For which values of $x \in \mathbb{R}$ does the determinant of the matrix $$ M = \begin{pmatrix} x & 0 & 1 & 2 \\ 2 & x & 0 & 1 \\ 1 & 2 & x & 0 \\ 0 & 1 & 2 & x \end{pmatrix}$$ vanish?
For clarification. I know you can just calculate the determinant but I'm looking for an elegant solution.
On
The circulant answer is pretty much the whole story. In this case ($4 \times 4$) you can package it in a more elementary way. If you want to know how one might think to look at these things, check out circulants.
The column vector $[1,1,1,1]^t$ is an eigenvector for $x+3$ and $[1,-1,1,-1]^t$ is an eigenvector for $x-1.$ That gives $x=-3$ and $x=1.$ Now consider the two vectors $u=[1,0,-1,0]^t$ and $v=[0,1,0,-1]^t.$ We have $Mu=(x+1)u-2v$ and $Mv=2u+(x+1)v$ this yields a $2\times 2$ matrix with determinant $x^2+2x+5$ and gives the other two solutions $-1 \pm \sqrt{-4}.$
An alternate approach to the last two solutions is to uses the eigenvectors $[1,\pm i,1,\mp i]^t.$
This is a circulant matrix and its eigenvalues are $$ \lambda_j=c_0+c_1w^{j}+c_2w^{2j}+c_3w^{3j}, \quad j=0,1,2,3 $$ where $w$ is the primitive fourth root of $1$, i.e., $w=i$, and $$ c_0=x,\quad c_1=0, \quad c_2=1, \quad c_3=2 $$ Thus $$ \lambda_j=x+i^{2j}+2i^{3j}=x+(-1)^j+2(-i)^j, \quad j=0,1,2,3. $$ Thus, the eigenvalues are $$ x+1+2, \quad x-1-2i, \quad x+1-2, \quad x-1+2i. $$ Hence the determinant vanishes iff $$ x=1, \quad x=-3 \quad\text{or}\quad x=1\pm 2i. $$