linear algebra - Determinant of complex matrix with almost constant lines

Let $0\neq c\in\mathbb{C}$. Take the matrix
$$A_C=\begin{pmatrix}
n&c&\dots&c&c \\
c&n&c &\dots & c\\

c &c & n &c &\dots\\
\vdots &\vdots&\vdots&\ddots & \vdots \\
c & \dots &\dots &c& n
\end{pmatrix}\in \mathbb{C}^{n \times n}.$$

I want to show that $A_C$ invertible. It seems like one can compute the determinant explicitly but I messed with it a bit.

How can I prove $\det A_C\neq 0$?

EDIT: We assume $c\neq n$.

Answer

This is a circulant matrix and as such has normalized eigenvectors
$$v_j=\frac{1}{\sqrt{n}}(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$
where
$$\omega_j=exp\left(\frac{2\pi i j}{n}\right)$$
The eigenvalues are
$$\lambda_j=n+c\omega_j+c\omega_j^2+\cdots+c\omega_j^{n-1}$$
Taking the product of the eigen values gives the determinant. Now since the $\omega_j$ are roots of unity we have
$$\sum_{k=1}^{n-1}\omega_j^k=-1$$
which gives

$\lambda_j=n-c$. Since we have assumed that $n\neq c$ no eigenvalue is equal to zero. Hence the determinant is non-zero.