A=⎣⎡211232112⎦⎤
1.1. det(A−λI)=∣∣2−λ1123−λ2112−λ∣∣=−λ3+7λ2−11λ+5=−(λ−5)(λ−1)2=0
Eigenvalues: λ1=5 and λ2=1
1.2.
λ1=5 : (A−5I)x=⎣⎡−3112−2211−3⎦⎤⎣⎡xyz⎦⎤=⎣⎡000⎦⎤ gives the eigenvector x1=⎣⎡111⎦⎤ .
Answer: 1.1. λ1=5 and λ2=1 ; 1.2. x1=⎣⎡111⎦⎤
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