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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$.The inverse of a diagonal matrix $A = 	ext{diag}(a, b, c)$ is given by

Vedclass · Accepted Answer

$\begin{bmatrix} \omega^2 & 0 & 0 \ 0 & \omega & 0 \ 0 & 0 & 1 \end{bmatrix}$

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$.The inverse of a diagonal matrix $A = 	ext{diag}(a, b, c)$ is given by $A^{-1} = 	ext{diag}(a^{-1}, b^{-1}, c^{-1})$.Here,$A = \begin{bmatrix} \omega & 0 & 0 \ 0 & \omega^2 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Therefore,$A^{-1} = \begin{bmatrix} \omega^{-1} & 0 & 0 \ 0 & (\omega^2)^{-1} & 0 \ 0 & 0 & 1^{-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 & 0 \ 0 & \frac{1}{\omega^2} & 0 \ 0 & 0 & 1 \end{bmatrix}$.Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.Thus,$A^{-1} = \begin{bmatrix} \omega^2 & 0 & 0 \ 0 & \omega & 0 \ 0 & 0 & 1 \end{bmatrix}$.

If $\omega$ is a complex cube root of unity and $A = \begin{bmatrix} \omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $A^{-1} = \dots$

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