 Solution

For a $2\times 2$ matrix, the characteristic polynomial is $t^2-\mathrm{tr}(A)t+\det A$. The given matrix trace and determinant therefore make it $t^2+1$. By the Cayley-Hamilton theorem, $A^2=-I$, and hence $A^4=I$. Since $A^2\ne I$, the order of a group element $A$ cannot be $1$ or $2$. If $A^3=I$, then $A=A^4(A^3{)}^{-1}=I$, also impossible. Thus the order is $4$.