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← JEE Main 2017 · 2 Apr · Shift 1
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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 63

Let ω be a complex number such that 2ω+1=z2\omega+1=z where z=−3z=\sqrt{-3}. If
∣1111−ω2−1ω21ω2ω7∣=3k,\begin{vmatrix}1&1&1\\1&-\omega^2-1&\omega^2\\1&\omega^2&\omega^7\end{vmatrix}=3k,
then k is equal to :

Choose one option
Answer and worked solution

Official answer: D

ω=−1+3i2\omega=\dfrac{-1+\sqrt3i}2 is a cube root of unity, so −ω2−1=ω-\omega^2-1=\omega and ω7=ω\omega^7=\omega:
∣1111ωω21ω2ω∣=3(ω2−ω)=3(−3i)=−3z.\begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix}=3(\omega^2-\omega)=3(-\sqrt3i)=-3z.
So k=−zk=-z.

Topics:Roots of unityDeterminants and properties
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