Let `omega` be a complex number such that `2omega+1=sqrt(3)i` . If `|[1,1,1],[1,-omega^(2)-1,omega^(2)],[1,omega^(2),omega^(7)]|=3k`, then k is equal to

Options

A. `-1`
B. 1
C. `-isqrt(3)`
D. `isqrt(3)`

Answer

C

Explanation

We have, `isqrt(3)=1+2omega rArr omega=(-1+isqrt(3))/(2)`. Clearly, `omega` is a cube root of unity such that `1+omega+omega^(2)=0` and `omega^(3)=1`.
Now, `|{:(1,1,1),(1,-omega^(2)-1,omega^(2)),(1,omega^(2),omega^(7)):}|=3k`
`rArr |{:(1,1,1),(1,-omega^(2)-1,omega^(2)),(1,omega^(2),omega):}|=3k`
`rArr |{:(3,1,1),(0,-omega^(2)-1,omega^(2)),(0,omega^(2),omega):}|=3k` Applying `C_(1) to C_(1)+C-(2)+C_(3)`
`rArr 3(-omega^(3)-omega-omega^(4))=3k`
`rArr 3(-1-2omega)=3k`
`rArr -1+1-isqrt(3)=k rArr k=-isqrt(3)`