Proving the equality: $D^{-1}+D^{-1}CPBD^{-1}=(D-CA^{-1}B)^{-1}$ where $P=(A-BD^{-1}C)^{-1}$

Question

I am trying to prove the following equality that I need to use as an intermediate step to solve one of my problems. The equality is the following:

$D^{-1}+D^{-1}CPBD^{-1}=(D-CA^{-1}B)^{-1}$, where $P=(A-BD^{-1}C)^{-1}$

All these matrices are $n\times n$ invertible matrices. Any help is much appreciated!

Answer 1

$$\left(D^{-1}+D^{-1}CPBD^{-1}\right)\cdot \left(D-CA^{-1}B \right) \ = \\ =\ I+D^{-1}CPB\ -\ D^{-1}CA^{-1}B \ - \ D^{-1}CPBD^{-1}CA^{-1}B\ =\ \\ =\ I+D^{-1}C\cdot\left(PA-I-PBD^{-1}C \right) \cdot A^{-1}B\ = \\ =\ I+D^{-1}C\cdot \big(\underbrace{P(A-BD^{-1}C)}_{=\,I}\ -\ I\big)\cdot A^{-1}B\ = \ I+0\ =\ I\,. $$ Similarly, multiplying from the left also gives $I$.