eigenvalues

Subtracting lambda from each of the diagonal elements and expanding the determinant along the top row gets you

$$(3-\lambda)\{(2-\lambda)(3-\lambda)-1\}-(3-\lambda)=0.$$

(3 - lambda) comes out naturally as a common factor.

If you choose to expand the whole lot though you get -19lambda rather than -20lambda.

Should be -19 rather than -20, further, (lambda-3) comes out naturally as a common factor when you expand the determinant.

edit: Aaah I found it 

I get   $$-\lambda^3+8\lambda^2-19\lambda+12=0$$

Multiply through by -1 (not essential - I just find it easier that way!)

$$\lambda^3-8\lambda^2+19\lambda-12=0$$

If we assume that, because you are doing it by hand, the solutions are integers, then consider

$$(\lambda-a)(\lambda-b)(\lambda-c)$$

The three constants must multiply together such that  $$abc=12$$

If they are each integers then we have either 1, 2 and 6 or 1, 3 and 4 (ok a couple of them might have negative signs attached!).

The sum of the three must also be 8, and since 1+2+6 is not 8, and 1+3+4 is, there is a good chance that the three eigenvalues are 1, 3 and 4.

Plugging each of these into the LHS of the first equation above does indeed give zero.

Thank you Alan, that was really helpfull :)

Now that I come to think of it,

for the eigenvalues it is always true that

$$\lambda_1 + \lambda_2 + \lambda_3 = trace (A)$$

and

$$\lambda_1*\lambda_2*\lambda_3 = det(A)$$

In this case

$$det(A) = 12\\ trace(A) = 8\\$$