Let $z = \frac{-1+\sqrt{3}i}{2}$, where $i = \sqrt{-1}$, and $r, s \in \{1, 2, 3\}$. Let $P = \begin{bmatrix} (-z)^{r} & z^{2s} \\ z^{2s} & z^{r} \end{bmatrix}$ and $I$ be the identity matrix of order 2. Then the total number of ordered pairs $(r, s)$ for which $P^{2} = -I$ is

Let z = (-1+√3i)/2, where i = √-1, and r, s ∈ {1, 2, 3}. Let P = (-z)ʳ z²ˢ z²ˢ zʳ and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P² = -I is

$\text{Given: } z = \dfrac{-1+\sqrt3 i}{2},\ i = \sqrt{-1},\ r,s \in \{1,2,3\}$

P = (-z)ʳ z²ˢ z²ˢ zʳ , I = I₂

$\text{To find: } \#\{(r,s) : P^2 = -I\}$

$z = \dfrac{-1+\sqrt3 i}{2} = \omega$

$\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0$

$A = (-z)^r,\quad B = z^{2s},\quad C = z^r$

P² = A² + B² B(A+C) B(A+C) C² + B²

$P^2 = -I \Rightarrow A^2 + B^2 = -1,\quad C^2 + B^2 = -1,\quad B(A+C) = 0$

$B = z^{2s} \ne 0 \Rightarrow A + C = 0$

$(-z)^r + z^r = 0 \Rightarrow (-1)^r z^r + z^r = 0$

$(-1)^r = -1 \Rightarrow r \text{ is odd} \Rightarrow r \in \{1, 3\}$

$C^2 + B^2 = -1 \Rightarrow z^{2r} + z^{4s} = -1$

$r = 3:\ z^6 + z^{4s} = 1 + z^{4s} = -1 \Rightarrow z^{4s} = -2 \ \text{impossible}$

$r = 1:\ z^2 + z^{4s} = -1 \Rightarrow z^{4s} = -1 - z^2 = z$

$4s \equiv 1 \pmod 3 \Rightarrow s \equiv 1 \pmod 3 \Rightarrow s = 1$

$\#\{(r,s)\} = 1, \quad (r,s) = (1,1)$

$\boxed{1}$