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If `omega` is a complex cube root of unity, then a root of the equation `|(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0`, isA. x = 1B. `x = omega`C. `x = omega^(2)`D. `x = 0` |
Answer» Correct Answer - D We have, `|(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0` `rArr |(x + 1 + omega + omega^(2),omega,omega^(2)),(x + 1 + omega + omega^(2),x + omega^(2),1),(x + 1 + omega + omega^(2),1,x + omega)| = 0 [("Applying"),(C_(1) rarr C_(1) + C_(2) + C_(3))]` `rArr (x + 1 + omega + omega^(2)) |(1,omega,omega^(2)),(1,x + omega^(2),1),(1,1,x + omega)| =0` `rArr x |(1,omega,omega^(2)),(0,x + omega^(2) - omega,1 - omega^(2)),(0,1 - omega,x + omega - omega^(2))| = 0 [("Using " R_(2) rarr R_(2) - R_(1)),(" "R_(3) rarr R_(3) - R_(1))]` `rArr x |(x + omega - omega^(2)) (x + omega^(2) - omega) - (1 - omega) (1 - omega^(2))| = 0` `rArr x = 0` is a root of the given equation |
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