Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2+x+1=0$. Then for y $\ne$ 0 in R, $\left|\begin{array}{ccc}y+1& \alpha & \beta \\ \alpha & y-\beta & 1\\ \beta & 1& y+\alpha \end{array}\right|$ 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
$\therefore \delta =\left|\begin{array}{ccc}y+1& \omega & {\omega }^2\\ \omega & y+{\omega }^2& 1\\ {\omega }^2& 1& y+\omega \end{array}\right|$
$R_1\to R_1+R_2+R_3$
$\Rightarrow \Delta =y\left|\begin{array}{ccc}1& 1& 1\\ \omega & y+{\omega }^2& 1\\ {\omega }^2& 1& y+\omega \end{array}\right|$
Expanding along $R_1,$ we get
$\Delta =y,y^2\Rightarrow D=y^3$