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Question

Let ω = 12 + i32 Then the value of the determinant ∣ ∣ ∣11111ω2ω21ω2ω4∣ ∣ ∣ is

A
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B
3ω(ω - 1)
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C
2
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D
3ω(1 - ω)
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Solution

The correct option is B 3ω(ω - 1)
= ∣ ∣ ∣31101ω2ω20ω2ω∣ ∣ ∣ (C1 C1+C2+C3) ( 1+ω+ω2=0) =3[ω.ωω4]=3(ω2ω)=3ω(ω1).

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