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Question

# Let ω be a complex cube root of unity with ω≠0 and P=[pij] be an n×n matrix with Pij=ωi+j. Then P2=0 when n is equal to

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 option is A 57 Here Pijn×n with pij=ωi+j ∴ When n = 1 P=[pij]1×1=[ω2]⇒ P2=[ω4]≠0 ∴ When n = 2 P=[pij]2×2=[P11P12P21P22]=[ω2ω3ω3ω4]=[ω211ω] P2=[ω211ω][ω211ω]⇒P2=[ω4+1ω2+ωω2+ω1+ω2]≠0 When n = 3 P=[pij]3×3=⎡⎢⎣ω2ω3ω4ω3ω4ω5ω4ω5ω6⎤⎥⎦=⎡⎢⎣ω21ω1ωω2ωω21⎤⎥⎦ P2=⎡⎢⎣ω21ω1ωω2ωω21⎤⎥⎦⎡⎢⎣ω21ω1ωω2ωω21⎤⎥⎦=⎡⎢⎣000000000⎤⎥⎦=0 ∴ P2 = 0, when n is a multiple of 3. P2≠0, when n is not a multiple of 3. ⇒ n = 57 is possible. ∴ n = 55, 58, 56 is not possible.

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