Thread: Eigenvalue/Eigenvector

I am loosing my mind. I cannot seem to factor the ^3-4^2-10-12.

My problem is to verify that is an eigenvector of A=(1,2,3), (3,2,1), (2,3,1).

det(-AI) = (-1)(-2)(-1)+(-2)(-1)(-2)+(-3)(-3)(-3)-(-2)(-2)(-3)-(-1)(-3)(-1)-(-1)(-2)(-3)
= ^3-4^2-10-12
= ???

since you are given the eigenvector, just calculate:



this tells you that 6 is an eigenvalue.

now divide the characteristic polynomial by λ-6.