TITLE: Is there a formula for the inverse of this bordered matrix?
QUESTION [5 upvotes]: Suppose I have a matrix $\mathbf{H}$ of size $n\times n$, and that I know its inverse $\mathbf{W}=\mathbf{H}^{-1}$. 
Then I add a column and a row to $\mathbf{H}$ to obtain a new matrix $\mathbf{G}$. That is $\mathbf{G}$ is given by
$$\mathbf{G}=\left(
\begin{array}{c|c}
  r_1 & \begin{array}{ccc} r_2 & \cdots & r_n \end{array} \\ \hline
\begin{array}{c} c_2 \\ \vdots \\ c_n \end{array}  & {\Huge{\mathbf{H}}}
\end{array}
\right)
$$
Is there a relation between $\mathbf{W}$ and $\mathbf{G}$ and $\mathbf{G}^{-1}$?

REPLY [5 votes]: The block matrix inverse formula is celebrated:
\begin{equation*}
G^{-1} = \begin{pmatrix}
a & x \\
y & H
\end{pmatrix}^{-1}
= \begin{pmatrix}
\xi^{-1}  & -\xi^{-1}xH^{-1} \\
-H^{-1}y\xi^{-1} & H^{-1} + H^{-1}y\xi^{-1}xH^{-1}
\end{pmatrix}
\end{equation*}
where $\xi = a - xH^{-1}y$, known as the Schur complement of $H$.