I found this question in a GRE math subject test from 1987. It's problem 61 at http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/GRE_8767_test.pdf. It boils down to the following.
Prove that if $A$ is a $2$x$2$ matrix with real entries not equal to $I$ or $-I$ such that $A=A^{-1},$ then $\textrm{tr}(A)=0.$
A nice proof might use the identity $\textrm{tr}(BC)=\textrm{tr}(CB).$ I know how to prove the result from looking at the matrix components, so I'd prefer to see a "coordinate-free" proof if possible.
We are given that $\,A=A^{-1}\,$ is a $2\times 2$ matrix and $\,A\notin \{I,-I\}.$
Now $\,A\,$ satisfies the equation $$A^2-\textrm{tr}(A)A+\det(A)I=0 \tag{1}$$ by the Cayley-Hamilton theorem, but $$A^2-I=AA^{-1}-I=I-I=0. \tag{2}$$ By property of determinants this implies that $$\det(A)^2 = 1\quad\text{ and }\quad \det(A)=\pm1. \tag{3}$$ Together $(1)$ and $(2)$ imply $$\textrm{tr}(A)A-(1+\det(A))I=0 \tag{4}$$ If $\,\textrm{tr}(A)=0\,$ then $\,\det(A)=-1\,$ and we are done.
If $\,\textrm{tr}(A)\ne0,\,$ then equation $(4)$ implies that $$A = \frac{1+\det(A)}{\textrm{tr}(A)}I. \tag{5}$$ From equation $(3)$, if $\,\det(A)=-1,\,$ then $\,A=0I\,$ which is not invertible, and thus $$\det(A)=1,\quad A=\frac2{\textrm{tr}(A)}I,\quad\text{ and } \quad \textrm{tr}(A)=\pm2.\tag{6}$$ However, this implies that $\,A=\pm I\,$ which contradicts the assumption $\,A\notin \{I,-I\}.$