Determinant of Combinatorial Matrix

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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:

$\det \left({C_n}\right) = x^{n-1} \left ({x + n y}\right)$

$\blacksquare$


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