Block inverse of symmetric matrices

Question

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.

Answer 1

Certainly, one can use the bordering method for this (a special case of the usual formula for block inversion):

$$\begin{pmatrix}\mathbf A&\mathbf \delta\\\mathbf \delta^\top&Z\end{pmatrix}^{-1}=\begin{pmatrix}\mathbf A^{-1}+\frac{\mathbf A^{-1}\mathbf \delta\mathbf \delta^\top\mathbf A^{-1}}{\mu}&-\frac{\mathbf A^{-1}\mathbf \delta}{\mu}\\-\frac{\mathbf \delta^\top\mathbf A^{-1}}{\mu}&\frac1{\mu}\end{pmatrix}$$

where $\mathbf \delta^\top=(X\quad Y)$ and $\mu=Z-\mathbf \delta^\top\mathbf A^{-1}\mathbf \delta$.