JEE Main 20199 Apr, Morning shiftMathematicsHardMatrices and Determinants245× since 2002Let α\alphaα and β\betaβ be the roots of the equation x² + x + 1 = 0. Then for y ≠\ne= 0 in R, ∣y+1αβαy+β1β1y+α∣\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|y+1αβαy+β1β1y+α is equal toAOption A. y(y² – 1)BOption B. y(y² – 3)COption C. y³DOption D. y³ – 1CheckSkip · show solution
