S = det ( A ) + det ( B ) k = 1 ∑ 7 det ( A + ω k − 1 B ) + k = 1 ∑ 7 det ( B + ω k − 1 A )
Let A , B ∈ M 7 ( C ) where M 7 ( C ) denotes a square matrix of order 7 × 7 having complex entities in it. Let ω = e 2 π i / 7 . Then find the value of S upto three correct places of decimals.
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Could you please explain c7=|B|.
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t 7 f ( t − 1 ) = g ( t ) , from which we have c 7 = g ( 0 ) = ∣ B ∣ .
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Note that f ( t ) = ∣ ∣ A + t B ∣ ∣ is a polynomial of degree 7 in t , so that f ( t ) = j = 0 ∑ 7 c j t j where c 0 = f ( 0 ) = ∣ A ∣ and c 7 = ∣ B ∣ . Then k = 1 ∑ 7 f ( ω k − 1 ) = j = 0 ∑ 7 c j k = 1 ∑ 7 ω j ( k − 1 ) = 7 ( c 0 + c 7 ) = 7 ( ∣ A ∣ + ∣ B ∣ ) since ω j is a primitive 7 th root of unity, so that ∑ k = 1 7 ω j ( k − 1 ) = 0 , for all 1 ≤ j ≤ 6 .
Similarly k = 1 ∑ 7 g ( ω k − 1 ) = 7 ( ∣ A ∣ + ∣ B ∣ ) where g ( t ) = ∣ ∣ A t + B ∣ ∣ , which makes the answer 7 + 7 = 1 4 . Of course we must assume that ∣ A ∣ + ∣ B ∣ = 0 .