Determinant of a block skew-symmetric matrix
By diagonalization in the $2\times 2$ case, we obtain the similarity relation $$ \pmatrix{A&-B\\B&A}=\pmatrix{-iI&iI\\I&I}\pmatrix{A-iB&0\\0&A+iB}\pmatrix{(i/2)I&(1/2)I\\-(i/2)I&(1/2)I}. $$ Whence $$ \det \pmatrix{A&-B\\B&A}=\det(A-iB)\det(A+iB)=\det(A^2+i(AB-BA)+B^2). $$ If $A$ and $B$ commute, this yields $\det(A^2+B^2)$.