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