Suppose to have a symmetric and positive definite matrix $\boldsymbol{\Sigma}$ and suppose to know its inverse $\boldsymbol{\Sigma}^{-1}$.
Let $\boldsymbol{\Sigma}_{+}= \left( \begin{array}{cc} \boldsymbol{\Sigma} & \mathbf{b} \\ \mathbf{b}^{T}& b_1 \\ \end{array}\right)$
and
$\boldsymbol{\Sigma}= \left( \begin{array}{cc} \boldsymbol{\Sigma}_{-} & \mathbf{c} \\ \mathbf{c}^{T}& c_1 \\ \end{array}\right)$
How can I find $\boldsymbol{\Sigma}_{-}^{-1}$ and $\boldsymbol{\Sigma}_{+}^{-1}$?
Let $d:=b_1-b^T\Sigma^{-1}b\neq 0$. The block inversion formula gives $$ \Sigma_+^{-1}=\pmatrix{\Sigma^{-1}+\frac{1}{d}\Sigma^{-1}bb^T\Sigma^{-1}&-\frac{1}{d}\Sigma^{-1}b\\-\frac{1}{d}b^T\Sigma^{-1}&\frac{1}{d}}. $$ Note that $d\neq 0$ is the necessary and sufficient condition for the invertibility of $\Sigma_+$. Actually, if $\Sigma$ is SPD, $\Sigma_+$ is SPD as well iff $d>0$.
Similarly, we can write $$ \Sigma^{-1}=\pmatrix{\Sigma_-^{-1}+\frac{1}{e}\Sigma_-^{-1}cc^T\Sigma_-^{-1}&-\frac{1}{e}\Sigma_-^{-1}c\\-\frac{1}{e}c^T\Sigma_-^{-1}&\frac{1}{e}}=:\pmatrix{X&x\\x^T&\xi}. $$ Since $\Sigma$ is SPD, $e>0$. We have $$ \Sigma_-^{-1}=X-\frac{1}{e}\Sigma_-^{-1}cc^T\Sigma_-^{-1}=X-e\left(-\frac{1}{e}\Sigma_-^{-1}c\right)\left(-\frac{1}{e}\Sigma_-^{-1}c\right)^T=X-\frac{1}{\xi}xx^T. $$