$I f α, β a r e t h e r o o t s o f x^{2} + x + 1 = 0, a n d S_{n} = α^{n} + β^{n} t h e n | \begin{matrix} 3 & 1 + S_{1} & 1 + S_{2} \\ 1 + S_{1} & 1 + S_{2} & 1 + S_{3} \\ 1 + S_{2} & 1 + S_{3} & 1 + S_{4} \end{matrix} | =$

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-27

-3

answer is B.

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Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s o f x^{2} + x + 1 = 0 t h e n \\ α + β = - 1, α β = 1 \\ G i v e n S_{n} = α^{n} + β^{n} \\ S_{1} = α + β = - 1 \\ S_{2} = α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 1 - 2 = - 2 \\ S_{3} = α^{3} + β^{3} = {(α + β)}^{3} - 3 α β (α + β) = - 1 - 3 (1) (- 1) = - 1 + 3 = 2 \\ S_{4} = α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2} = 1 - 2 (1) = - 1 \\ | \begin{matrix} 3 & 1 + S_{1} & 1 + S_{2} \\ 1 + S_{1} & 1 + S_{2} & 1 + S_{3} \\ 1 + S_{2} & 1 + S_{3} & 1 + S_{4} \end{matrix} | = | \begin{matrix} 3 & 0 & 0 \\ 0 & 0 & 3 \\ 0 & 3 & 0 \end{matrix} | = 3 (0 - 9) = - 27 \end{array}$