IMC 2007 · Day 2 · P10

Notice that $A = B^2$, with $b_{ij} = \begin{cases} 1 & \text{if } i - j \equiv \pm 1 \pmod n \\ 0 & \text{otherwise} \end{cases}$. So it is sufficient to find $\det B$.

To find $\det B$, expand the determinant with respect to the first row, and then expad both terms with respect to the first column.

$$\det B = \begin{vmatrix} 0 & 1 & & & 1 \\ 1 & 0 & 1 & & \\ & 1 & 0 & \ddots & \\ & & \ddots & \ddots & 1 \\ 1 & & & 1 & 0 \end{vmatrix} = - \begin{vmatrix} 1 & 1 & & \\ & 0 & \ddots & \\ & \ddots & \ddots & 1 \\ 1 & & 1 & 0 \end{vmatrix} + \begin{vmatrix} 1 & 0 & 1 & \\ & 1 & \ddots & \\ & & \ddots & 1 \\ 1 & & 1 & 0 \end{vmatrix}$$ $$= - \left( \begin{vmatrix} 0 & 1 & & \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ 1 & 0 & 1 & \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & & & \\ & \ddots & & \\ & & \ddots & \\ 1 & 0 & 1 & \ddots \end{vmatrix} \right) + \left( \begin{vmatrix} 1 & 0 & 1 & \\ & \ddots & \ddots & \\ 1 & 0 & \ddots & 1 \\ & 1 & & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 1 & & \\ 1 & 0 & \ddots & \\ & \ddots & \ddots & 1 \\ & & 1 & 0 \end{vmatrix} \right) = -(0 - 1) + (1 - 0) = 2,$$ since the second and the third matrices are lower/upper triangular, while in the first and the fourth matrices we have $\mathrm{row}_1 - \mathrm{row}_3 + \mathrm{row}_5 - \cdots \pm \mathrm{row}_{n-2} = \bar{0}$.

So $\det B = 2$ and thus $\det A = 4$.