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Question

Let ω be a complex cube root of unity with ω1 and P=[pij] be a n × n matrix with pij=ωi+j. Then P20, when n=

A
57
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B
55
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C
58
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D
56
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Solution

The correct options are
A 55
C 58
D 56
P=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ω2ω3ω4....ωn+1ω3ω4ω5....ωn+2ω4ω5ω6....ωn+3................ωn+1ωn+2......ω2n⎥ ⎥ ⎥ ⎥ ⎥ ⎥
P=ω2ω3ω4.....ωn+1⎢ ⎢ ⎢ ⎢ ⎢ ⎢1ωω2....ωn11ωω2....ωn11ωω2....ωn1................1ωω2...ωn1⎥ ⎥ ⎥ ⎥ ⎥ ⎥
P=ωn(n+3)2⎢ ⎢ ⎢ ⎢ ⎢ ⎢1ωω2....ωn11ωω2....ωn11ωω2....ωn1................1ωω2...ωn1⎥ ⎥ ⎥ ⎥ ⎥ ⎥
Now, P2=ωn2+3n⎢ ⎢ ⎢ ⎢ ⎢ ⎢1ωω2....ωn11ωω2....ωn11ωω2....ωn1................1ωω2...ωn1⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎢ ⎢ ⎢ ⎢ ⎢ ⎢1ωω2....ωn11ωω2....ωn11ωω2....ωn1................1ωω2...ωn1⎥ ⎥ ⎥ ⎥ ⎥ ⎥
Since, P2O
1+ω+ω2+....ωn10
1ωn1ω0
If n is a multiple of 3, then 1ωn=0
So, n must not be a multiple of 3.
Hence, n cannot be 57.

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