Solution
= Solution
For a $2\times2$ <matrix>, the <characteristic polynomial> is $t^2-\operatorname{tr}(A)t+\det A$. 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 \b[the order is $4$].