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

$z = \dfrac{-1+\sqrt{3}\,i}{2},\quad i=\sqrt{-1},\quad r,s\in\{1,2,3\}$

P = (-z)ʳ z²ˢ z²ˢ zʳ

$I = \text{identity matrix of order } 2$

$\text{count } (r,s)\in\{1,2,3\}^2 \text{ with } P^2 = -I$

$z = \dfrac{-1+\sqrt{3}\,i}{2} \Rightarrow z^2+z+1=0$

$z^3 = 1,\quad z \ne 1$

$a = (-z)^r,\quad b = z^{2s},\quad c = z^r$

P = a b b c P² = a²+b² b(a+c) b(a+c) b²+c²

$P^2 = -I \Rightarrow b(a+c) = 0$

$b = z^{2s} \ne 0 \Rightarrow a + c = 0$

$(-z)^r + z^r = 0 \Rightarrow (-1)^r = -1 \Rightarrow r = 1 \text{ or } 3$

$a = -c \Rightarrow a^2 = c^2 = z^{2r},\quad b^2 = z^{4s}$

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

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

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

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

$z^{4s} = -2 \ \text{(impossible: } |z^{4s}|=1\text{)}$

$(r,s)=(1,1):\ a=-z,\ b=z^2,\ c=z$

P² = z²+z 0 0 z+z² = -1 0 0 -1

$\text{only } (r,s) = (1,1)$

$\boxed{1}$