If $a_{r}=\cos \frac{2r\pi}{9}+i \sin \frac{2r\pi}{9}, r=1,2,3 \ldots, i=\sqrt{-1}$, then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to:

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Complex Numbers

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

, then the determinant

a_{1} a_{4} a_{7} a_{2} a_{5} a_{8} a_{3} a_{6} a_{9}

is equal to:

Answer: (C)

Writing

ω = e^{i 2 π /9}

, each

a_{r} = ω^{r}

, and the entry in row

i

, column

j

ω^{3 (i - 1) + j} = ω^{3 (i - 1)} ω^{j}

. Every row is thus the vector

(ω, ω^{2}, ω^{3})

times a constant, so the rows are proportional and the determinant is

0

. Among the listed expressions,

a_{1} a_{9} - a_{3} a_{7} = ω^{10} - ω^{10} = 0

, so the determinant equals

a_{1} a_{9} - a_{3} a_{7}

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option