# 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:

$b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$

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

The sum of all the elements of $C_n^{-1}$ is:

$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \dfrac n {x + n y}$

## Proof

All $n^2$ elements of $C_n^{-1}$ have a term $\dfrac {-y} {x \paren {x + n y} }$.

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

Hence:

 $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}$ $=$ $\ds n^2 \dfrac {-y} {x \paren {x + n y} } + n \dfrac {x + n y} {x \paren {x + n y} }$ $\ds$ $=$ $\ds \dfrac {-n^2 y + n \paren {x + n y} } {x \paren {x + n y} }$ $\ds$ $=$ $\ds \dfrac {-n^2 y + n x + n^2 y} {x \paren {x + n y} }$ $\ds$ $=$ $\ds \dfrac {n x} {x \paren {x + n y} }$ $\ds$ $=$ $\ds \dfrac n {x + n y}$

$\blacksquare$