Determinant of block matrix when $CD^T=DC^T$ When $CD^T=DC^T$ and $D$ is invertible we have:
$$\left(\begin{array}{cc} A & B\\ C & D\\\end{array}\right)\times\left(\begin{array}{cc} D^T & 0\\ -C^T & D^{-1}\\\end{array}\right)=\left(\begin{array}{cc} AD^T-BC^T & BD^{-1}\\ 0 & I\\\end{array}\right)$$
then $\det\left(\begin{array}{cc} A & B\\ C & D\\\end{array}\right)=\det(AD^T-BC^T)$.
Then I want to consider when $D$ is noninvertible. 
It's bad that I cannot use the continuity of determinant since there is the condition $CD^T=DC^T$ (unless there are always a series of $D_i$ converges to $D$ and $CD_i^T=D_iC^T\forall i$, which is unlikely).

We can use $D_k:=D+k^{-1}C$: for $k$ large enough this matrix is invertible (because the set of invertible matrices is open) and then we can use continuity of the determinant.
Since 
\begin{align}
CD_k^T&=C(D^t+k^{-1}C^T)\\
&=CD ^T+k^{-1}CC^T\\
&=DC^T+k^{-1}CC^T\\
&=(D+k^{-1}C)C^T\\
&=D_k C^T,
\end{align}
we can use the previous case.
