Let assume that I have an $n\times n$ symmetric matrix $A$ and I know $A^{-1}$. Now, I have a new matrix $$M = \begin{pmatrix} A & b \\ b^T & c \end{pmatrix},$$ where $b$ is a vector and $c$ is a single element which is $0$.
My goal is to find $\sum(M^{-1}) - \sum(A^{-1})$ in an efficient way (in low computational cost without inverting $M$), where $\sum(A)$ refers to sum of all the elements of A.
From bordering method, I know that $M^{-1}$ could be found easily: $$M^{-1} = \begin{pmatrix} A^{-1}+\frac{1}{k}A^{-1}bb^TA^{-1} & -\frac{1}{k}A^{-1}b \\ -\frac{1}{k}b^tA^{-1} & \frac{1}{k} \end{pmatrix},$$ where $k=c-b^TA^{-1}b$.
Any help will be greatly appreciated. Thanks.
With $d=[\delta_1,\ldots,\delta_n]^T:=A^{-1}b$, we have $$ M^{-1}=\pmatrix{A^{-1}+\frac{1}{k}dd^T & -\frac{1}{k}d\\-\frac{1}{k}d^T&\frac{1}{k}} $$ and hence the difference between the sums of the entries of $M^{-1}$ and $A^{-1}$ is the same as the sum of entries of $$ \frac{1}{k}\pmatrix{dd^T & d \\ d^T & 1}. $$ Therefore, $$ \Sigma(M^{-1})-\Sigma(A^{-1})=\frac{1}{k}\left[\Sigma(dd^T)-2\Sigma(d)+1\right] =\frac{1}{k}(\sigma^2-2\sigma+1)=\frac{1}{k}(\sigma-1)^2, $$ where $$ \sigma:=\sum_{i=1}^n\delta_i. $$