Sum of Elements in Inverse of Cauchy Matrix

Theorem
Let $C_n$ be the Cauchy matrix of order $n$ given by:


 * $C_n = \begin{bmatrix}

\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_m + y_1} & \dfrac 1 {x_m + y_2 } & \cdots & \dfrac 1 {x_m + y_n} \\ \end{bmatrix}$

Let $C_n^{-1}$ be its inverse, from Inverse of Cauchy Matrix:
 * $b_{i j} = \dfrac {\displaystyle \prod_{k \mathop = 1}^n \left({x_j + y_k}\right) \left({x_k + y_i}\right)} {\displaystyle \left({x_j + y_i}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \left({y_i - x_k}\right)}\right)}$

The sum of all the elements of $C_n^{-1}$ is:
 * $\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$

Proof
See Invertible Matrix Sum of Elements.