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. $-3$

Answer 1

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)=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 =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, and

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