# Block inverse of symmetric matrices

Let us assume we have a symmetric $n \times n$ matrix $A$. We know the inverse of $A$. Let us say that we now add one column and one row to $A$, in a way that the resulting matrix ($B$) is an $(n+1) \times (n+1)$ matrix that is still symmetric.

For instance,

$A = \begin{pmatrix}a & b \\b & d \\\end{pmatrix}$

and

$B = \begin{pmatrix}a & b & X \\b & d & Y \\X & Y & Z\end{pmatrix}$

Given that I know $A^{-1}$, is there any way of using this information to find $B^{-1}$ without having to compute this latter inverse from scratch? If an exact solution is not possible, approximations would also help.

Thanks,
Bruno

P.S. in case it makes any difference, both $A$ and $B$ are covariance matrices.

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The Schur Complement might be what you are looking for. en.wikipedia.org/wiki/Schur_complement –  copper.hat Aug 14 '12 at 3:34
I don't know if there exists a solution that just involves $A^{-1}$, but I have seen something similar. –  Tunococ Aug 14 '12 at 3:36
Cool, thanks for the pointers! I'll take a look at those links. –  Bruno Aug 14 '12 at 14:09

$$\begin{pmatrix}\mathbf A&\mathbf \delta\\\mathbf \delta^\top&Z\end{pmatrix}^{-1}=\begin{pmatrix}\mathbf A^{-1}+\frac{\mathbf A^{-1}\mathbf \delta\mathbf \delta^\top\mathbf A^{-1}}{\mu}&-\frac{\mathbf A^{-1}\mathbf \delta}{\mu}\\-\frac{\mathbf A^{-1}\mathbf \delta}{\mu}&\frac1{\mu}\end{pmatrix}$$
where $\mathbf \delta^\top=(X\quad Y)$ and $\mu=Z-\mathbf \delta^\top\mathbf A^{-1}\mathbf \delta$.
$\mu$ is the Schur Complement of $A$. –  copper.hat Aug 14 '12 at 3:48