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}$

Then its inverse $C_n^{-1} = \sqbrk b_n$ can be specified as:
 * $b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$

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

Proof
From the definition of the combinatorial matrix:
 * $C_n = x \mathbf I_n + y \mathbf J_n$

where:
 * $\mathbf I_n$ is the unit matrix of order $n$
 * $\mathbf J_n$ is the square ones matrix of order $n$.

From Square of Ones Matrix we have $\mathbf J_n^2 = n \mathbf J_n$.

Hence:

So we have found a specification for the matrix which, when multiplied by $C_n$, yields $\mathbf I_n$.

By using the identities $\mathbf I_n = \sqbrk {\delta_{i j} }_n$ and $\mathbf J_n = \sqbrk 1_n$ we obtain the stated result:
 * $b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$