Prove that if rank $(\begin{smallmatrix} A &B\\ C&D\end{smallmatrix})=\operatorname{rank}(A)$, then $D=CA^{-1}B$ . Let $A$ be an invertible $n \times n$ matrix with entries from a field $F$. Prove that if rank $\begin{pmatrix} A &B\\ C&D\end{pmatrix}=\operatorname{rank}(A)$, then $D=CA^{-1}B$ .
 A: Hint
It is easy to show
$$\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}\to\begin{bmatrix}
A&0\\
0&D-CA^{-1}B
\end{bmatrix}$$
so we have
$${\rm rank}\begin{bmatrix}
A&B\\
C&D
\end{bmatrix}={\rm rank}(A)+{\rm rank}{(D-CA^{-1}B)}$$
and hence
$$\rm{rank(D-CA^{-1}B)}=0$$
then you can solve your problem 
A: Hint. Using block row operations, write the matrix as a product of two block triangular matrices, one of which is invertible. (To get an idea of what the operation should be, pretend first that $a, b, c, d$ are all scalars and determine what row operation would eliminate $c$. Then perform the analogous operation, but by blocks.)
A: Here is another way.
Note that $\begin{bmatrix} I & 0 \\ -C A^{-1} & I \end{bmatrix}  \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A & B \\ 0 & -CA^{-1}B+D \end{bmatrix}$. The first matrix is invertible,
so the rank of the last matrix is the same as the rank of the middle matrix.
We are given that the rank of the middle matrix is the same as the rank
of $A$, hence we must have $-CA^{-1}B+D = 0$, otherwise the rank of the
last matrix would be strictly greater than the rank of $A$.
