[Math] Determinant of a $4n \times 4n$ block matrix where every block is singular

Question

I have a 4n$\times$4n matrix, which can be written as
\begin{pmatrix}
0 & A &B &C \cr
D& 0& E & F \cr
G& H & 0 & J \cr
K& L& M& 0
\end{pmatrix}

each entry being an n$\times$n matrix with vanishing determinant. Is there a rule for checking if the full matrix has zero determinant? How about the special case
\begin{pmatrix}
0 & A &B &C \cr
-A^T & 0& E & F \cr
-B^T & E^T & 0 & J \cr
-C^T & F^T & J^T & 0
\end{pmatrix}

still with vanishing determinants for each n$\times$n matrix?

(The n is the dimension of an SU group — I can probably work out the SU(2) or n=3 case by brute force, but I would like to know if there is some method that does not require explicit calculation.)

Many thanks in advance for any help or suggestion.

Answer 1

It would be nice if the rule for determinants for $2\times2$ matrices generalized to the case of $2n\times 2n$ matrices:

$\det \begin{pmatrix} A & B \cr C & D \end{pmatrix} =\det A \det D - \det B\det C$,

but this is sadly not true.

Nonetheless, the familiar Laplace expansion theorem for minors of order $n-1$ does have a generalization to minors of any order, including, in this case, minors of order $2n$ of a $4n \times 4n$ matrix, see http://www.proofwiki.org/wiki/Laplace's_Expansion_Theorem

This might help.