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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 complex cube root of unity, then
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where $\omega$ is complex cube root of unity, then
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Verified Answer
The correct answer is:
$A^{-1} $does not exits
$\begin{aligned} \text { We have, } \mathrm{A} &=\left[\begin{array}{ccc}1 & \frac{1}{\omega} & \frac{1}{\omega^{2}} \\ \frac{1}{\omega} & \frac{1}{\omega^{2}} & 1 \\ \frac{1}{\omega^{2}} & 1 & \frac{1}{\omega}\end{array}\right] \\|\mathrm{A}| &=1\left\{\frac{1}{\omega^{3}}-1\right\}-\frac{1}{\omega}\left(\frac{1}{\omega^{2}}-\frac{1}{\omega^{2}}\right)+\frac{1}{\omega^{2}}\left(\frac{1}{\omega}-\frac{1}{\omega^{4}}\right) \\ &=1(1-1)-0+\frac{1}{\omega^{2}}\left(\frac{1}{\omega}-\frac{1}{\omega}\right) \quad\left[\because \omega^{3}=1 \text { and } \because \omega^{4}=\omega\right] \\|\mathrm{A}| &=0 \Rightarrow \mathrm{A}^{-1} \text { does not exist } \end{aligned}$
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