Golden Matrix

Since $\Phi^2 - \Phi - I= 0$ , the minimum polynomial of $\Phi$ divides $t^2 - t - 1$ . Since $\Phi$ is not a multiple of the identity, its minimum polynomial is not linear, and hence the minimum polynomial of $\Phi$ is $t^2 - t - 1$ . Thus $\Phi$ has two distinct eivenvalues, and hence is diagonalisable, so we can write $\Phi \; = \; S \left(\begin{array}{cc} \alpha_+ & 0 \\ 0 & \alpha_- \end{array} \right) S^{-1}$ for some nonsingular matrix $S$ , where $\alpha_\pm = \tfrac12(1 \pm \sqrt{5})$ are the roots of $t^2 - t - 1 = 0$ (note that $\alpha_+ = \phi$ ). Thus $\mathrm{Tr}\,\Phi \; = \; \alpha_+ + \alpha_- = 1 \hspace{2cm} \mathrm{det}\,\Phi \; = \; \alpha_+\alpha_- = -1$ making the answer $1 \times -1 = \boxed{-1}$ .

This can be solved easily with Cayley Hamilton Theorem. Since $\Phi$ satisfies the equation $x^2-x-I_2=0$ We see that the characteristic equation of the $2 *2$ Matrix $\Phi$ (The fact that it is 2*2 matrix can be easily noticed from the fact that the equation has an $I_2$ in it) is $x^2-x-1=0$

Now it can be easily proven for a Characteristic equation $ax^2+bx+c=0$ of any $2*2$ matrix $A$ ,it is true that $\operatorname {det} (A) =\frac ca$ while $\operatorname {tr} (A) =\frac {-b}{a}$

Hence the answer follows.