I have the following matrix
$$A=\begin{pmatrix} b & a & 0 & 0 & \cdots & 1\\ a & b & a & 0 & \cdots & 0\\ 0 & a & b & a & \cdots & 0\\ 0 & 0 & a & b & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & a & b & a & 0\\ 0 & 0 & 0 & a & b & a\\ 1 & 0 & 0 & 0 & a & b \end{pmatrix}$$
That can be written as $A=T+u_{1}v_{1}^{T}+u_{2}v_{2}^{T}$, where T is tridiagonal. I have to compute the inverse of this matrix in terms of $T^{-1}$. I had thought that I could extend the Sherman-Morris formula, but the procedure for obtaining it cannot be applied to this case. In the statement of the problem there is a hint. It says that it is useful writting $A=L(\mathbb{I}+stuff)U$ [sic]. Where L (U) is a lower (upper) triangular matrix. I have thought that those LU could be the LU factorization of A, but when U compute $L^{-1}(u_{1}v_{1}^{T}+u_{2}v_{2}^{T})U^{-1}$ the stuff term becomes a singular matrix. Do you have any tip to solve this problem?.
You have reduced the problem to that of finding the inverse of $I + L^{-1}(u_1v_1^T + u_2v_2^T)U^{-1}$. With that in mind, apply the Woodbury matrix identity to compute $(I + PQR)^{-1}$, where $$ P = \pmatrix{L^{-1}u_1 & L^{-1}u_2}, \quad Q = I_{2 \times 2}, \quad R = \pmatrix{v_1^TU^{-1}\\v_2^TU^{-1}}. $$