Solving a Matrix Equation: Choosing the Right Answer

[Yankel]

Hello,

I have this question, I need to choose the correct answer:

A is a square matrix such that

\[A^{2}+A+I=0\]a) \[A^{-1}=A\]

b) \[A^{-1}=A^{2}\]

c) It is not possible to say if A is invertible

d) A is not invertible

e) \[A^{-1}=A+I\]I got that

\[-A^{2}-A=I\]

and thus

\[A(-A-I)=0\]

and thus

\[A^{-1}=(-A-I)\]

and answer which doesn't exist, am I wrong ?

Thanks !

 

[HallsofIvy]

Yes, you are wrong! You seem to be under the impression that if B is the inverse of A then AB= 0. That is not the case- it is AB= I.
From A^2+ A+ I= 0, -(A^2+ A)= I, A(-(A- I))= I.

 

[Yankel]

Putting 0 was a typing mistake. You seemed to be doing what I did, your answer also doesn't appear as an option

 

[Ackbach]

Perhaps the original equation was $A^{2}+A-I=0$? Your working (except for the typo) seems correct to me.

 

[Yankel]

no, the question is as I wrote it, so you guys agree with me that none of the possible answers is correct ?

 

[Ackbach]

no, the question is as I wrote it, so you guys agree with me that none of the possible answers is correct ?

As stated, I would agree that none of the answers are correct.

 

[Prove It]

I think the answer must be c), there is never any guarantee that an arbitrary square matrix is invertible.

 

[Deveno]

I think the answer must be c), there is never any guarantee that an arbitrary square matrix is invertible.

Any square matrix that satisfies a polynomial with a non-zero constant term *is* invertible.

This is equivalent to saying the minimal polynomial for such a matrix has non-zero constant term, that is: it does not have 0 as an eigenvalue and thus has trivial kernel.