Determinant of non-triangular block matrix 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 
$$Z = \begin{bmatrix}
    0 & 0 & 0 & 1 & 1 & 1 \\
    0 & 0 & 1 & -9 & 0 & 1 \\
    0 & 1 & 1 & 0 & 0 & -1 \\
    1 & -9 & 0 & -1 & 2 & 0 \\
    1 & 0 & 0 & 2 & 1 & 0 \\
    1 & 1 & -1 & 0 & 0 & 1
  \end{bmatrix}$$
 A: There is no such rule to calculate determinant easy as in the case the zero block is in the top right or bottom left corner. 
Here you can see all rules you can apply on block matrices https://en.wikipedia.org/wiki/Determinant.
Instead, you can transform your matrix with Gaussian transformations to an upper triangular matrix and just multiply elements on diagonal.
A: Let us assume that all matrices are square and $n \times n$. Permuting the columns,
$$\det \begin{bmatrix} \mathrm O_n & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} = \det \begin{bmatrix} \mathrm A & \mathrm O_n\\ \mathrm B & \mathrm A^{\top}\end{bmatrix} \cdot \underbrace{\det \begin{bmatrix} \mathrm O_n & \mathrm I_n\\ \mathrm I_n & \mathrm O_n\end{bmatrix}}_{= \pm 1} = \pm \left( \det (\mathrm A) \right)^2$$
A: Let $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm B \in \mathbb R^{n \times n}$. Assuming that $\mathrm B$ is invertible, we use the Schur complement
$$\begin{bmatrix} \mathrm O_m & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} \begin{bmatrix} \mathrm I_m & \mathrm O_{m \times n}\\ -\mathrm B^{-1} \mathrm A^{\top} & \mathrm I_n\end{bmatrix} = \begin{bmatrix} -\mathrm A \mathrm B^{-1} \mathrm A^{\top} & \mathrm A\\ \mathrm O_{n \times m} & \mathrm B\end{bmatrix}$$
and, taking the determinant, we obtain
$$\det \begin{bmatrix} \mathrm O_m & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} \cdot \underbrace{\det \begin{bmatrix} \mathrm I_m & \mathrm O_{m \times n}\\ -\mathrm B^{-1} \mathrm A^{\top} & \mathrm I_n\end{bmatrix}}_{=1} = \det \begin{bmatrix} -\mathrm A \mathrm B^{-1} \mathrm A^{\top} & \mathrm A\\ \mathrm O_{n \times m} & \mathrm B\end{bmatrix} = \color{blue}{(-1)^m \det (\mathrm A \mathrm B^{-1} \mathrm A^{\top}) \, \det(\mathrm B)}$$
