Let $t \in [0,1/6]$, $n$ be a positive integer, then how to compute the determinant of the matrices as
$$A=\left[ \begin{matrix} 1-6t&2t&t&t&0&0&\dots&t&t\\ 2t&1-6t&t&t&0&0&\dots&t&t\\ t&t&1-6t&2t&t&t&\dots&0&0\\ t&t&2t&1-6t&t&t&\dots&0&0\\ 0&0&t&t&1-6t&2t&\dots&0&0\\ 0&0&t&t&2t&1-6t&\dots&0&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots \\ t&t&0&0&0&0&\dots&1-6t&2t\\ t&t&0&0&0&0&\dots&2t&1-6t \end{matrix} \right]_{2n \times 2n}?$$
Let $V=\frac1{\sqrt{2}}\pmatrix{1&1\\ -1&1}$. Then \begin{cases} V^\ast\pmatrix{1-6t&2t\\ 2t&1-6t}V&=\pmatrix{1-8t&0\\ 0&1-4t},\\ V^\ast\pmatrix{t&t\\ t&t}V&=\pmatrix{0&0\\ 0&2t}. \end{cases} Hence $\det(A)=(1-8t)^n\det(C)$, where $C$ is the $n\times n$ circulant matrix $$ \pmatrix{ 1-4t&2t&0&\cdots&0&2t\\ 2t&\ddots&\ddots&\ddots&&0\\ 0&\ddots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&0\\ 0&&\ddots&\ddots&\ddots&2t\\ 2t&0&\cdots&0&2t&1-4t}. $$ Using the determinant formula for circulant matrices, you may obtain the final answer easily.