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

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)

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

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 $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ 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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$3 ω$

$3 ω (ω - 1)$

$3 ω^{2}$

$3 ω (1 - ω)$

answer is B.

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Detailed Solution

Using $1 + ω^{2} = - ω, ω^{4} = ω$ and applying

$C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get

$\begin{array}{l} Δ = |\begin{array}{ccc} 1 & 0 & 0 \\ 1 & ω - 1 & ω^{2} - 1 \\ 1 & ω^{2} - 1 & ω - 1 \end{array}| \\ = (ω - 1)^{2} - {(ω^{2} - 1)}^{2} \\ = (ω + ω^{2} - 2) (ω - 1 - ω^{2} + 1) \\ = (- 3) (ω - ω^{2}) = 3 ω (ω - 1) \end{array}$