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Let P=[ cos...

Question

Let $P = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{9} & sin \frac{π}{9} - sin \frac{π}{9} & cos \frac{π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$ and $α, β, γ$ be non-zero real numbers such that $α P^{6} + β P^{3} + γ I$ is the zero matrix. Then, $(α^{2} + β^{2} + γ^{2})^{(α - β) (β - γ) (γ - α)}$ is

A

π

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B

\frac{π}{2}

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C

0

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D

1

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Solution

The correct option is D $1$
$P = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{9} & sin \frac{π}{9} - sin \frac{π}{9} & cos \frac{π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$

$P^{2} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{9} & sin \frac{π}{9} - sin \frac{π}{9} & cos \frac{π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$ $⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{9} & sin \frac{π}{9} - sin \frac{π}{9} & cos \frac{π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$ = $⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{2 π}{9} - sin \frac{- 2 π}{9} & 2 cos \frac{π}{9} sin \frac{π}{9} - 2 sin \frac{π}{9} cos \frac{π}{9} & cos \frac{2 π}{9} - sin \frac{2 π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{2 π}{9} & sin \frac{2 π}{9} - sin \frac{2 π}{9} & cos \frac{2 π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$

On solving,

$P^{3} = P^{2} . P = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{3 π}{9} & sin \frac{3 π}{9} - sin \frac{3 π}{9} & cos \frac{3 π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$ $= ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{3} & sin \frac{π}{3} - sin \frac{π}{3} & cos \frac{π}{3} \end{matrix} ⎤ ⎥ ⎥ ⎦$

by $P^{6} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{6 π}{9} & sin \frac{6 π}{9} - sin \frac{6 π}{9} & cos \frac{6 π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$ $= ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{2 π}{3} & sin \frac{2 π}{3} - sin \frac{2 π}{9} & cos \frac{2 π}{9} \end{matrix} ⎤ ⎥ ⎥ ⎦$

Given, $α p^{6} + β p^{3} + γ I = 0$ matrix

$∴ α cos \frac{2 π}{3} + β cos \frac{π}{3} + γ = - 0 \Rightarrow α - β = 2 γ$

and $α sin \frac{2 π}{3} + β sin \frac{π}{3} = 0 \Rightarrow α + β = 0$

On solving, $2 α = 2 γ \Rightarrow α = γ$
and $β = - α = - γ$

$∴ (α^{2} + β^{2} + γ^{2})^{(α - β) (β - γ) (γ - α)} = (α^{2} + β^{2} + γ^{2})^{0}$ $= 1$
Option D

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