Determinant of complex matrix with almost constant lines Let $0\neq c\in\mathbb{C}$. Take the matrix 
$$A_C=\begin{pmatrix}
n&amp;c&amp;\dots&amp;c&amp;c \\
c&amp;n&amp;c &amp;\dots &amp; c\\
c &amp;c &amp; n &amp;c &amp;\dots\\
\vdots &amp;\vdots&amp;\vdots&amp;\ddots &amp; \vdots \\
c &amp; \dots &amp;\dots &amp;c&amp; n
\end{pmatrix}\in \mathbb{C}^{n \times n}.$$

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 $c\neq n$.
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.
See if its so take it in eigenvalues. We know that sum of eigenvalues is the trace of matrix  as $n \neq  c $ is not equal to $0$ so we can easily conclude that $det(A) \neq 0$ you can also do that as $det(A)=π \lambda_i$ and as $n\neq 0$ so the determinant is also non-zero and its done.