How to find the inverse of a symmetric tridiagonal Toeplitz matrix? Let $a, b>0$, and the matrix $A_{n \times n}$ and such
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix}$$
Find the inverse $A^{-1}$.

My idea: $$A^{-1}=\dfrac{A^{*}}{|A|}$$
and let $|A|=D_{n}$, then we have
$$D_{n}=aD_{n-1}-b^2D_{n-2}$$
so
$$D_{n}=\begin{cases}
(n+1)\left(\dfrac{a}{2}\right)^n,a^2=4b^2\\
\dfrac{(a+\sqrt{a^2-4b^2})^{n+1}-(a-\sqrt{a^2-4b^2})^{n+1}}{2^{n-1}\sqrt{a^2-4b^2}}&a^2\neq 4b^2
\end{cases}$$
and then I fell very ugly, do you have other methods?
Thank you

Notice that
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix} = a \begin{bmatrix}
1&b/a&0&\cdots&0&0\\
b/a&1&b/a&\cdots&0&0\\
0&b/a&1&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b/a&1&b/a\\
\cdots&\cdots&\cdots&\cdots&b/a&1
\end{bmatrix} =: aX.$$
So, $A^{-1} = \frac{1}{a} X^{-1}$.
To find $X^{-1}$, apply this answer. You can also find some references here.
In case you want to search further, your matrix is tridiagonal (a special kind of band) Toeplitz matrix.
