Is there any matrix $2\times 2$ such that $A\neq I$ but $ A^3=I$?

Question

Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?

In my opinion: no. Thank you very much.

Answer 1

Rotation by $2\pi/3$ in the plane. Find the $2 \times 2$ matrix that gives you this linear transformation.

Answer 2

Since you don't specify what field the entries of this matrix have to come from, I could just take a diagonal matrix whose entries are $1$ and $\omega$ where $\omega=e^{2\pi i /3}$ is a primitive cube root of 1 in the complex numbers.

I guess you want real or integer entries though. If $A^3=I$ then the eigenvalues of $A$, that is, the roots of the characteristic polynomial, have to be third roots of unity. A primitive third root of unity satisfies $x^2+x+1=0$, so you could look for a matrix over the integers with that as a characteristic polynomial....