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Question

# Let ω be a complex cube root of unity with ω≠1 and P=[pij] be a n × n matrix with pij=ωi+j. Then P2≠0, 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 56P=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ω2ω3ω4....ωn+1ω3ω4ω5....ωn+2ω4ω5ω6....ωn+3................ωn+1ωn+2......ω2n⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦P=ω2ω3ω4.....ωn+1⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1ωω2....ωn−11ωω2....ωn−11ωω2....ωn−1................1ωω2...ωn−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦P=ωn(n+3)2⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1ωω2....ωn−11ωω2....ωn−11ωω2....ωn−1................1ωω2...ωn−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ Now, P2=ωn2+3n⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1ωω2....ωn−11ωω2....ωn−11ωω2....ωn−1................1ωω2...ωn−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1ωω2....ωn−11ωω2....ωn−11ωω2....ωn−1................1ωω2...ωn−1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦Since, P2≠O⇒1+ω+ω2+....ωn−1≠0⇒1−ωn1−ω≠0If n is a multiple of 3, then 1−ωn=0So, n must not be a multiple of 3.Hence, n cannot be 57.

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