Let ω be a complex number such that 2ω+1=z,when z=−3 if |1111−ω2−1ω21ω2ω7|=3k then k is equal to

Let Δ=1111−ω2−1ω21ω2ω7=1111ωω21ω2ω∵1+ω+ω2=0 and ω3=1Applying C1→C1+C2+C3, then we getΔ=3110ωω20ω2ω ∵1+ω+ω2=0=3(1+2ω)=−3z=3k( given ) (∵1+2ω=z)∴k=−z

Let ω be a complex number such that 2ω+1=z,when z=−3 if |1111−ω2−1ω21ω2ω7|=3k then k is equal to

Let Δ=1111−ω2−1ω21ω2ω7=1111ωω21ω2ω∵1+ω+ω2=0 and ω3=1Applying C1→C1+C2+C3, then we getΔ=3110ωω20ω2ω ∵1+ω+ω2=0=3(1+2ω)=−3z=3k( given ) (∵1+2ω=z)∴k=−z

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Let $ω$ be a complex number such that $2 ω + 1 = z$ ,when $z = \sqrt{- 3} if | \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} | = 3 k$ then k is equal to

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Detailed Solution

Let $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{array}| = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{array}|$
$(∵ 1 + ω + ω^{2} = 0 and ω^{3} = 1)$
Applying $C_{1} \to C_{1} + C_{2} + C_{3}$ , then we get
$\begin{array}{l} Δ = |\begin{array}{ccc} 3 & 1 & 1 \\ 0 & ω & ω^{2} \\ 0 & ω^{2} & ω \end{array}| (∵ 1 + ω + ω^{2} = 0) \\ = 3 (1 + 2 ω) \\ = - 3 z = 3 k (given) (∵ 1 + 2 ω = z) \\ ∴ k = - z \end{array}$