Q.

If (ω≠1)  is a cube root of unity, then |11+ω1+ω21+ω211+ω1+ω1+ω21|

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a

0

b

1

c

–4

d

2

answer is C.

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Detailed Solution

|11+ω1+ω21+ω211+ω1+ω1+ω21| Apply c1→c1+c2+c3 ,then determinant is=|21+ω1+ω2211+ω21+ω21|R2→R2-R1 , R3→R3-R1=2|21+ω1+ω20−ωω−ω20−ω+ω2−ω2|                 since 1+ω+ω2=0=2(ω3+ω2−ω3−ω3+ω4)=2(ω2−1+ω)=−4
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