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

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

• try multiply A in both sides in some equations. A * A^-1 = I. – J Tg Sep 1 '17 at 7:49
• Hint: Consider that $I_n$ = $AA^{-1}$=$A^{-1}A$.Now get to work on the algebra. – Mathemagician1234 Sep 1 '17 at 7:51
• multiplying $A$ both sides one gets characteristic equation $A^2-2A+I=0$ – Arun Sep 1 '17 at 7:54
• $$(x+\dfrac1x)^2=x^2+2+\dfrac1{x^2}.$$ $$(x+\dfrac1x)^3=?$$ – Jyrki Lahtonen Sep 1 '17 at 7:59
• It is easy to eliminate three of these by simple substitution into the original equation – Mark Bennet Sep 1 '17 at 9:02

## 2 Answers

Guide:

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

Also, try to cube both sides.

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

• @Alexander :It is almost a real solution , – Khosrotash Sep 1 '17 at 8:06