# How to proceed with following determinant inequality

Let $$A,B\in M_{n}(\mathbb{R})$$ be such that $$B^2=I_n$$ and $$A^2=AB+I_n$$. Prove that $$\det(A)\leq\left(\frac{1+\sqrt{5}}{2}\right)^n$$

I have been able to show that $$AB=BA$$, $$B=A-A^{-1}$$ and $$A^4-3A^2+I=0$$. Now from this how I can approach the problem.

If $$A^4-3A^2+I=0$$ than any eigenvalue of $$A$$ should satisfy $$\lambda^4-3\lambda^2+1=0$$. This is a biquadratic equation, solving which we get 4 roots: $$\pm\frac{1+\sqrt{5}}{2}, \pm\frac{1-\sqrt{5}}{2}$$. $$det(A)$$ is a product of its eigenvalues, so it will be biigest when all of them will be equal to $$\frac{1+\sqrt{5}}{2}$$.