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Matrices and Determinants question

2019 · 9 Apr · Shift 1 · Q37
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  5. /2019 · 9 Apr · Shift 1 · Q37

Matrices and Determinants question

2019 · 9 Apr · Shift 1 · Q37

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of the equation x2 + x + 1 = 0. Then for y eee 0 in R, ∣y+1αβαy+β1β1y+α∣\left| {\begin{matrix} {y + 1} & \alpha & \beta \\ \alpha & {y + \beta } & 1 \\ \beta & 1 & {y + \alpha } \\ \end{matrix} } \right|​y+1αβ​αy+β1​β1y+α​​ is equal to
  1. A
    y(y2 – 1)
  2. B
    y(y2 – 3)
  3. C
    y3
  4. D
    y3 – 1
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