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(A) Given $A = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determi

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$A^{2}$

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(A) Given $A = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A| = \omega \cdot \omega - 0 \cdot 0 = \omega^{2}$.Next,find the adjoint of $A$ for a diagonal matrix $\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$ is $\begin{bmatrix} b & 0 \ 0 & a \end{bmatrix}$.Thus,$	ext{adj}(A) = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{\omega^{2}} \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} = \begin{bmatrix} \frac{\omega}{\omega^{2}} & 0 \ 0 & \frac{\omega}{\omega^{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 \ 0 & \frac{1}{\omega} \end{bmatrix}$.Since $\omega^{3} = 1$,we have $\frac{1}{\omega} = \omega^{2}$.So,$A^{-1} = \begin{bmatrix} \omega^{2} & 0 \ 0 & \omega^{2} \end{bmatrix}$.Now,check the options:$A^{2} = \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^{2} & 0 \ 0 & \omega^{2} \end{bmatrix}$.Thus,$A^{-1} = A^{2}$.

If $\omega$ is a complex cube root of unity and $A=\begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $A^{-1}=$

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