products. If ${\bf F} = A{\bf e}_1+B{\bf e}_2 + C {\bf e}_3$, then: $$\begin{bmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \end{bmatrix} = \begin{bmatrix} -B_z + C_y \\ A_z - C_x \\ -A_y + B_x \end{bmatrix}$$ is exactly the coordinates of $\nabla \times {\bf F}$ in the basis $\{ {\bf e}_i\}_{i=1}^3$. We check this: $$\begin{vmatrix} {\bf e}_1 & {\bf e}_2 & {\bf e}_3 \\ \partial_x & \partial_y & \partial_z \\ A & B & C \end{vmatrix}  = (C_y - B_z){\bf e}_1 + (A_z - C_x){\bf e}_2 + (B_x - A_y){\bf e}_3.$$
To elaborate a bit more: the curl $\nabla\times$ is an operator that takes a vector field $\bf F$ and spits another one: $\nabla \times {\bf F}$. The matrix there is just a representation. If you write the curl, say, in spherical coordinates, you'll have another matrix with $\partial_r,\partial_\theta$ and $\partial_\phi $, that will do the same job: take the spherical coordinates of a vector field to the spherical coordinates of its curl.
A: One thing to keep in mind is that while for matrices we have $(AB)C=A(BC)$ and as you said above with corssproduct, if we have on the other hand
$$V\times\nabla\times W$$
This can only mean one thing, namely $V\times\nabla\times W=V\times(\nabla\times W)$
because $V\times\nabla$ is not defined and hence meaningless, same goes for your matrix
$$A[\nabla]C=A([\nabla]C)$$
Only for the same reason, taking any different order is simply non-sensical and undefined.
A: Over all these years, I believe I have finally solved this question in my head:
The answer is a sounding YES: cross products can be written as matrixs.
Although I had found it psychologically discomforting, there is nothing wrong with the matrix representation. The key is: algorithms.
As I writing the curl (vector) into matrices, what I am doing exactly is: encoding the algorithms of curl (vector) into matrices representation. 
Hence what is forbidden in the old algorithm does not have to be forbidden in the new matrix algorithm. The question is HOW to encode the old algorithms into matrix? 
What is obvious right (and makes my calculation in electrodynamics class simpler):
$A\times{(B}\times{C})\rightarrow\begin{bmatrix} 0 & -A_z & A_y \\ A_z & 0 & -A_x \\ -A_y & A_x & 0 \end{bmatrix}\begin{bmatrix} 0 & -B_z & B_y \\ B_z & 0 & -B_x \\ -B_y & B_x & 0 \end{bmatrix}\begin{bmatrix} C_x \\ C_y \\ C_z \end{bmatrix}$
On the other hand, it is not so easy to encode: 
$(A\times{B})\times{C}$
As we know:
$ [A \times B] \rightarrow \begin{bmatrix} 0 & -A_z & A_y \\ A_z & 0 & -A_x \\ -A_y & A_x & 0 \end{bmatrix}\begin{bmatrix} B_x \\ B_y \\ B_z \end{bmatrix}$
but this results in a vector! Column vector can't "time" another column vector! Hence, we will have to write a new matirx to represent this whole thing acting on vector C: $$(A \times B)\times $$
Which requires a different matrix. Therefore, the contradiction does not stand. 
