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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
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B
2
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C
4
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D
3
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Solution

The correct option is B $$-2$$

|AλI|λ3+3λ2(λ1)2(λ+2)=0
=0
|A−λI|=0⟹−λ3+3λ−2=0⋯(i)⟹(λ−1)2(λ+2)=0
So 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, and

det(A)=det(A)= Product of eigen values =λ1λ2λ3=2=λ1⋅λ2⋅λ3=−2.

Or, you can apply directly Vita's formula in (i) which gives:

Sum of roots of polynomial =Tr(A)=01=0=Tr(A)=−0−1=0, and

Product of roots of polynomial =det(A)=21=2=det(A)=−−2−1=−2.


Applied Mathematics

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