<s>[INST] Questions concerning $ \det ({}^A_{C\,}{}^B_D) = \det ({}^D_{B\,}{}^C_A)$ Let $A, B, C, D $ be $n \times n $ matrices. Using Schur complements I have found that 
$$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} = 
\begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix} 
\begin{pmatrix} I & 0 \\ C & I \end{pmatrix} 
\begin{pmatrix} I & A^{-1}B \\ 0 & D-CA^{-1}B \end{pmatrix}  $$
and
$$ \begin{pmatrix} D & C \\ B & A \end{pmatrix} = 
\begin{pmatrix} I & CA^{-1} \\ 0 & I \end{pmatrix} 
\begin{pmatrix} D-CA^{-1}B & 0 \\ 0 & A \end{pmatrix} 
\begin{pmatrix} I & 0 \\ A^{-1}B & I \end{pmatrix},  $$
 from which the determinant equality follows, as long as $ A^{-1} $ exists. However, how do I tackle this when $A$ is singular? Here, I could just switch decompositions, but then I will get the same problem when $D$ is singular. Maybe one could derive two more decompositions using Schur complements, involving $B^{-1}$ and $C^{-1}$, respectively, and then one could say that the equality holds if at least one of the submatrices is nonsingular? Then, if all four submatrices are singular, the determinant must be zero -- from which equality follows trivially.
Also, does the equality hold when $A, B, C, D$ are not necessarily square but of matching sizes? Here, it doesn't seem like the decompositions will be valid, as $A$ or $D$ aren't necessarily square matrices (although then $B$ and $C$ must be).
 [/INST] One approach is to take advantage of the continuity of the determinant.  In particular, if $A$ is singular, we have
$$
\det \pmatrix{A &B\\C&D} = \lim_{t \to 0} \det\pmatrix{A + tI & B\\C&D}
= \lim_{t \to 0} \det \pmatrix{D&C\\B&A + tI} = 
\det \pmatrix{D&C\\B&A}
$$
Alternatively, you could have noted that
$$
\pmatrix{D&C\\B&A} = \pmatrix{0&I\\I&0} \pmatrix{A&B\\C&D} \pmatrix{0&I\\I&0}
$$
from there, we would only need to show that
$$
\det \pmatrix{0&I_n\\I_n&0} = (-1)^n
$$
</s>