Let ω$$=$$−$$12+32i$$ Then the value of the determinant Δ$$=1111$$−$$1$$−ω$$2$$ω$$21$$ω$$2$$ω$$4$$ is

Correct option is 23ω (ω – 1)

Questions

Let $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ Then the value of the determinant $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}|$ is

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3ω

3ω (ω – 1)

3ω2

3ω(1−ω)

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detailed solution

Correct option is B

Using 1+ω2=−ω,ω4=ω and applying C2→C2−C1,C3→C3−C1 we get Δ=1001ω−1ω2−11ω2−1ω−1=(ω−1)2−ω2−12=ω+ω2−2ω−1−ω2+1=(−3)ω−ω2=3ω(ω−1)

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Let ω=−12+32i Then the value of the determinant Δ=1111−1−ω2ω21ω2ω4 is

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Let $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ Then the value of the determinant $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}|$ is