Given the invertible matrix $A$ so that $A+A^{-1}=2I_n$

Question

Given the invertible matrix $A$ so that $A+A^{-1}=2I_n$,

which of the following equalities stand true?

1)$A=3I_n$

2)$A^3+A^{-3}=2I_n$

3)$A=-A$

4)$A^2+A^{-2}=I_n$

5)$A-A^{-1}=2I_n$

I know the formula for $A^{-1}$, but I'm not sure if and how should I use it here or what else should I apply.

Could I have some hints on how to approach this? Thank you

Answer 1

Guide:

First move would be to consider let $A=I$, that would eliminate a few options.

Also, try to cube both sides.

Answer 2

Hint :$$A+A^{-1}=2I \to (A+A^{-1}=2I)^3$$simplify then to what you want $$A^3+(A^{-1})^3+AAA^{-1}+AA^{-1}A+A^{-1}AA+AA^{-1}A^{-1}+A^{-1}AA^{-1}+A^{-1}A^{-1}A\\=A^3+A^{-3}+3A+3A^{-1}=8I\\\to \\A^3+A^{-3}+3\underbrace{(A+A^{-1})}_{2I}=8I$$