TITLE: Invertibility of the Schur Complement
QUESTION [5 upvotes]: Suppose that 
$$
M = \begin{bmatrix}A & B \\ C & D\end{bmatrix}.
$$
I know that if $D$ and $M\setminus D$ (where $M\setminus D$ is the Schur Complement of $D$ in $M$) are invertible, then $M$ is invertible. 
However, suppose that we know that $M$ and $D$ are invertible, but we know nothing about the invertibility of $A$. Can we say that $M\setminus D$ is invertible?

REPLY [3 votes]: Yes, if you take the determinants, you obtain with 
$$\operatorname{det}(M)=\operatorname{det}(M/ D)\cdot\operatorname{det}(D)
$$ 
therefore if $\operatorname{det}(M)$ is non-zero then $\operatorname{det}(M/ D)$ non-zero.