JEE Main 2019 · 9 April · Shift I

JEE Main 2019 Mathematics question on Matrices and Determinants

Let α\alpha and β\beta be the roots of the equation x2+x+1=0x^2 + x + 1 = 0. Then, for y≠0y \neq 0 in R\mathbf{R},
∣y+1αβαy+β1β1y+α∣\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{vmatrix}
is equal to
  1. A.
    y(y2−1)y(y^2-1)
  2. B.
    y(y2−3)y(y^2-3)
  3. C.
    y3−1y^3-1
  4. D.
    y3y^3
Show answer

Correct option: D

[JEE Main 2019 · 9 April · Shift I]

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