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

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.

• This looks like an anti-circulant matrix to me. The determinant of such a matrix is known in terms of roots of unity. You might want to try that. Oct 23, 2022 at 21:47

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}
• The second is $R_1\mapsto \sum_{k=1}^{n}R_k$, the fifth is $R_k\mapsto -R_n+R_k, k=1,…,n-1$, and the sixth is because the $n$th column only has the $-1$.