I find myself in trouble with easy-looking linear algebra task.
Let $A, B, C, D$ be square $n\times n$ matrices such that $C\cdot D^T = D\cdot C^T$. Show that $\begin{vmatrix}A & B\\ C & D \end{vmatrix} = \vert A\cdot D^T - B\cdot C^T\vert$.
For square matrices we indeed have $\begin{vmatrix}A & B\\ C & D \end{vmatrix} = \vert AD - BC \vert$, but how to proceed further? I tried to plug $D = C D^T (C^T)^{-1}$ and $C = DC^T(D^T)^{-1}$ into this, but wasn't able to simplify it to desired form, although it seems the only straightforward way.
In line 4, your equality is absolutely false, except when $A,B,C,D$ commute. Let $U=\begin{pmatrix}A&B\\C&D\end{pmatrix}$.
Case 1. $A,D$ are invertible. Then $\det(U)=\det(A)\det(D-CA^{-1}B)$
cf. https://en.wikipedia.org/wiki/Determinant
$\det(U)=\det(AD)\det(I-(D^{-1}C)A^{-1}B)$. Here $D^{-1}C=C^TD^{-T}$.
$\det(U)=\det(AD^T)\det(I-C^T(D^{-T}A^{-1}B))=\det(AD^T)\det(I-(D^{-T}A^{-1}B)C^T)$.
Finally $\det(U)=\det(AD^T-BC^T)$.
Case 2. $A$ or $D$ is non-invertible. Proceed by continuity.