The value of beginvmatrixa1 -a2 -a3 a4 - a5 -a6 11a7 -a8 -a9 lend vmatrix is- where ar-cos 2 r -i sin 2 r -1 - 9A- cos 5 -B- cos9 -7- i sin9 -7C- 0D- 2

The correct option is C 0a_r = (cos 2r π + i sin 2r π)^1/9 = e^i 2 r π/9 = e^i 2 π/9 amp; e^i 4 π/9 amp; e^i 6 π/9 e^i 8 π/9 amp; e^i 10 π/9 amp; e^i ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

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, where $a_{r} = (c o s 2 r π + i s i n 2 r π)^{1 / 9}$

c o s 5 π

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c o s \frac{9 π}{7} + i s i n \frac{9 π}{7}

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Solution

$a_{r} = (c o s 2 r π + i s i n 2 r π)^{1 / 9} = e^{i \frac{2 r π}{9}}$
$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{8 π}{9}} & e^{i \frac{10 π}{9}} & e^{i \frac{12 π}{9}} e^{i \frac{14 π}{9}} & e^{i \frac{16 π}{9}} & e^{i \frac{18 π}{9}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{2 π}{9}} & e^{i \frac{4 π}{9}} & e^{i \frac{6 π}{9}} e^{i \frac{14 π}{9}} & e^{i \frac{16 π}{9}} & e^{i \frac{18 π}{9}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0$

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Similar questions

Q. 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} ∣ ∣ ∣ ∣

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

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The value of ∣∣ ∣∣a1a2a3a4a5a6a7a8a9∣∣ ∣∣ is, where ar=(cos 2r π+i sin 2r π)1/9

ar=(cos 2rπ+i sin 2rπ)1/9=ei2rπ9 =∣∣ ∣ ∣ ∣∣ei2π9ei4π9ei6π9ei8π9ei10π9ei12π9ei14π9ei16π9ei18π9∣∣ ∣ ∣ ∣∣=∣∣ ∣ ∣ ∣∣ei2π9ei4π9ei6π9ei2π9ei4π9ei6π9ei14π9ei16π9ei18π9∣∣ ∣ ∣ ∣∣=0

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, where $a_{r} = (c o s 2 r π + i s i n 2 r π)^{1 / 9}$