linear algebra - Let $A=(a_{ij})$ be a $10 times 10$ matrix such that $a_{ij}=1$ for $i neq j$ and $a_{ii}=alpha +1$, Where $alpha >0$

Let $A=(a_{ij})$ be a $10 \times 10$ matrix such that $a_{ij}=1$ for $i \neq j$ and $a_{ii}=\alpha +1$, Where $\alpha >0$ . let $\lambda$ and $\mu$ be the largest and smallest eigenvalues of $A,$ respectively. If $\lambda+ \mu =24,$ then $\alpha$ equals

My Idea:

By the given information the matrix is of the form all non-diagonal elements are $1$ and diagonal elements are $\alpha +1$ but i don't how to processed further

Answer

Hint: Write $A=U+\alpha I$, where $U$ is the matrix having $1$ in all entries. Then the eigenvalues of $A$ are of the form $\lambda+\alpha$, where $\lambda$ is an eigenvalue of $U$.