# Finding eigenvalues of a matrix using λ

$$\begin{pmatrix} -2 & 0 & 1 \\ -6 & -2 & 0 \\ 19 & 5 & -4 \\ \end{pmatrix}$$

The main I have is if I'm doing my algebra right because the back of my book has λ = -8 (Sorry I misread back of the book) I still don't know how they got one answer.

I end up getting:

(-2 - λ)[(-2 - λ)(-4 - λ)] + 1[- 30 -( 19 (-2 - λ))] = 0

after simplifying:

λ^3 + 8λ^2 + λ - 8 = 0

Factoring:

(λ + 1)(λ - 1)(λ + 8)

What am I doing wrong? I see what I did wrong I forgot the negative when I added -16 to 8.

-

I got -8, $\pm i$ with Mathematica. The characteristic polynomial is

$-8 - x - 8 x^2 - x^3$

(or you could do all positive coefficients, but it just means you messed up the sign of the -8) By the way, I don't think $\lambda + 8$ is a factor of your polynomial, so you made 2 different mistakes.

If you throw away the negative sign and factor out an x+8, you are left with $x^2 + 1$.

By the way, if the book got 8 and 2: 1. The book is wrong, or 2. You typed it in wrong, or 3. I typed it in wrong but I checked a few times so I don't think so.

-
Sorry I the book got -8 but I'm still not getting the right answer –  user9121 Apr 13 '11 at 1:39
Like I said, it looks like you were right at λ^3 + 8λ^2 + λ - 8 = 0 except for the -8, which should be +8. So, look at all the constant terms in the previous step. There is -2*-2*-4-30-19*-2=-16-30+38=-8. And, then, you moved everything to the other side of the equation (I can tell because you have $+\lambda^3$, for example, so the -8 becomes +8 when you do this. So, maybe your problem was just that you forgot to change the sign when you moved it to the other side. –  Graphth Apr 13 '11 at 2:13