Inverse of Cauchy Matrix

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


 * $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_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{bmatrix}$

Then its inverse $C_n^{-1} = \left[{b}\right]_n$ can be specified as:


 * $\begin{bmatrix} b_{ij} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \left({x_j + y_k}\right) \left({x_k + y_i}\right)} {\displaystyle \left({x_j + y_i}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \left({y_i - x_k}\right)}\right)} \end{bmatrix}$