Finding spectral radius of matrix without computing characteristic polynomial

Question

Let

$$A=\begin{pmatrix}3&1&1\\1&2&1\\0&1&2\end{pmatrix}.$$

I am asked to find its spectral radius, i.e., $$\rho(A) = \max \left\{ |\lambda| : \lambda \text{ eigenvalue of }A \right\}$$ without using the characteristic polynomial. I have tried Gershgorin circles, bounding by norms ($\rho(A)=\inf_{\|\cdot\|}{\| A \|}$), etc. I only got $\rho(A)\leq4$ with these methods. Any ideas?

(Practice exam of Numerical-Methods for Algebra, 2nd Grade in Mathematics).

Answer 1

I found a solution, I post it in order to mark the question as answered.

As shown here, since $\sum_{i=1}^3a_{ij}=4$ for $j=1,2,3$, we have that 4 is eigenvalue of $A$. Therefore, $$\rho(A)\geq4$$ and we conclude $\rho(A)=4$.