Equality of determinants of two square matrices

Question

From some abstract considerations I know that $$\det \pmatrix{A & -\bar{B} \\ B & \bar{A}}= \det \pmatrix{\bar{A} & \bar{B} \\ -B & A}$$ for $A,B$ complex square matrices satisfying $A^*=A$ and $B^T=-B$. Since the first one is hermitian one can take bar inside and obtain "almost" the second one but with $\bar{B}$ and $-B$ exchanged. Is there a clever shuffling showing the equality?

Answer 1

Let $n$ denote the size of $A$ and $B$, let $I$ denote the identity matrix of size $n$. We find that in general, $$ \det \pmatrix{A & B\\C & D} = \left(\det \pmatrix{0&I\\I&0}\right)^2\pmatrix{A & B\\C & D} = \\ \det \left(\pmatrix{0&I\\I&0}\pmatrix{A & B\\C & D}\pmatrix{0&I\\I&0} \right) = \\ \det \pmatrix{D & C\\B&A} $$