how to find: $A$ is invertible Suppose matrices $A\in K^{m \times m}$, $B\in K^{m \times n}$, $C\in K^{n \times m}$, $D\in K^{n \times n}$. 
I need to show that if $A$ and $D-CA^{-1}B$ are invertible, then $\begin{pmatrix}
  A & B \\ 
  C & D
 \end{pmatrix} \in K^{(m+n) \times (m+n)}$ is also invertible. 
I know what invertible means: $A$ is invertible if $A A' = E$ where $E$ has dimension of $A$. 
Is there a trick to know if it is invertible before calculating many matrix elements? 
Thanks in advance!
 A: Notice 
$$\begin{pmatrix}A & B\\C & D\end{pmatrix} \begin{pmatrix}I & -A^{-1}B\\0 & I\end{pmatrix} = \begin{pmatrix}A & 0\\C & D - CA^{-1}B\end{pmatrix}$$
we have 
$$\det\begin{pmatrix}A & B\\C & D\end{pmatrix} = \det(A)\det( D - CA^{-1}B) \ne 0$$
whenever $A$ and $D - CA^{-1}B$ invertible.
A: You are actually defining the Schur complement. There is a very standard way of dealing with this question (you can find it anywhere on the web).
A: Note that a matrix $P$ is invertible iff $Px=0$ has only trivial solution. Now consider 
$
\begin{bmatrix}
A & B \\
C & D
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}.
$
As $A$ is invertible, $x_1 = -A^{-1}Bx_2$. And substituting this in the next equation, we get $(D - CA^{-1}B)x_2=0$, which has only trivial solution if $D - CA^{-1}B$ is invertible. Hence the original system of equations has only trivial solution.
A: (Improved my comment into an answer)
It might be useful do observe that
$$
\begin{bmatrix}
A & B \\
C & D
\end{bmatrix}
\begin{bmatrix}
A' & B' \\
C' & D'
\end{bmatrix}
=
\begin{bmatrix}
AA'+BC' & AB'+BD' \\
CA'+DC' & DB'+DD'
\end{bmatrix}
$$
so that you have 4 equations to solve whose variables are the matrices $A',B',C',D'$:
$$
\begin{cases}
AA'+BC' = E_m \\
AB'+BD' = 0 \\
CA'+DC' = 0 \\
DB'+DD' = E_n
\end{cases}
$$
Since $A$ is invertible, from the first equation we deduce that
$$
A' = A^{-1}(E_m-BC')
$$
so that the system turns into
$$
\begin{cases}
A'=A^{-1}-A^{-1}BC' \\
AB'+BD' = 0 \\
CA^{-1} + \big(D - CA^{-1}B\big) C' = 0 \\
DB'+DD' = E_n
\end{cases}
$$
Now, let $M=\big(D - CA^{-1}B\big)^{-1}$ so that from the third equation it follows that $C'=-MCA^{-1}$, and so on...
