Determinant 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 the determinant of $C_n$ is given by:
 * $\det \left({C_n}\right) = x^{n-1} \left ({x + n y}\right)$

Proof
Take the determinant $\det \left({C_n}\right)$:


 * $\det \left({C_n}\right) = \begin{vmatrix}

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

Subtract column $1$ from columns $2$ to $n$.

From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:


 * $\det \left({C_n}\right) = \begin{vmatrix}

x + y & -x & -x & \cdots & -x \\ y & x & 0 & \cdots & 0 \\ y & 0 & x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & 0 & 0 & \cdots & x \end{vmatrix}$

Add rows $2$ to $n$ to row $1$.

Again, from Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:


 * $\det \left({C_n}\right) = \begin{vmatrix}

x + n y & 0 & 0 & \cdots & 0 \\ y & x & 0 & \cdots & 0 \\ y & 0 & x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & 0 & 0 & \cdots & x \end{vmatrix}$

This is now the determinant of a (lower) triangular matrix.

From Determinant of Triangular Matrix, it follows immediately that:
 * $\det \left({C_n}\right) = x^{n-1} \left ({x + n y}\right)$