Let $\omega$ be a complex number such that $2 \omega+1=z$ where $z=\sqrt{-3}$. If $\left|\begin{array}{rrr}1 & 1 & 1 \\ 1 & -\omega^{2}-1 & \omega^{2} \\ 1 & \omega^{2} & \omega^{7}\end{array}\right|=3 \mathrm{k}$, then k is equal to :

JEE Main · 2017 · 02 Apr 2017 · Offline · Complex Numbers

Let

ω

be a complex number such that

2 ω + 1 = z

where

z = - 3

. If

111 1 - ω^{2} - 1 ω^{2} 1 ω^{2} ω^{7} = 3 k

,
then k is equal to :

Answer: (B)

Since

2 ω + 1 = - 3

, we have

ω = \frac{- 1 + i 3}{2}

, a primitive cube root of unity, so

1 + ω + ω^{2} = 0

and

ω^{3} = 1

. These give

- ω^{2} - 1 = ω

and

ω^{7} = ω

, so the determinant reduces to

111 1 ω ω^{2} 1 ω^{2} ω

. Expanding along the first row gives

(ω^{2} - ω^{4}) - (ω - ω^{2}) + (ω^{2} - ω) = 3 (ω^{2} - ω)

(using

ω^{4} = ω

). Hence

3 k = 3 (ω^{2} - ω)

, so

k = ω^{2} - ω = \frac{- 1 - i 3}{2} - \frac{- 1 + i 3}{2} = - i 3 = - - 3 = - z

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option