I think to this question for two days :
Let $A$ be a $3\times3$ real matrix such that $\det(A) = 1$ and $A^{-1}= A^T$. Prove that one of the eigenvalues is equal to $1$.
I used the fact that determinant of $A$ is the product of its eigenvalues and then I wrote many equations that I couldn't solve the question.
Let $\lambda \in \mathbb C$ be an eigenvalue with eigenvector $0 \neq x \in \mathbb C^n$. Then $$(x,x) = (A^tAx,x) = (Ax,Ax) = \lvert\lambda \rvert^2 (x,x).$$ Thus $\lvert \lambda \rvert = 1$ for all eigenvalues. Since $A$ is real, eigenvalues come in conjugate pairs, so either all three are real, or two are complex and one is real, but the two which are complex form a conjugate pair. Together with $\det A = 1$, do you see why this implies that one of the eigenvalues is $1$?