We have the following determinant property
$$\det \begin{bmatrix} U & O \\ V & W \end{bmatrix} = \det(U) \cdot \det(W)$$
where $U \in R^{n\times n}$, $V \in R^{m\times n}$, $W \in R^{m\times m}$ and $O \in R^{n\times m}$ (the zero matrix).
Now suppose the zero block appears in the top left corner instead. Does there in that case also exist a rule to calculate the determinant of the matrix more easily?
The matrices I am thinking of here are of the form
$$Z = \begin{bmatrix} O & A \\ A^T & B \end{bmatrix}$$
with all matrices conformable. An example would be
$$Z = \begin{bmatrix} 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & -9 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & -1 \\ 1 & -9 & 0 & -1 & 2 & 0 \\ 1 & 0 & 0 & 2 & 1 & 0 \\ 1 & 1 & -1 & 0 & 0 & 1 \end{bmatrix}$$