<s>[INST] Matrices operations My professors asked to find the inverse of M:
$$ M=
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
$$
Therefore, to find $M^{-1}$, we calculate:
$$
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
\begin{bmatrix}
C&D\\
E&F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
-A&I_q\\
\end{bmatrix}
$$
How do I go from
$$
\begin{bmatrix}
C&D\\
AC+E&AD+F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
0&I_q\\
\end{bmatrix}
$$
to 
$$
\begin{bmatrix}
C&D\\
E&F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
-A&I_q\\
\end{bmatrix}
$$
What operations did my professor use to get rid of $AC$ and $AD$ in the first matrix to just get $C$, $D$, $E$, $F$?
 [/INST] Assuming the matrix equation
$\begin{bmatrix} C&D\\AC+E&AD+F\\\end{bmatrix} =
\begin{bmatrix} I_p&0\\ 0&I_q\\\end{bmatrix} \tag 1$
is to be interpreted in terms of matrix blocks, we immediately have
$C = I_p, \tag 2$
$D = 0, \tag 3$
$AC + E = 0, \tag 4$
$AD + F = I_q; \tag 5$
then
$A + E = AI_p + E =  AC + E = 0, \tag 6$
whence
$E = -A, \tag 7$
and
$F = A(0) + F = AD + F = I_q; \tag 8$
substituting (2), (3), (7), and (8) into
$\begin{bmatrix} C & D \\ E & F \end{bmatrix} \tag 9$
yields
$\begin{bmatrix} C & D \\ E & F \end{bmatrix} = \begin{bmatrix} I_p & 0 \\ -A & I_q \end{bmatrix}. \tag{10}$
</s>