Question

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

• A. $y^3$
• B. $y^3-1$
• C. $y(y^2 - 1)$
• D. $y(y^2 - 3)$

Answer 1

Answer: (A)

Roots of the equation $x^2 + x + 1 =0$ are
$\alpha = \omega$ and $\beta = \omega^2$
where $\omega, \omega^2$ are complex cube roots of unity
$\therefore \ \delta = \begin {vmatrix} y+1 & \omega & \omega^2 \\ \omega & y+\omega^2 & 1 \\ \omega^2 & 1 & y+\omega \end {vmatrix}$
$R_1 \rightarrow R_1 + R_2 + R_3$
$\Rightarrow \ \ \Delta = y \begin {vmatrix} 1 & 1 & 1 \\ \omega & y+\omega^2 & 1 \\ \omega^2 & 1 & y+\omega \end {vmatrix}$
Expanding along $R_1,$ we get
$\Delta = y, y^2 \Rightarrow \ \ D \ = \ y^3$