Determinant of non-all-square block matrix I have the following matrix 
$M=\begin{pmatrix} 
O_{n \times m} & | &A_{n \times n} \\ \hline B_{m \times m} &|& C_{m \times n}
\end{pmatrix}$
where $O$ is the all zero matrix.
I can't say that $\det M=-\det (O \times C -B \times A)$ because $A$ and $B$ don't have the same size 
and i can't say $\det(M)=\det(O)\times \det(C)- \det(B) \times \det(A)$ because $O$ and $C$ are not square.
These are the only guesses i have because i'm expecting something like  $\det(M)=(-1)^{nm} \det(B) \times \det(A)$. 
 A: By moving the last $n$ columns of $M$ across the first $m$ colums of $M$, we get
$$
\det\overbrace{\begin{pmatrix} 
O_{n \times m} &|& A_{n \times n}\\ \hline B_{m \times m} &|& C_{m \times n}
\end{pmatrix}}^M
=(-1)^{mn}\det\overbrace{\begin{pmatrix} 
A_{n \times n} &|& O_{n \times m}\\ \hline C_{m \times n} &|& B_{m \times m}
\end{pmatrix}}^N
$$
Since $O_{n\times m}$ is all zeros, to get a non-zero product in the Leibniz formula, the factors in the first $n$ rows of $N$ must come from from $A_{n\times n}$, leaving no terms to come from $C_{m\times n}$. Therefore, the terms in the bottom $m$ rows of $N$ must come from $B_{m\times m}$.
Thus, the only permutations that give rise to non-zero terms in the Leibniz formula are ones that are products of a permutation on $A_{n\times n}$ and a permutation on $B_{m\times m}$. Since the main diagonals of $A_{n\times n}$ and $B_{m\times m}$ lie on the main diagonal of $N$, the sign of one of these permutations is the product of the signs of the permutations on $A_{n\times n}$ and $B_{m\times m}$. Therefore,
$$
\det(N)=\det(A_{n\times n})\det(B_{m\times m})
$$
which gives
$$
\det(M)=(-1)^{mn}\det(A_{n\times n})\det(B_{m\times m})
$$
