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(B) $A = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$We know that $A = \omega I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \

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$\omega A$

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(B) $A = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$We know that $A = \omega I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Then,$A^{50} = (\omega I)^{50} = \omega^{50} I^{50}$.Since $I^n = I$ for any positive integer $n$,we have $A^{50} = \omega^{50} I$.We know that $\omega^3 = 1$. Therefore,$\omega^{50} = (\omega^3)^{16} \cdot \omega^2 = (1)^{16} \cdot \omega^2 = \omega^2$.Thus,$A^{50} = \omega^2 I = \begin{bmatrix} \omega^2 & 0 \ 0 & \omega^2 \end{bmatrix}$.Since $A = \omega I$,we have $I = \frac{1}{\omega} A = \omega^2 A$ (as $\frac{1}{\omega} = \omega^2$).Therefore,$A^{50} = \omega^2 (\omega^2 A) = \omega^4 A = \omega A$ (since $\omega^3 = 1$).Alternatively,$A^{50} = \omega^{50} I = \omega^2 I$. Since $A = \omega I$,then $\omega A = \omega^2 I$.Hence,$A^{50} = \omega A$.

If $\omega$ is a complex cube root of unity and $A=\begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $A^{50}$ is equal to

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