Finding the eigenvalues of a 3x3

[stryker105]

Homework Statement

Find the eigenvalues of:
|13 -30 0|
|1 0 0|
|0 1 0|

Homework Equations

Equation for the eigenvalues: det(A-λI)=0

Cofactor expansion = det A = a11(a22a33-a23a32)+a12(a23a31-a21a33)+a13(a21a32-a22a31)

The Attempt at a Solution

|13-λ -30 0 |
| 1 -λ 0 |
| 0 1 -λ|

(13 -λ)(λ$^{2}$)-30(-λ)+0 = 0
13λ$^{2}$-λ$^{3}$+30
-λ$^{3}$+13λ$^{2}$+ 30λ
-λ(λ$^{2}$-13λ-30)
(λ+2)(λ-15)

Eigenvalues = 0, 2, 15

However, this not the correct answer according the the software I'm using. Can anyone see what I'm doing wrong?

 

[micromass]

The -λ in bold should be just λ

(13 -λ)(λ$^{2}$)-30(-λ)+0 = 0

 

[I like Serena]

Hmm, I'd say that the -30(-λ) should really be --30(-λ).

 

[micromass]

Hmm, I'd say that the -30(-λ) should really be --30(-λ).

Yes, I agree that that is the most common way of seeing it. But I guess that the OP wanted to follow his formula:

Cofactor expansion = det A = a11(a22a33-a23a32)+a12(a23a31-a21a33)+a13(a21a32-a22a31)

In that case, it is $-30\lambda$. But it is a weird way of doing things, agreed.

 

[HallsofIvy]

Expanding on the last column gives $-\lambda(\lambda^2- 13\lambda+ 30)$, as I Like Serena and micromass are saying, not $-\lambda(\lambda^2- 13\lambda- 30)$.

That factors as $-\lambda(\lambda- 10)(\lambda- 3)= 0$.