If ω is an imaginary cube root of unity, then the value of [ 1 ω^2 1-ω^4 ω 1 1+ω^5 1 ω ω^2 ] is

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MARKS Content Team · Accepted Answer

Correct answer: (B) ω^2-4. Given, ω^3=1 and 1+ω+ω^2=0 = | 1 ω^2 1-ω^4 ω 1 1+ω^5 1 ω ω^2 |= | 1 ω^2 1-ω ω 1 1+ω^2 1 ω ω^2 | Expand with respect to R_1 = (ω^2-ω-ω^3 )-ω^2 (ω^3-1-ω^2 ) +(1-ω) (ω^2-1 ) = (ω^2-ω-1 )-ω^2 (1-1-ω^2 ) +(1-ω) (ω^2-1 ) =ω^2-ω-1+ω^4+ω^2-1-ω^3+ω =2 ω^2-2+ω-1=2 ω^2+ω-3 =ω^2+ (ω+ω^2 )-3 =ω^2-1-3=ω^2-4

If $\omega$ is an imaginary cube root of unity, then the value of $\left[\begin{array}{ccc}1 & \omega^{2} & 1-\omega^{4} \\ \omega & 1 & 1+\omega^{5} \\ 1 & \omega & \omega^{2}\end{array}\right]$ is

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