# A closed-form expression for the inverse of a block-matrix

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.

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 $$Y=-x^{-1}M{\bf1},\qquad X^T=-b{\bf1}^TA^{-1}.$$