evaluate the determinant of a matrix whose rows are all permutations of each other

Question

Let $n\ge 1$. Consider an $n\times n$ matrix where the $k$th row of the matrix has entries $k, k+1,\cdots, n, 1,2,\cdots, k-1$ in that order. Find the determinant of this matrix. For instance when $n=2$ the matrix is $\begin{pmatrix}1 & 2\\ 2 & 1\end{pmatrix}$

Let $R_i$ be the ith row and $C_i$ be the ith column. Performing the operation $C_i\mapsto C_i - C_{1}$ for $2\leq i\leq n$ yields a matrix whose first column consists of the entries $1,2,\cdots, n$, and the kth column has $-(n-1)$ as the $n-k+2$th entry and $1$'s elsewhere. Then perform the operations $R_i\mapsto R_i-R_1$ for $2\leq i\leq n$ to get a matrix where the first column equals $(1,1,2,\cdots, n-1)^T$ and the kth column starts with a 1, has a $-1$ in the $n-k+2$th entry and zeroes elsewhere.

I'm not sure how to proceed from here. It might be possible to factor out a common factor to compute the determinant.

Edit: here's a link to a relevant post about anti-circulant matrices: Eigenvalues of anti-circulant matrix.

Answer 1

Denote the matrix by $A$. Then $$\begin{aligned} \det A=&\left| \begin{matrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{matrix}\right| \\ =\dfrac{1}{2}n(n+1)&\left| \begin{matrix} 1 & 1 & 1 & \cdots & 1 & 1 \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{matrix}\right| \\ =\dfrac{1}{2}n(n+1)&\left| \begin{matrix} 1 & 0 & 0 & \cdots & 0 & 0 \\ 2 & 1 & 2 & \cdots & n-2 & -1 \\ 3 & 1 & 2 & \cdots & -2 & -1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ n-1 & 1 & 2-n & \cdots & -2 & -1 \\ n & 1-n & 2-n & \cdots & -2 & -1 \end{matrix}\right| \\ =\dfrac{1}{2}n(n+1)&\left| \begin{matrix} 1 & 2 & \cdots & n-2 & -1 \\ 1 & 2 & \cdots & -2 & -1 \\ \vdots & \vdots & & \vdots & \vdots \\ 1 & 2-n & \cdots & -2 & -1 \\ 1-n & 2-n & \cdots & -2 & -1 \end{matrix}\right| \\ =\dfrac{1}{2}n(n+1)&\left| \begin{matrix} n & n & \cdots & n & 0 \\ n & n & \cdots & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ n & 0 & \cdots & 0 & 0 \\ 1-n & 2-n & \cdots & -2 & -1 \end{matrix}\right| \\ =\dfrac{1}{2}n(n+1)&\cdot(-1)^{\frac{1}{2}n(n-1)}n^{n-2}=(-1)^{\frac{1}{2}n(n-1)}\cdot\dfrac{n+1}{2}\cdot n^{n-1}.\\ \end{aligned}$$