For θ=2π5,letB=[bij] be a square matrix of order 2 such that bij=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩cosθ,i=jcos(jπ2+θ),i>jcos(iπ2−θ),i<j⎫⎪
⎪
⎪⎬⎪
⎪
⎪⎭ The trace of matrix B5 is equal to
A
-2
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B
-1
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C
1
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D
2
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Solution
The correct option is D 2 B=[cosθsinθ−sinθcosθ] B2=[cosθsinθ−sinθcosθ][cosθsinθ−sinθcosθ]=[cos2θ−sin2θcosθsinθ+sinθcosθ−cosθsinθ−sinθcosθ−sin2θ+cos2θ]=[cos2θsin2θ−sin2θcos2θ],Similarly,B5=[cos5θsin5θ−sin5θcos5θ] So, Tr(B5)=2cos5θ at θ=2π5=2×1=2