<s>[INST] Proving an equality of matrices I am asked to prove that $$\begin{pmatrix} \\ A & B\\ C &D\end{pmatrix}^{-1}=\begin{pmatrix} M & -MBD^{-1} \\ -D^{-1}CM & D^{-1}+D^{-1}CMBD^{-1} \end{pmatrix}$$
Where $M=(A-BD^{-1}C)^{-1}$.
Unfortunately, I have no idea what to do about it.
 [/INST] I guess rather than verifying the inverse stated in the assignment, you should derive it. Let $$
    \begin{pmatrix} A & B \cr C & D \end{pmatrix}^{-1} = \begin{pmatrix} U & V \cr W & X \end{pmatrix}
$$
We have:
$$
    \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix} = \begin{pmatrix} A & B \cr C & D \end{pmatrix} \cdot \begin{pmatrix} U & V \cr W & X \end{pmatrix} = \begin{pmatrix} A U + B W & AV + B X \cr CU + D W & CV + DX \end{pmatrix}
$$
Thus $AV = -BX$ and $CU = -DW$, giving $X = -B^{-1} A V$ and $W=-D^{-1} C U$. Substituting back into block-diagonals:
$$
    A U - B D^{-1} C U = \left(A  - B D^{-1} C U\right) U= 1 \qquad C V-D B^{-1} A V = \left(C - D B^{-1} A \right) V= 1
$$
Hence 
$$
    U = \left(A  - B D^{-1} C\right)^{-1} \qquad
    V = \left(C - D B^{-1} A\right)^{-1}
$$
Now $$U B D^{-1} = \left(A  - B D^{-1} C\right)^{-1} B D^{-1} = \left( D B^{-1} \left(A  - B D^{-1} C\right)\right)^{-1} = -V $$
</s>