Finding eigenvalues of a $3 \times 3$ matrix.

Question

An interval in which all the eigenvalues of the matrix $$A = \begin{pmatrix} 35 & 40 & 13 \\ 11 & 13 & 8 \\ 12 & 4 & 71 \end{pmatrix}$$ lie, is

(a) $[-5,-17]$

(b) $[-18,88]$

(c) $[89,100]$

(d) $[90,117]$

It can be solved by finding characteristic polynomial and then finding it's zeros but that's a quite tedious calculation. Is there any easier way to solve this problem? Any help in this regard will be highly appreciated.

Thanks in advance.

Answer 1

Hint:

The sum of the eigenvalues is $tr(A)=35+13+71=119$.

You can use this to eliminate three of the answers.

Answer 2

As the determinant is $2329$ and is known to be the product of eigenvalues, this rules out options (a),(c) and (d).