Question

# If $$A$$ is $$3\times 3$$ square matrix whose characteristic polynomial equations is $$\lambda^{3}-3\lambda^{2}+4=0$$ then roots of $$polynomial$$ is ?

A
0
B
2
C
4
D
3

Solution

## The correct option is B $$-2$$|A−λI|⟹−λ3+3λ−2⟹(λ−1)2(λ+2)=0=0|A−λI|=0⟹−λ3+3λ−2=0⋯(i)⟹(λ−1)2(λ+2)=0So the eigen values are given by: λ1=1, λ2=1, λ3=−2λ1=1, λ2=1, λ3=−2.Now Tr(A)=Tr(A)= Sum of eigen values =λ1+λ2+λ3=0=λ1+λ2+λ3=0, anddet(A)=det(A)= Product of eigen values =λ1⋅λ2⋅λ3=−2=λ1⋅λ2⋅λ3=−2.Or, you can apply directly Vita's formula in (i) which gives:Sum of roots of polynomial =Tr(A)=−0−1=0=Tr(A)=−0−1=0, andProduct of roots of polynomial =det(A)=−−2−1=−2=det(A)=−−2−1=−2.Applied Mathematics

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