your determinant does not look correct. i get you should have:
(λ-3)(λ)(λ+1) + (1)(-5)(-4) + (1)(12)(2) - (-4)(λ)(1) - (-5)(2)(λ-3) - (λ+1)(1)(12)
= λ^3 - 2λ^2 - 3λ + 20 + 24 + 4λ + 10λ - 30 - 12λ - 12
= λ^3 - 2λ^2 - λ + 2
I am to find the eigenvalues and corresponding eigenvectors of the following matrix:
[3 -1 -1]
[-12 0 5]
[4 -2 -1]
Here is what I have so far:
det(I - A) = det [-3 1 1], [12 -5], [-4 2 +1] = (-3)()(+1)
I know that this is not completely correct, but not sure how to finish. Thanks in advance.
your determinant does not look correct. i get you should have:
(λ-3)(λ)(λ+1) + (1)(-5)(-4) + (1)(12)(2) - (-4)(λ)(1) - (-5)(2)(λ-3) - (λ+1)(1)(12)
= λ^3 - 2λ^2 - 3λ + 20 + 24 + 4λ + 10λ - 30 - 12λ - 12
= λ^3 - 2λ^2 - λ + 2
First we will find the characteristic polynomial-
Thus, our eigenvalues are:
Now, I'll find eigenvector for eigenvalue - (And you will do the rest...)
The eigenvectors that belongs to eigenvalue creating the solution for homogeneous system , in other words:
Now, you just need to solve it...
The solution is your eigenvector that belongs to eigenvalue .
λ=2: (λI-A)=0
[(-1,1,1), (12,2,-5), (-4,2,3)] [x1, x2, x3] = 0
-x1 + x2 + x3 = 0
x1 = 2
x2 = 1
x3 = 1
λ=1: (λI-A)=0
[(-2, 1, 1), (12, 1, -5), (-4, 2, 2)] [x1, x2, x3] = 0
-2x1 + x2 + x3 = 0
x1 = 1
x2 = 1
x3 = 1
λ=-1: (λI-A)=0
[(-4, 1, 1), (12, -1, -5), (-4, 2, 0)] [x1. x2. x3] = 0
-4x1 + x2 + x3 = 0
x1 = 1
x2 = 2
x3 = 2
Are these eigenvector correct?
I alos need help on finding the eigenvector and want some clarity.
Are you suppose to just multiply the 3x3 and the 3x1 matrix to get a 3x1 matrix or are you suppose to take the parametric vector form?
Also shouldn't the solution be (x1, y1, z1)^T?
ANd what is x1, y1, and z1? Are they our respective eigenvalues?
And what do you do when there is a repeated eigenvalue? Does that mean there will be two of the same eigenvectors?