Finding eigenvalues of a 3 x 3 matrix

[dink87522]

I have a 3 x 3 matrix

A =
(0 -1 -3)
(2 3 3)
(-2 1 1)

Let & represent lambda here.

I am trying to find the eigenvalues of A.

I start off by taking the characteristic equation of A and end up with -&[(&-3)(&-1) -3] + (2& - 8) - 3(-2& + 8)
yet can't then get that factored down. Am I missing something basic here?

 

[Dickfore]

$$\left|\begin{array}{ccc}<br /> -\lambda & -1 & -3 \\<br /> <br /> 2 & 3 - \lambda & 3 \\<br /> <br /> -2 & 1 & 1 - \lambda<br /> \end{array}\right| = 0$$

$$-\lambda (3 - \lambda) (1 - \lambda) + 6 - 6 - 6 (3 - \lambda) + 3 \lambda + 2 (1 - \lambda) = 0$$

$$-\lambda(3 - 4 \lambda + \lambda^{2}) - 18 + 6 \lambda + 3 \lambda + 2 - 2 \lambda = 0$$

$$-\lambda^{3} + 4 \lambda^{2} - 3 \lambda + 7 \lambda - 16 = 0$$

$$\lambda^{3} - 4 \lambda^{2} - 4 \lambda + 16 = 0$$

This can be factored.

 

[Simon_Tyler]

Dickfore has (of course) got the right answer.

The final results can easily be checked with, e.g., Wolfram|Alpha.
http://www.wolframalpha.com/input/?i=eigenvalues+{{0,+-1,+-3},{2,+3,+3},{-2,+1,+1}}"
Factor cubic