Prove that the matrix
$$\begin{bmatrix} 1 & 1 &\cdots &1 \\ \cos a& \cos 2a &\cdots &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2na \\ \vdots & \vdots & \ddots &\vdots \\ \cos(n-1)a &\cos 2(n-1)a &\cdots &\cos(n-1)na \end{bmatrix}$$
whereas \( a= \dfrac{2\pi}{n} \) is invertible.
Solution:
$$\begin{bmatrix} 1 & 1 &\cdots &1 \\ \cos a& \cos 2a &\cdots &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2na \\ \vdots & \vdots & \ddots &\vdots \\ \cos(n-1)a &\cos 2(n-1)a &\cdots &\cos(n-1)na \end{bmatrix}$$
whereas \( a= \dfrac{2\pi}{n} \) is invertible.
Solution:
