I have a question regarding a 3x3 matrix and its eigenvalues. The matrix is $A= \begin{bmatrix}1 & 1 & 2\\-2 & -1 & 2 \\ -1 & -2 & 3\end{bmatrix}$

What I have attempted is to let $det(A-\lambda I)=0$, which resulted in the cubic, $-\lambda ^3+3\lambda ^2 -7\lambda +11 =0$

I believe I have worked out the correct polynomial, but without the help of a computer or the equation for cubic roots I am unsure on how to solve this equation. Normally my method would involve guessing a solution (a factor of $\lambda^0$, in this case 11) and then doing polynomial division. In this case the polynomial has one real root and a complex conjugate.

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    $\begingroup$ Not much you can do here: your cubic is fine. Just use Cardano's formula en.wikipedia.org/wiki/Cubic_function $\endgroup$ – Pierpaolo Vivo Feb 28 '16 at 8:56
  • $\begingroup$ There's no general method, to solve a cubic. You've to ultimately get creative in coming up with solutions. This'll help even when you try to solve differential equations. $\endgroup$ – Isomorphic Feb 28 '16 at 8:57
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    $\begingroup$ @Iota I disagree with your statement. There is a general formula (Cardano's formula, lined above), as much as there is a quadratic formula to solve a quadratic equation. It is just more tedious and cumbersome for a cubic than it is for a quadratic equation, but the roots of a cubic can always be found algebraically. In some cases, you can be 'creative' and find shortcuts, but this is by no means the only way to find the full set of roots. $\endgroup$ – Pierpaolo Vivo Feb 28 '16 at 9:15
  • $\begingroup$ @PierpaoloVivo Sorry, I didn't know the formula. But I still think being creative at finding solutions is a much more deep learning than any formula to find zeroes. $\endgroup$ – Isomorphic Feb 28 '16 at 10:44

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