Let $\bf A$ be an $n \times n$ non-singular matrix over $\mathbb{F}$. 
Let $x$ be a non-zero element of $\mathbb{F}$.
Suppose that ${\bf 1}\_{n}$ is a symbol for the all-one vector of length $n$ over $\mathbb{F}$. 
Now consider the following $(n+1) \times (n+1)$ matrix and assume that $\bf B$ is a invertible matrix over $\mathbb{F}$. 
$$
{\bf B}=
\left(
 \begin{array}{cc}
x &{\bf 1}^T\_{n} \\
 {\bf 1}\_{n} & {\bf A}
 \end{array}
 \right).
 $$ 


My question: Is there a closed-form expression for the inverse of $\bf B$, denoted with ${\bf B}^{-1}$?


Thanks for any help.



A closed-form expression for the inverse of a block-matrix
Say that
$$B^{-1}=:\begin{pmatrix} b & X^T \\ Y & M \end{pmatrix}.$$
Then using *Schur's complement formula* (thanks to Nathaniel), $b=(x-{\bf1}^TA^{-1}{\bf1})^{-1}$ and $M=(A-x^{-1}{\bf11}^T)^{-1}$. From this, you can compute the vectors
$$Y=-x^{-1}M{\bf1},\qquad X^T=-b{\bf1}^TA^{-1}.$$


