Suppose
$\textbf{M}=\begin{pmatrix}\textbf{A}&\textbf{B}\\\\ \textbf{C}&\textbf{D}\end{pmatrix}.$
We can compute $\textbf{M}^{-1}$ using (see here):
$\textbf{M}^{-1}=\begin{pmatrix}\textbf{A}^{-1}+\textbf{A}^{-1}\textbf{B}\textbf{S}^{-1}\textbf{C}\textbf{A}^{-1}&-\textbf{A}^{-1}\textbf{B}\textbf{S}^{-1}\\\\ -\textbf{S}^{-1}\textbf{C}\textbf{A}^{-1}&\textbf{S}^{-1}\end{pmatrix}$, where $\textbf{S} = \textbf{D}-\textbf{C}\textbf{A}^{-1}\textbf{B}$
How would one get $\textbf{S}^{-1}$ (i.e. inverting the Schur complement of $\textbf{A}$)?
One way would be to use Woodbury matrix identity, but we would end up with "the chicken or the egg dilemma", since $\textbf{S}^{-1}= \textbf{D}^{-1}+\textbf{D}^{-1}\textbf{C}\textbf{T}^{-1}\textbf{B}\textbf{D}^{-1}$ and $\textbf{T}^{-1}= \textbf{A}^{-1}+\textbf{A}^{-1}\textbf{B}\textbf{S}^{-1}\textbf{C}\textbf{A}^{-1}$ (i.e. we need $\textbf{T}^{-1}$ to compute $\textbf{S}^{-1}$ and we need $\textbf{S}^{-1}$ to compute $\textbf{T}^{-1}$).
Is there a way around that?