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If ar = cos2r...

Question

If $a_{r} = (cos 2 r π + i sin 2 r π)^{\frac{1}{9}}$ , then the value of $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣$ is:

A

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B

1

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C

0

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D

2

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Solution

The correct option is C $0$
Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣$
$a_{r} = (cos 2 r π + i sin 2 r π)^{\frac{1}{9}} = e^{i \frac{2 r π}{9}}$
$\Rightarrow Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{8 π}{9}} & e^{i \frac{10 π}{9}} & e^{i \frac{12 π}{9}} e^{i \frac{14 π}{9}} & e^{i \frac{16 π}{9}} & e^{i \frac{18 π}{9}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
Taking $e^{i \frac{6 π}{9}}$ common from $R_{2}$ , we have:
$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣ = e^{i \frac{6 π}{9}} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{14 π}{9}} & e^{i \frac{16 π}{9}} & e^{i \frac{18 π}{9}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣ = 0$
( $∵ R_{1}$ and $R_{2}$ are identical.)

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Similar questions

If $α a n d β$ are the zeros of $2 x^{2} + 5 x - 9$ then the value of $α β$ is

(a) $\frac{- 5}{2}$ (b) $\frac{5}{2}$ (c) $\frac{- 9}{2}$ (d) $\frac{9}{2}$

a_{r} = (cos 2 r π + i sin 2 r π)^{\frac{1}{9}}

, then the value of

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣

a_{r} = {(cos 2 r π + i sin 2 r π)}^{1 / 9}

, then the value of

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣

a_{r} = cos \frac{2 r π}{9} + i sin \frac{2 r π}{9}, r = 1, 2, 3, . . ., i = \sqrt{- 1},

then the determinant

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣

a_{r} = (cos 2 r π + i sin 2 r π)^{\frac{1}{9}}

, then the value of

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} a_{4} & a_{5} & a_{6} a_{7} & a_{8} & a_{9} \end{matrix} ∣ ∣ ∣ ∣

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