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If w is a complex cube root of unity and A =

, then A 98 is equal to
(a) w I (b) w 2 (c) A2 (d) A itself

Dear Student,
Please find below the solution to the asked query:

A=ω00ω⇒A=ω1001⇒A=ωI, where I is identity matrix⇒A98=ω98I98But INatural Number=I⇒I98=I⇒A98=ω98I=ω96.ω2IAs 96 is multiple of 3, hence ω96=1 As ω3n=1, n∈Z ⇒A98=ω2I=ω21001⇒A98=ω200ω2.....iA2=ω00ωω00ω=ω2+00+00+00+ω2A2=ω200ω2...iiBy i and iiA98=A2...OptionC

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Hello Khushraj,it's answer is (i) Because, IAI= w^2 and IAI^98=w^196 but w^3= 1 and when we divide 196 by 3 only 1 is left as remainder so, at last we left with= (w^3)^65 * w therefore = 1*w= w correct answer = w (i)
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