Let z = frac{-1 + sqrt{3}i}{2}, where i = sqr | vedclass

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(C) Given $z = \frac{-1}{2} + i \frac{\sqrt{3}}{2} = \omega$, where $\omega$ is the cube root of unity. Thus, $\omega^3 = 1$ and $1 + \omega + \omega^

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(C) Given $z = \frac{-1}{2} + i \frac{\sqrt{3}}{2} = \omega$, where $\omega$ is the cube root of unity. Thus, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.$P = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix}$.$P^2 = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix} \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \ \omega^{2s} & \omega^r \end{bmatrix} = \begin{bmatrix} (-1)^{2r} \omega^{2r} + \omega^{4s} & (-1)^r \omega^{r+2s} + \omega^{r+2s} \ (-1)^r \omega^{r+2s} + \omega^{r+2s} & \omega^{4s} + \omega^{2r} \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$.For the off-diagonal elements to be zero: $((-1)^r + 1) \omega^{r+2s} = 0$. Since $\omega 
eq 0$, we must have $(-1)^r + 1 = 0$, which implies $r$ must be odd. Thus, $r \in \{1, 3\}$.Case $1$: $r = 1$. The diagonal elements give $\omega^2 + \omega^{4s} = -1$. Since $1 + \omega + \omega^2 = 0$, we have $\omega^2 + \omega = -1$. Thus, $\omega^{4s} = \omega$, which means $4s \equiv 1 \pmod 3$, so $s \equiv 1 \pmod 3$. For $s \in \{1, 2, 3\}$, $s = 1$.Case $2$: $r = 3$. The diagonal elements give $\omega^6 + \omega^{4s} = -1$. Since $\omega^3 = 1$, $1 + \omega^{4s} = -1$, so $\omega^{4s} = -2$. This is impossible as $|\omega^{4s}| = 1$.Thus, the only valid pair is $(r, s) = (1, 1)$. The total number of pairs is $1$.

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

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