Let omega be a complex cube root of unity wit | vedclass

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(A) Given $P = \begin{bmatrix} \omega^{1+1} & \omega^{1+2} \ \omega^{2+1} & \omega^{2+2} \end{bmatrix} = \begin{bmatrix} \omega^2 & \omega^3 \ \omeg

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$57$

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(A) Given $P = \begin{bmatrix} \omega^{1+1} & \omega^{1+2} \ \omega^{2+1} & \omega^{2+2} \end{bmatrix} = \begin{bmatrix} \omega^2 & \omega^3 \ \omega^3 & \omega^4 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix}$.Calculate $P^2 = \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix} \begin{bmatrix} \omega^2 & 1 \ 1 & \omega \end{bmatrix} = \begin{bmatrix} \omega^4+1 & \omega^2+\omega \ \omega^2+\omega & 1+\omega^2 \end{bmatrix} = \begin{bmatrix} \omega+1 & -1 \ -1 & -\omega \end{bmatrix} = \begin{bmatrix} -\omega^2 & -1 \ -1 & -\omega \end{bmatrix} = -P$.Since $P^2 = -P$,we have $P^3 = P^2 \cdot P = (-P) \cdot P = -P^2 = -(-P) = P$.Thus,$P^3 = P$. For $P^k = P$,$k$ must be an odd integer greater than or equal to $3$. Since $P^3 = P$,$P^5 = P^3 \cdot P^2 = P \cdot (-P) = -P^2 = P$. In general,$P^k = P$ for any odd $k \geq 3$.Checking the options,$k$ must be odd. Among the given options,$57$ is the only odd number.

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be a $2 \times 2$ matrix with $p_{ij} = \omega^{i+j}$. For $P^2 \neq 0$,if $P^k = P$,then $k$ is equal to

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