Determinant of $2 \times 2$ block matrix whose diagonal blocks are zero $\Bbb A$ is an $n\times n$ matrix, and $\Bbb B$ is an $m × m$ matrix.$\space$What is the determinant of matrix $\Bbb C$?
\begin{equation*}
\mathbb{C}=
\begin{pmatrix}
\begin{array}{@{}c|c@{}}
  \begin{matrix}
  0
  \end{matrix}
  & \mathbb{A} \\
\hline
  \mathbb{B} &
  \begin{matrix}
 0
  \end{matrix}
\end{array}
\end{pmatrix}
\end{equation*}
I thought it could be just
($\det\Bbb A\cdot\det\Bbb B$)
or ($-\det\Bbb A\cdot\det\Bbb B$),
but I'm not sure, as it seems too easy.

This determinant can be computed by the Laplace expansion theorem (the generalized form). Fix the first $1, 2, \ldots, n$ rows, and let columns range over $(j_1, j_2, \ldots, j_n) \in \{1, 2, \ldots, n + m\}$. Since the square minor is non-zero only if $(j_1, j_2, \ldots, j_n) = (m + 1, m + 2, \ldots, m + n)$,  it follows that
\begin{align*}
\det C = & C\begin{pmatrix}1 & 2 & \cdots & n \\
m + 1 & m + 2 & \cdots & m + n\end{pmatrix}(-1)^{1 + \cdots + n + m + 1 + \cdots + m + n}C\begin{pmatrix}n + 1 & n + 2 & \cdots & n + m \\
1 & 2 & \cdots & m \end{pmatrix} \\
= & (-1)^{n(n + 1) + mn}\det(A)\det(B).
\end{align*}
