If the elements of matrix A are the reciproca | vedclass

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(D) Given the matrix $M = \left[\begin{array}{ccc}1 & \omega & \omega^{2} \ \omega & \omega^{2} & 1 \ \omega^{2} & 1 & \omega\end{array}ight]$.The

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$A^{-1}$ does not exist

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(D) Given the matrix $M = \left[\begin{array}{ccc}1 & \omega & \omega^{2} \ \omega & \omega^{2} & 1 \ \omega^{2} & 1 & \omega\end{array}ight]$.The matrix $A$ consists of the reciprocals of the elements of $M$,so $A = \left[\begin{array}{ccc}1 & \frac{1}{\omega} & \frac{1}{\omega^2} \ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \ \frac{1}{\omega^2} & 1 & \frac{1}{\omega}\end{array}ight]$.Using the property $\omega^3 = 1$,we can write $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.Thus,$A = \left[\begin{array}{ccc}1 & \omega^2 & \omega \ \omega^2 & \omega & 1 \ \omega & 1 & \omega^2\end{array}ight]$.Now,calculate the determinant $|A|$:$|A| = 1(\omega^3 - 1) - \omega^2(\omega^4 - \omega) + \omega(\omega^2 - \omega^2)$$|A| = 1(1 - 1) - \omega^2(\omega - \omega) + \omega(0)$$|A| = 0 - 0 + 0 = 0$.Since the determinant of matrix $A$ is $0$,the matrix $A$ is singular,and therefore $A^{-1}$ does not exist.

If the elements of matrix $A$ are the reciprocals of elements of matrix $\left[\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right]$,where $\omega$ is a complex cube root of unity,then:

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