Properties of 4 by 4 Matrices Define
$    
A=\begin{pmatrix}
x_1 & x_2 & 0 & 0\\
0 &1&0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}, 
B=\begin{pmatrix}
1 & 0 & 0 & 0\\
x_3 &x_4&x_5&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix},
C=\begin{pmatrix}
1 & 0 & 0 & 0\\
0 &1&0&0\\
0&x_6&x_7&x_8\\
0&0&0&1
\end{pmatrix},$
$
D=\begin{pmatrix}
1 & 0 & 0 & 0\\
0 &1&0&0\\
0&0&1&0\\
0&0&x_9&x_{10}
\end{pmatrix}$.  
Let  $W$ be a product of above matrices. (For example $ W $ can be $D^2ABDC$ or $CAD$ (any product of above matrices)). 
Now define reverse of $W$ as $Re(W)$. For example if $W=D^2ABDC$ then $Re(W)=CDBAD^2$, If $W=CAD$ then $Re(W)=DAC$.
I want to prove $\det(W+Re(W))-\det(W-Re(W))$ is divisible by $4$. In other words $\det(W+Re(W))-\det(W-Re(W))\equiv 0 \mod 4$.
Any comment really appreciated.
Note (Grumpy Parsnip): A generalized phenomenon seems to be true here. One can define $n\times n$ matrices of a similar form. E.g. $A_2=\left(\begin{array}{cc} x_1&x_2\\0&1\end{array}\right)$ and $B_2=\left(\begin{array}{cc} 1&0\\x_3&x_4\end{array}\right)$. Then it seems that for $2m\times 2m$ matrices, 
$$\det(W+Re(W))+(-1)^{m-1}\det(W-Re(W))=0\mod 4.$$
Perhaps the $2\times 2$ case is easier to understand. I'll award a bounty for a solution to the $2\times 2$ case. (Hopefully the OP, who is a friend of mine, will not mind my edits.)

My impulse is to prove it by induction (on the number of factors in $W$).  Clearly the result holds for any single-factor products.  So, assuming that the result holds for $W$, show that $\det(AW+WA)\equiv \det(AW-WA)\mod 4$, and similarly for $B,C$, and $D$.  Just doing the algebra by hand, I showed that it's true at least for $A$.  I assume the others would work out somewhat similarly.  At the very least, the calculation for $D$ would be symmetric.  
