Show $\pmatrix{A & B\\ C & I}$ is nonsingular given that $A-BC$ is nonsingular. I'm having some trouble showing that the block matrix $$D = \pmatrix{A & B\\ C & I}$$ is nonsingular, given that $A-BC$ is nonsingular.
I have gotten close with the following by saying let $$T = \pmatrix{(A-BC)^{-1} & -B(A-BC)^{-1}\\-C(A-BC)^{-1} & A(A-BC)^{-1}}.$$ (Basically, I'm trying to extend the formula for the inverse of a $2\times2$ real matrix to block matrices.) 
Then $$\begin{align}DT &= \pmatrix{A & B\\ C & I}\pmatrix{(A-BC)^{-1} & -B(A-BC)^{-1}\\-C(A-BC)^{-1} & A(A-BC)^{-1}} \\ &= \pmatrix{A(A-BC)^{-1}-BC(A-BC)^{-1} & -AB(A-BC)^{-1}+BA(A-BC)^{-1}\\ C(A-BC)^{-1}-C(A-BC)^{-1} & -CB(A-BC)^{-1} + A(A-BC)^{-1}}\\ &=\pmatrix{(A-BC)(A-BC)^{-1} & (BA - AB)(A-BC)^{-1}\\ O & (A - BC)(A-BC)^{-1}}\\ &=\pmatrix{I & (BA - AB)(A-BC)^{-1}\\ O & I}.\end{align}$$
I manage to get almost everything, except the top-right block. I want it to be $O$, but I'm not sure what information I have that makes it equal $O$. I don't necessarily know that $A$ and $B$ commute, i.e., $AB = BA$, which would imply the top-right block would zero out. Can anyone see if I made a mistake? If there is no mistake, can anyone give a hint on what to do to try and zero out the top-right block?
 A: $$\pmatrix{I&-B\\O&I}\pmatrix{A&B\\C&I}
=\pmatrix{A-BC&O\\C&I}.$$
The first matrix is clearly invertible. The last is iff $A-BC$ is.
A: We have 
$$
\pmatrix{A&B\\C&I}\pmatrix{I&0\\-C&I}=\pmatrix{A-BC&B\\0&I}
$$
and $$\text{det}\left(\pmatrix{A-BC&B\\0&I}\right)=det(A-BC)\neq0$$
A: Block matrix determinant formula: if $D$ is invertible,
$$ \det \pmatrix{A & B\cr C & D\cr} = \det(D) \det(A-BD^{-1} C)$$
Take $D=I$.
A: Curiously enough, this question can be answered quite simply from first principles, without reference to determinants; we merely need to show that any vector mapped to $0$ by the matrix
$ \mathscr M = \begin{bmatrix} A & B \\ C & I \end{bmatrix} \tag 1$
is itself $0$ ; so let $\mathscr X$ be a vector such that
$\mathscr M \mathscr X = \begin{bmatrix} A & B \\ C & I \end{bmatrix}\mathscr X = 0. \tag 2$
If the size of $A$ is $p$ and the size of $I$ is $q$, then the size of $\mathscr M$ is $p + q$ whilst $B$ is $p \times q$ and $C$ is $q \times p$.
We can then write $\mathscr X$ in terms of two vectors $\mathbf x$ and $\mathbf y$, where $\dim \mathbf x = p$ and $\dim \mathbf y = q$, thusly:
$\mathscr X = \begin{pmatrix} \mathbf x \\ \mathbf y \end{pmatrix}; \tag 3$
we have:
$\mathscr M \mathscr X =  \begin{bmatrix} A & B \\ C & I \end{bmatrix}\begin{pmatrix} \mathbf x \\ \mathbf y \end{pmatrix} = \begin{pmatrix} A\mathbf x + B\mathbf y \\ C\mathbf x + I\mathbf y \end{pmatrix} = \begin{pmatrix} A\mathbf x + B\mathbf y \\ C\mathbf x + \mathbf y \end{pmatrix}; \tag 4$
since
$\mathscr M \mathscr X = 0, \tag 5$
we have from (4) that
$A\mathbf x + B\mathbf y = 0, \tag 6$
and
$C\mathbf x + \mathbf y = 0, \tag 7$
whence
$A\mathbf x = -B\mathbf y \tag 8$
and
$C\mathbf x = -\mathbf y; \tag 9$
therefore,
$A\mathbf x = -B\mathbf y = B(-\mathbf y) = B(C\mathbf x) = BC\mathbf x, \tag{10}$
or
$(A - BC)\mathbf x = 0; \tag{11}$
since we are given that $A - BC$ is nonsingular, we conclude that
$\mathbf x = 0, \tag{12}$
and from (9) that
$\mathbf y = -C\mathbf x = -C(0) = 0 \tag{13}$
as well; so
$\mathscr X = 0; \tag{14}$
since the only solution to (2) is (14), we conclude that $\mathscr M$ is itself a nonsingular matrix.
A: Let's try finding the inverse; so be it
$$
E=\begin{pmatrix} P & Q \\ R & S\end{pmatrix}
$$
Then
$$
DE=\begin{pmatrix} A & B \\ C & I\end{pmatrix}
\begin{pmatrix} P & Q \\ R & S\end{pmatrix}=
\begin{pmatrix} AP+BR & AQ+BS \\ CP+R & CQ+S\end{pmatrix}
$$
We need $R=-CP$; this leads to $AP-BCP=I$ that yields $P=(A-BC)^{-1}$.
Next we need $S=-CQ+I$, so we get
$$
AQ+BS=AQ-BCQ+B=0
$$
that is, $(A-BC)Q+B=0$ or
$$
Q=-(A-BC)^{-1}B
$$
In conclusion, the inverse is
$$
\begin{pmatrix}
(A-BC)^{-1} & -(A-BC)^{-1}B \\
-C(A-BC)^{-1} & I+C(A-BC)^{-1}B
\end{pmatrix}
$$
