Let us consider the matrix $A \in \mathbb R^{2N \times 2N}$ defined as \begin{equation} A = \begin{pmatrix} I & I+\Lambda_{12} \\ I + \Lambda_{21} & I\end{pmatrix}, \end{equation} where $I$ is the $N\times N$ identity matrix, and the linear system \begin{equation} A\begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}, \end{equation} where $v_1$, $v_2$, $b_1$ and $b_2$ are vectors of $\mathbb R^N$. Assume that $\Lambda_{12}$ and $\Lambda_{21}$ are tridiagonal and such that $A$ is invertible.
Is it possible to derive a bound on $\|v_1 - v_2\|$ as a function of $\|\Lambda_{12}\|$, $\|\Lambda_{21}\|$, $\|\Lambda_{12} \pm \Lambda_{21}\|$ (or their inverses) and $\|b_1 - b_2\|$? If yes, under which conditions?
Update: Case $N = 1$
In the case $N = 1$ it is possible to find explicitly \begin{align} v_1 &= \frac1{\det A}(b_1-(1+\Lambda_{12})b_2)\\ v_2 &= \frac1{\det A}(b_2-(1+\Lambda_{21})b_1), \end{align} where $\det A = -(\Lambda_{12} + \Lambda_{21} + \Lambda_{12}\Lambda_{21})$. After a couple of applications of the triangle inequality, one gets \begin{equation} |v_1 - v_2| \leq \frac{(2+|\Lambda_{12}|)\,|b_1-b_2|+|\Lambda_{12} - \Lambda_{21}|\, |b_1|}{|\det A|}. \end{equation} Does this generalize to higher dimensions ?
I answer my own question. Finding $v_1$ and $v_2$ by substitution, and doing a couple of triangle inequalities, one finds \begin{align} \|v_1 - v_2\| &\leq \|Q_2^{-1} \|\,\|\Lambda_{12}-\Lambda_{21}\|\,\|b_2\| + \|Q_2^{-1} - Q_1^{-1}\|\,\|\Lambda_{21}\|\,\|b_1\| \\ &\quad + \|Q_2^{-1}+Q_1^{-1}+Q_2^{-1}\Lambda_{21}\|\,\|b_2-b_1\|, \end{align} where \begin{align} Q_1 &= \Lambda_{21}+\Lambda_{12}+\Lambda_{21}\Lambda_{12}, \\ Q_2 &= \Lambda_{21}+\Lambda_{12}+\Lambda_{12}\Lambda_{21}. \end{align} Note that in case $N = 1$, we have $Q_1 = Q_2$ and therefore this bound reduces to the one given in the question itself. Moreover, working a bit more, one gets $$ \|Q_2^{-1} - Q_1^{-1}\| \leq \|\Lambda_{12}\Lambda_{21} - \Lambda_{21}\Lambda_{12} \|\,\|Q_1^{-1}\|\,\|Q_2^{-1}\|, $$ which may (or may not) be more explicit.