|λE-A| =
|λ-2 2 0|
|2 λ-1 2|
|0 2 λ|
按第1行直接展开，
|λE-A| = (λ-2)[λ(λ-1)-4] - 2(2λ)
= λ^3-3λ^2-6λ+8 = (λ-1)(λ^2-2λ-8) = (λ-1)(λ+2)(λ-4)
|λE-B| =
|λ-1 2 2|
|2 λ-2 0|
|2 0 λ|
按第3行直接展开，
|λE-A| = 2(-2)(λ-2) + λ[(λ-1)(λ-2)-4]
= λ^3-3λ^2-6λ+8 = (λ-1)(λ+2)(λ-4)
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