Question

已知矩陣A=,向量α=.
(1)求A的特征值λ₁,λ₂和對應的特征向量α₁,α₂.
(2)計算A⁵α的值.

Answer 1

(1) 當λ₁=2時,解得α₁=(t≠0),當λ₂=3時,解得α₂=(t≠0)
(2) A⁵α=+=(t≠0)

(1)矩陣A的特征多項式為
λ²-5λ+6,由λ²-5λ+6=0,解得λ₁=2,λ₂=3.
當λ₁=2時,解得α₁=(t≠0),當λ₂=3時,解得α₂=(t≠0).
(2)由α=mα₁+nα₂得解得m=3,n=1.
則A⁵α=A⁵(3α₁+α₂)
=3(A⁵α₁)+A⁵α₂
=3(α₁)+α₂
=3×2⁵α₁+3⁵α₂
=+=(t≠0).