Matrices operations

Question

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 $C$, $D$, $E$, $F$?

Answer 1

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

$\begin{bmatrix} C & D \\ E & F \end{bmatrix} = \begin{bmatrix} I_p & 0 \\ -A & I_q \end{bmatrix}. \tag{10}$

Answer 2

Hint: $$ \begin{bmatrix} C&D\\ AC+E&AD+F\\ \end{bmatrix} = \begin{bmatrix} C&D\\ E&F\\ \end{bmatrix} + \begin{bmatrix} 0&0\\ AC&AD\\ \end{bmatrix} $$