Here are some formula for matrix inversion.
Lemma 1: For a block matrix $latex M=\begin{pmatrix}A & B \\C &D\end{pmatrix}$,
$latex M^{-1}=\begin{pmatrix}(A-B D^{-1} C)^{-1}& -A^{-1}B(D-CA^{-1}B)^{-1}\\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}\end{pmatrix}$
$latex =\begin{pmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}& -A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{pmatrix}$,
where $latex S=D-CA^{-1}B$ is basically the Schur's
complement of block $latex A$.
Proof: Let $latex M^{-1}=\begin{pmatrix}E&F\\G&H\end{pmatrix}$, $latex M M^{-1}=1$ gives us
$latex AE+BG=I$
$latex AF+BH=0$
$latex CE+DG=0$
$latex CF+DH=I$
From the four equations, we have
$latex E=(A-BD^{-1}C)^{-1}$
$latex F=-A^{-1}B(D-CA^{-1}B)$
$latex G=-D^{-1}C(A-BD^{-1}C)^{-1}$
$latex H=(D-CA^{-1}B)^{-1}$
And similarly from $latex M^{-1}M=I$, we have
$latex E=(A-BD^{-1}C)^{-1}$
$latex F=-(A-BD^{-1}C)^{-1}BD^{-1}$
$latex G=-(D-CA^{-1}B)^{-1}CA^{-1}$
$latex H=(D-CA^{-1}B)^{-1}$
Together, they show the first inequality. Also note that $latex AE+BG=I$ and thus $latex E=A^{-1}-A^{-1}BG$. Substituting $latex G$ into above shows the second equality. $latex \Box$.
Lemma 2 (Matrix Inversion Formula/Woodbury Matrix Identity):
$latex (A+BDC)^{-1}=A^{-1}-A^{-1}B(D^{-1}+CA^{-1}B)^{-1}CA^{-1}$
Proof: From the previous proof, we have $latex E=(A-BD^{-1}C)^{-1}=A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}$. We get the identity immediately as we flip the sign of $latex B$. $latex \Box$