# Is there a formula for the inverse of this bordered matrix?

Suppose I have a matrix $\mathbf{H}$ of size $n\times n$, and that I know its inverse $\mathbf{W}=\mathbf{H}^{-1}$.

Then I add a column and a row to $\mathbf{H}$ to obtain a new matrix $\mathbf{G}$. That is $\mathbf{G}$ is given by $$\mathbf{G}=\left( \begin{array}{c|c} r_1 & \begin{array}{ccc} r_2 & \cdots & r_n \end{array} \\ \hline \begin{array}{c} c_2 \\ \vdots \\ c_n \end{array} & {\Huge{\mathbf{H}}} \end{array} \right)$$ Is there a relation between $\mathbf{W}$ and $\mathbf{G}$ and $\mathbf{G}^{-1}$?

• Nice use of nested arrays to format the expanded matrix! – hardmath Aug 9 '15 at 21:47

The block matrix inverse formula is celebrated: \begin{equation*} G^{-1} = \begin{pmatrix} a & x \\ y & H \end{pmatrix}^{-1} = \begin{pmatrix} \xi^{-1} & -\xi^{-1}xH^{-1} \\ -H^{-1}y\xi^{-1} & H^{-1} + H^{-1}y\xi^{-1}xH^{-1} \end{pmatrix} \end{equation*} where $\xi = a - xH^{-1}y$, known as the Schur complement of $H$.
• Such is also called a "bordered matrix" in the literature (i.e. adding a "border" of new row and column to an existing matrix $H$). – hardmath Aug 9 '15 at 21:37