Question

Let α and β be the roots of the equation x² + x + 1 = 0. Then for y ≠ 0 in R, |(y+1) α β |
| α (y+β) 1|
| β 1 (y+α) |
is equal to:
(A) y(y² – 3) (B) y³ – 1
(C) y³
(D) y(y² – 1)

Answer 1

$\textbf{Given:}$

$\text{ \alpha and \beta are roots of the equation x^2+x+1=0 }$

$\text{Then,}$

$\alpha+\beta=-1$

$\alpha\beta=1$

$\text{Now,}$

$\left|\begin{array}{ccc}y+1&\alpha&\beta\\\alpha&y+\beta&1\\\beta&1&y+\alpha\end{array}\right|$

$=\left|\begin{array}{ccc}y+\alpha+\beta+1&y+\alpha+\beta+1&y+\alpha+\beta+1\\\alpha&y+\beta&1\\\beta&1&y+\alpha\end{array}\right|$ $R_1\implies\,R_1+R_2+R_3$

$=\left|\begin{array}{ccc}y&y&y\\\alpha&y+\beta&1\\\beta&1&y+\alpha\end{array}\right|$

$\text{Expanding along first row, we get}$

$=y[y^2+(\alpha+\beta)y+\alpha\beta-1]-y[\alpha\,y+{\alpha}^2-\beta]+y[\alpha-\beta\,y-{\beta}^2]$

$=y[y^2+(-1)y+1-1]-\alpha\,y^2-{\alpha}^2y+\beta\,y+\alpha\,y-{\beta}^2y+\alpha\,y$

$=y^3-y^2-y^2(\alpha+\beta)-y({\alpha}^2+{\beta}^2)+(\alpha+\beta)y$

$=y^3-y^2+y^2-y[(\alpha+\beta)^2-2\alpha\,\beta]-y$

$=y^3-y^2+y^2-y[1-2]-y$

$=y^3+y-y$

$=y^3$

$\therefore\bf\left|\begin{array}{ccc}y+1&\alpha&\beta\\\alpha&y+\beta&1\\\beta&1&y+\alpha\end{array}\right|=y^3$

$\implies\text{Option (C) is correct}$