There is a triagonal block matrix $M$ of form:
$$ M = \begin{bmatrix} A & B^T & 0 & 0 & \cdots & 0 & 0 \\ B & A & B^T & 0 & \cdots & 0 & 0 \\ 0 & B & A & B^T & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & B & A \end{bmatrix} $$ where $A, B$ are real-valued square matrices of the same size.
Also, $A$ is positive definite and symmetric. Later makes $M$ symmetric as well.
My interest is in closed-form solution for elements of $M^{-1}$.
From "Explicit inverses of some tridiagonal matrices" C.M. da Fonseca, J. Petronilho, I am aware of the closed-form solution for tridiagonal toeplitz matrix of form:
$$ T = \begin{bmatrix} a & b & 0 & 0 & \cdots & 0 & 0 \\ c & a & b & 0 & \cdots & 0 & 0 \\ 0 & c & a & b & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & c & a \end{bmatrix} $$ where $a, b, c$ - scalars.
$$ (T^{-1})_{ij} = \begin{cases} (-1)^{i+j}\frac{b^{j-i}}{\left(\sqrt{bc}\right)^{j-i+1}}\frac{U_{i-1}(d)U_{n-j}(d)}{U_{n}(d)} \quad \text{if} \quad i \le j \\ (-1)^{i+j}\frac{c^{i-j}}{\left(\sqrt{bc}\right)^{i-j+1}}\frac{U_{j-1}(d)U_{n-i}(d)}{U_{n}(d)} \quad \text{if} \quad i \gt j \end{cases} $$ where $d = \frac{a}{2\sqrt{bc}}$, $U_{k}(x)$ - Chebyshev polynomials of the second kind.
Don't see a way to extend it to block matrices though. Does anybody know how it can be done? or any alternative way?