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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
58
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C
56
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D
55
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Solution

## The correct option is D 55P=[Pij]n×n P2≠0 Pij=ωi+j P=⎡⎢ ⎢ ⎢ ⎢⎣ω21ωω21...1ωω21ω...ωω21ωω2.....................⎤⎥ ⎥ ⎥ ⎥⎦n×n P2=⎡⎢ ⎢ ⎢ ⎢⎣ω21ωω21...1ωω21ω...ωω21ωω2.....................⎤⎥ ⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢ ⎢⎣ω21ωω21...1ωω21ω...ωω21ωω2.....................⎤⎥ ⎥ ⎥ ⎥⎦ (ω4+1+ω2)+(ω4+1+ω2)+...=0 Only when n is a multiple of 3. ∴n can be 55,58,56 (p2≠0)

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