Give a general formula in terms of $n$ for the determinant of the following matrix.

Question

Let $M_n$ denote the $n$ x $n$ matrix over $\mathbb{R}$ of which the entry in the $i$-th row and the $j$-th column equals $1$ if $|i-j|\leq 1$ and $0$ otherwise. For example: $M_6=$ \begin{pmatrix} 1& 1& 0& 0& 0& 0\\ 1& 1& 1& 0& 0& 0\\ 0& 1& 1& 1& 0& 0\\ 0& 0& 1& 1& 1& 0\\ 0& 0& 0& 1& 1& 1\\ 0& 0& 0& 0& 1& 1 \end{pmatrix} Give (with proof) a general formula in terms of $n$ for the determinant of $M_n$. Thanks in advance.

Answer 1

solve the recurrence relations $D_n = D_{n-1} - D_{n-2}$ with the initial condition $D_1 = 1 \mbox{ and} D_2 = 0.$ try $D_n = \lambda^n.$ the indicial equation is $\lambda^2 - \lambda + 1 = 0$ whose roots are $\lambda = {1 \pm i\sqrt 3 \over 2}.$ sso $D_n = k (\cos(n\pi/3 + \phi).$ requiring $D_2 = 0$ gives $\phi = -\pi/6$ and $D_! = 1$ shows $k = \frac{2}{\sqrt 3}.$

so the answer is $$D_n = \frac{2}{\sqrt 3}\cos(n\pi/3 - \pi/6).$$

i hope i did not make too many errors.