Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $x^2+x+1=0$. Then for $y\neq 0$ in R, $\begin{vmatrix} y+1 & \alpha & \beta\ \alpha & y+\beta & 1\ \beta & 1 & y+\alpha\end{vmatrix}$ is equal to?

  1. $y^3$
  2. $y^3-1$
  3. $y(y^2-1)$
  4. $y(y^2-3)$
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A Correct answer
Explanation

Given x^2+x+1=0, the roots are omega and omega^2, where alpha+beta = -1 and alpha*beta = 1. Substituting these into the determinant and simplifying yields y^3.

AI explanation

Add the second and third rows to the first row to simplify the matrix. The new first row becomes (y + 1 + alpha + beta, y + 1 + alpha + beta, y + 1 + alpha + beta). Since alpha and beta are roots of x^2 + x + 1 = 0, their sum is alpha + beta = -1. The first row therefore becomes (y, y, y), allowing you to factor out y. Subtracting the new first row from the second and third rows reduces the matrix to a lower triangular form with ones on the diagonal, leaving the determinant as y multiplied by 1 multiplied by 1. The value of the determinant is y^3.

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