Block matrix with multiple inverses? In this Wikipedia article on block matrices, there are two expressions for the inverse of a block matrix. These two expressions depend on the invertibility of one of the diagonal sub matrices and the others Schur complement.
My question is, if all of the invertibility conditions hold, namely $\mathbf{A}$, $\mathbf{D}$, $\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B}$, and $\mathbf{A}-\mathbf{B}\mathbf{D}^{-1}\mathbf{C}$ are all invertible, are the two expressions for the inverse equivalent?
 A: They are of course equivalent, as the inverse of a square matrix is unique, but you may also view their equality as a variant of the identity
$$
(I-XY)^{-1}=I+X(I-YX)^{-1}Y\tag{1}
$$
where $X$ and $Y$ are any two rectangular matrices such that $I-XY$ and $I-YX$ are invertible. If you have learnt about the method of universal identities, $(1)$ naturally arises from expanding $(I-XY)^{-1}$ as a formal power series
\begin{aligned}
(I-XY)^{-1}
&=I+XY+XYXY+XYXYXY+\cdots\\
&=I+X(I+YX+YXYX+\cdots)Y\\
&=I+X(I-YX)^{-1}Y.\\
\end{aligned}
You may also prove $(1)$ directly:
\begin{aligned}
&(I-XY)\left[I+X(I-YX)^{-1}Y\right]\\
&=I-XY+X(I-YX)^{-1}Y-XYX(I-YX)^{-1}Y\\
&=I-XY+X(I-YX)(I-YX)^{-1}Y\\
&=I-XY+XY\\
&=I.
\end{aligned}
At any rate, if we put $X=BD^{-1}$ and $Y=CA^{-1}$ into $(1)$, we obtain
\begin{aligned}
(I-BD^{-1}CA^{-1})^{-1}&=I+BD^{-1}(I-CA^{-1}BD^{-1})^{-1}CA^{-1},\\
A(A-BD^{-1}C)^{-1}&=I+B(D-CA^{-1}B)^{-1}CA^{-1},\\
(A-BD^{-1}C)^{-1}&=A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}.\\
\end{aligned}
