<s>[INST] Determinant of $ n \times n$ matrix and its characteristic polynomial. Suppose, $M_4, M_5,..M_n$ is as follows then determinant and characteristic polynomial of $M_n$. $M_4=\left(
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
\end{array}
\right),M_5=\left(
\begin{array}{ccccc}
 0 & 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 & 1 \\
 1 & 0 & 0 & 0 & 1 \\
 1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 0 & 0 \\
\end{array}
\right),M_6=\left(
\begin{array}{cccccc}
 0 & 0 & 1 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 & 1 & 1 \\
 1 & 0 & 0 & 0 & 1 & 1 \\
 1 & 1 & 0 & 0 & 0 & 1 \\
 1 & 1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 1 & 0 & 0 \\
\end{array}
\right), \quad M_8=\left(
\begin{array}{cccccccc}
 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\
 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\
\end{array}
\right)$
 [/INST] If $J$ is the $n\times n$ matrix with all entries 1, then $M_n=J-I-C_n$, where $C_n$ is the adjacency matrix of the cycle. The all-ones vector is an eigenvector for $C_n$ with eigenvalue 2. Since $C_n$ is symmetric, if $\lambda\ne2$ is an eigenvalue for $C_n$ with eigenvector $z$, then $z$ must be orthogonal to the all-ones vector. Hence $Jz=0$ and therefore
$$
 M_nz = (J-I-C_n)z = -(1+\lambda)z.
$$
Therefore the eigenvalues of $M_n$ are $n-3$ and the numbers $-1-\lambda$, where $\lambda$ runs over the eigenvalues of $C_n$ not equal to 2, i.e., the numbers $2\cos(2\pi k/n)$ for $k=1,\ldots,n-1$.
</s>