Sum of Elements in Inverse of Combinatorial Matrix

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


 * $C_n = \begin{bmatrix}

x + y & y & \cdots & y \\ y & x + y & \cdots & y \\ \vdots & \vdots & \ddots & \vdots \\ y & y & \cdots & x + y \end{bmatrix}$

Let $C_n^{-1}$ be its inverse, from Inverse of Combinatorial Matrix:
 * $b_{i j} = \dfrac {-y + \delta_{i j} \left({x + n y}\right)} {x \left({x + n y}\right)}$

where $\delta_{i j}$ is the Kronecker delta.

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

Proof
All $n^2$ elements of $C_n^{-1}$ have a term $\dfrac {-y} {x \left({x + n y}\right)}$.

Further to this, the $n$ elements on the main diagonal contribute an extra $\dfrac {x + n y} {x \left({x + n y}\right)}$ to the total.

Hence: