Let `omega` be a complex number such that `2omega +1=z" where " z=sqrt(-3)`
`" If " |{:(1,,1,,1),(1,,-omega^(2)-1,,omega^(2)),(1,,omega^(2),,omega^(7)):}|=3k` then k is equal to

Options

A. `-z`
B. z
C. `-1`
D. 1

Explanation

Given, `2 omega + 1 = z`
`rArr 2omega + 1 = sqrt(-3) " " [because z = sqrt(-3)]`
`rArr omega = (-1+sqrt(3)i)/(2)`
Since, `omega` is cube root of unity.
`therefore omega^(2) = (-1-sqrt(3)i)/(2) " and "omega^(3n) = 1`
Now, `|{:(1 ," "1,1),(1,-omega^(2)-1,omega^(2)),(1," "omega^(2),omega^(2)):}| = 3k`
`rArr |{:(1 ,1,1),(1,omega,omega^(2)),(1,omega^(2),omega):}| = 3k " " [because 1 + omega + omega^(2) = 0 " and "omega^(7) = (omega^(3))^(2)*omega =omega]`
On applying `R_(1) to R_(1) + R_(2) + R_(3)`, we get
`|{:(3 ,1+omega + omega^(2),1+omega + omega^(2)),(1," "omega," "omega^(2)),(1," "omega^(2)," "omega):}| = 3k`
`rArr |{:(3 ,0,0),(1,omega,omega^(2)),(1,omega^(2),omega):}| = 3k`
`rArr 3(omega^(2)-omega^(4)) = 3k`
`rArr (omega^(2) -omega) = k`
`therefore k = ((-1-sqrt(3)i)/(2)) - ((-1+sqrt(3)i)/(2)) = -sqrt(3)i = -z`