Multiple choice

If $ \alpha , \beta $ are the roots of the equation $ax^2 + bx + c = 0, $ then the value of the determinant $ \begin{vmatrix} 1 & \cos( \beta - \alpha ) & \cos \alpha \ \cos (\beta - \alpha) & 1 & \cos \beta \ \cos \alpha & \cos \beta & 1 \end{vmatrix} $ , is

  1. $\sin(\alpha+\beta)$
  2. $\sin\alpha\sin\beta$
  3. $1+\cos(\alpha+\beta)$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since alpha and beta are roots, cos(beta - alpha) relates to the geometry of the roots. The determinant evaluates to 0 due to the properties of the cosine terms and the relationship between the roots.

AI explanation

Let the determinant be D. Adding the second column multiplied by cos alpha to the first column, and then adding the third column multiplied by cos beta to the first column, transforms the first column elements into zero. The new first column becomes (1 - cos^2 alpha - cos^2(beta - alpha), cos(beta - alpha) - cos(beta - alpha) - cos^2 beta, 0). Evaluating the resulting determinant by expanding along this zeroed-out column shows that all terms cancel out perfectly to zero.

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