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Q.

Elements of a matrix A of order 10  × 10 are defined as aij=ωi+j (where ω is imaginary cube root of unity), then trace (A) of the matrix is

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a

0

b

ω3

c

3

d

ω2

answer is D.

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Detailed Solution

tr(A)=i=jaij =(a11+a22+a33+...+a1010) =(ω2+ω4+ω6+...+ω20) =ω2(1+  ω2+  ω4+...+  ω18) =ω2[(1+ω+ω2)+...+(1+ω+ω2)+1] =ω2×1 =ω2

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