I have received this question to solve, but, quite frankly, I have no idea how to do it. Can anyone help?
Let $A,B,C,D \in R^{n \times n}$ Show that if $B,D,A-BD^{-1}C$, and $C-DB^{-1}A$ are nonsingular, then $$\begin{bmatrix} A & B\\[0.3em] C & D\\[0.3em] \end{bmatrix}^{-1} = \begin{bmatrix} (A-BD^{-1}C)^{-1} & (C-DB^{-1}A)^{-1}\\[0.3em] -D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}-D^{-1}C(C-DB^{-1}A)^{-1}\\[0.3em] \end{bmatrix}$$
I don't know whether you can apply the same rules as with numerical matrices or not, that's where the root of my problem is. Thanks in advance!
Multiplication of block matrices works the same as that of ordinary matrices, except that you have to be careful of the order of things because matrices don't commute. Thus $$ \left[ \matrix{A & B\cr C & D\cr} \right] \left[ \matrix{E & F\cr G & H\cr} \right] = \left[ \matrix{AE+BG & AF + BH\cr CE +DG & CF + DH\cr} \right]$$
So multiply $\left[ \matrix{A & B\cr C & D\cr} \right]$ by its alleged inverse (after you get that corrected), and check that you can simplify the result to $\left[\matrix{I & 0\cr 0 & I}\right]$.