Eigenvalue perturbation under sparse perturbations

Question

Let $A \in \{0,1\}^{n \times n}$ be an irreducible matrix whose entries are in $\{0,1\}$, and let $\lambda_1(A)$ be the eigenvalue with the largest magnitude. By Perron–Frobenius theorem, we know that $\lambda_1(A) \in \mathbb{R}$.

Now, define the matrix $B = \{0,1\}^{n \times n}$ to be a matrix satisfying the following conditions:

1. $B$ has at most $t$ ones, where $t$ should be regarded as a small constant, e.g. 2.
2. If $B_{i,j} = 1$, then $A_{i,j} = 0$.
3. $A+B$ is irreducible.

I am interested in estimating $\lambda_1(A+B)$. The way I had in mind is the following (basically using some perturbation theory). First, define the function $f : [0,1] \rightarrow \mathbb{C}$ by $f(\epsilon) = \lambda_1(A + \epsilon B)$. Then, it might be possible to write $f$ as a power series around $0$, and use this series up to some order as an estimation.

All of this is pretty standard, but it is usually used in order to estimate $f$ around $0$, whereas I am interested in estimating $f$ at $1$. The reason I thought this method might still work is that $B$ is very sparse, and so even when $\epsilon = 1$, the perturbation is still "small" in some sense. Are there any known results of this type?

EDIT: as shown in an answer, the conjecture below is wrong. The question that still remains is estimating $\lambda_1(A+B)$.

What seem to be the case is (and if $A$ and $B$ are symmetric, it seems even more plausible) that $\lambda_1(A+B) \approx \lambda_1(A) + O(\frac{t}{n})$.

Answer 1

The conjecture is not true. Let $A_{ij}=1$ iff $i=j\pm 1$. Let $B$ be diagonal, with one nonzero entry in the $\lfloor n/2\rfloor$-th position. Then, $\lambda_1(A)<2$. (If we had instead taken $A_{ij}=1$ iff $i=j\pm 1 \mod n$ then $\lambda_1(A)=2$). Meanwhile, $\lambda_1(A+B)$ is monotonically increasing in $n$, and one may verify explicitly that $\lambda_1(A+B)$ is $>2$ for large enough $n$, indeed fairly small $n$ suffice.