How to invert a block tridiagonal matrix?

Question

I'm dealing with the inversion of a pretty large block-matrix, whereby I am only interested in a particular block of the inverse. I would like to avoid the inversion of the whole matrix and I wonder whether there exist a convenient inversion rule for my case (as with, e.g., block diagonal matrices or Toeplitz matrices).

The matrix has the form $M = \begin{pmatrix} A_1 & -I_n & 0 & ... & ... & 0 \\ -I_n & A_2 & -I_n & 0 & ... & 0 \\ 0 & -I_n & A_3 & -I_n & ... & 0 \\ & &... &... & ...& \\ & & &... & ...& ... \\ & & & &-I_n & A_T\end{pmatrix} $,

where $A_i$ has dimension $n \times n~ \forall i$. The block of $M^{-1}$ that I am interested in is the last $n \times n$ block (corresponding to $A_T$ in $M$). I would be grateful for any suggestions! Thanks in advance!

Answer 1

Inverting the matrix is probably unnecessary. In most cases, factorization is preferred. Consider reading John Cook's post on this topic.

Answer 2

Sorry, I think I didn't spend enough time on the topic before I posted this question. In this case, indeed I have to invert the matrix. The solution (based on https://iopscience.iop.org/article/10.1088/1749-4699/5/1/014009/meta) is the following:

Let $M_{i}:=M_{[(i-1)n+1:n],[(i-1)n+1:n]}$, such that, e.g. $M_T = A_T$. Then I obtain my desired block of the inverse as follows:

$(M^{-1})_{T}=[A_T-Y_T]^{-1}$, where $Y_i$ is defined recursively as $Y_1 = 0$, $Y_i = (A_{i-1}-Y_{i-1})^{-1}$ for $i=2:T$.