TITLE: Invert a symmetric partition matrix
QUESTION [2 upvotes]: How do I invert this matrix?
$$ X = \begin{bmatrix} 1 & \mathbf{x'} \\ \mathbf{x} & I \end{bmatrix} $$
If $X$ is $n \times n$, then $I$ is the $(n-1) \times (n-1)$ identity and $x \in \mathbb R^{n-1}$ is a vector.
Thanks

REPLY [2 votes]: With the Shermann-Morrison formula and the section "Block matrix inversion" at 
https://en.wikipedia.org/wiki/Block_matrix we can infer the following.
$X^{-1} = \begin{pmatrix} 1 - \frac{x^\intercal x}{1-x^\intercal x} & -x^\intercal - \frac{x^\intercal x}{1-x^\intercal x} x^\intercal \\ -x - \frac{x^\intercal x}{1-x^\intercal x} x & I_{n-1} + \frac{1}{1-x^\intercal x} x x^\intercal\end{pmatrix}$
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