I was working on a problem involving a mixed matrix of Grassmann variables and complex numbers. Essentially I found two expressions for a so called "superdeterminant" of the below matrix $G$. At the moment I have the following resultant equality:
$ \dfrac{det(D - CA^{-1}B)}{det(A)} = \dfrac{det(D)}{det(A - BD^{-1}C)} $
where the matrices $A, B, C, D$ are constituent parts of the matrix $ G = \left( \begin{array}{cc} A & B \\ C & D \\ \end{array} \right)$
such that $A \in \mathbb{C}_{n \times n}, D \in \mathbb{C}_{m \times m}$ and B, C are matrices of Grassmann variables of size $n \times m$ and $m \times n$ respectively. Using Sylvester's determinant theorem on the LHS, so that the dimensions agree, I have that:
$det(\mathbb{I}_{n \times n} - A^{-1}BD^{-1}C) = \dfrac{1}{det(\mathbb{I}_{n \times n} - A^{-1}BD^{-1}C)}$
after cancelling the ratio of $ \dfrac{det(D)}{det(A)} $ from both sides.
So, I have a few questions regarding this result. Firstly, I conclude that $ det(\mathbb{I}_{n \times n} - A^{-1}BD^{-1}C) = \sqrt{1} $.
However, what does this say about the $det(A^{-1}BD^{-1}C)$? Can I conclude anything directly about this matrix product from the result? Is the determinant $ det(\mathbb{I}_{n \times n} - A^{-1}BD^{-1}C)$ always its own inverse? If so, why is this the case?
NB: the definition I've been using for the "superdeterminant" is the inverse of that given as the Berezinian i.e. $ \dfrac{1}{det(G)} $ is my definition of the superdeterminant.