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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

  1. A \(y\,({y^2} - \,3)\)
  2. B \({y^3} - \,1\)
  3. C \(y^3\)
  4. D \(y\,({y^2} - \,1)\)
Verified Solution

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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