Let alpha be a root of the equation x^{2}+x+1 | vedclass

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(A) Given the equation $x^{2}+x+1=0$,its roots are the complex cube roots of unity,$\omega$ and $\omega^{2}$. Let $\alpha = \omega$. Then $\alpha^{2}

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$A^{3}$

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(A) Given the equation $x^{2}+x+1=0$,its roots are the complex cube roots of unity,$\omega$ and $\omega^{2}$. Let $\alpha = \omega$. Then $\alpha^{2} = \omega^{2}$ and $\alpha^{4} = \omega^{4} = \omega$.The matrix $A$ becomes $A = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^{2} \ 1 & \omega^{2} & \omega \end{bmatrix}$.Calculating $A^{2} = A \cdot A = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^{2} \ 1 & \omega^{2} & \omega \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^{2} \ 1 & \omega^{2} & \omega \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix}$.Now,$A^{4} = A^{2} \cdot A^{2} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I_{3}$.Since $A^{4} = I_{3}$,we have $A^{31} = A^{28} \cdot A^{3} = (A^{4})^{7} \cdot A^{3} = I_{3}^{7} \cdot A^{3} = A^{3}$.

Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4} \end{bmatrix}$,then the matrix $A^{31}$ is equal to

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