JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(\alpha \) and \(\beta \) be the roots of the equation \(x^2 + x + 1 = 0.\) Then for \(y \ne 0\) in \(R,\) \(\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|\) is equal to
- A \(y\,({y^2} - \,3)\)
- B \({y^3} - \,1\)
- C \(y^3\)
- D \(y\,({y^2} - \,1)\)
Answer & Solution
Correct Answer
(C) \(y^3\)
Step-by-step Solution
Detailed explanation
Roots of the equation \({x^2} + x + 1 = 0\) are \(\alpha = \omega \) and \(\beta = {\omega ^2}\) where \(\omega ,{\omega ^2}\)are complex cube roots of unity…
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