In this problem all blocks are $n\times n$.
(a) $A = \begin{bmatrix}I & 0 \\ C & I\end{bmatrix}$, where $C$ is $n \times n$ and $I$ is the $n \times n$ identity. What is $A^{−1}$?
(b) Now consider a general case $A = \begin{bmatrix}M & 0 \\ C & N\end{bmatrix}$, where $M$ and $N$ are invertible $n \times n$ matrices. What is $A^{-1}$?
Right now I am using the determinant method: In (a), I find the $\det(A)$ and $\operatorname{adj}(A)$ and get the inverse as $A^{-1} = \begin{bmatrix}I & 0 \\ -C & I\end{bmatrix}$; For (b) I am a bit confused; Also I am not sure if using the determinant method is right.
You can search an inverse $B$ of the matrix $A$ in the following form: $$ B=\begin{pmatrix}B_1&B_2\\ B_3&B_4 \end{pmatrix} $$ where $B_i$ is a $n\times n $ matrix (i=1,2,3,4).
You must have $AB=I$, which gives: $$ B_1=M^{-1}, B_2=0,B_3=-N^{-1}CM^{-1}, B_4=N^{-1}. $$ So, $$ A^{-1}=\begin{pmatrix}M^{-1}&0\\ -N^{-1}CM^{-1}&N^{-1} \end{pmatrix} $$