Question

If $\alpha$ and $\beta$ are non-real numbers satisfying $x^3-1=0,$ then the value of $\left|\begin{array}{ccc}\lambda +1& \alpha & \beta \\ \alpha & \lambda +\beta & 1\\ \beta & 1& \lambda +\alpha \end{array}\right|$ is

• A. 0
• B. ${\lambda }^3$
• C. $x^2+2bx+4=0$
• D. $x^2-bx+1=0$

Answer 1

Answer: (B)

It is given that $\alpha$ and $\beta$ are the non-real roots of the equation $x^3-1=0$ . We have, $x^3-1=0$ $\Rightarrow$ $x^3=1$ $\Rightarrow$ $x=1,\omega ,{\omega }^2$ Hence, $\alpha =\omega$ and $\beta ={\omega }^2$ Now, $\left|\begin{array}{ccc}\lambda +1& \alpha & \beta \\ \alpha & \lambda +\beta & 1\\ \beta & 1& \lambda +\alpha \end{array}\right|$ $=\left|\begin{array}{ccc}\lambda +1+\alpha +\beta & \lambda +1+\alpha +\beta & \lambda +1+a+\beta \\ \alpha & \alpha +\beta & 1\\ \beta & 1& \lambda +\alpha \end{array}\right|$ $[\because R_1\to R_1+R_2+R_3]$ $=(\lambda +1+\alpha +\beta )\left|\begin{array}{ccc}0& 0& 1\\ \alpha -\lambda -\beta & \lambda +\beta -1& 1\\ \beta -1& 1-\lambda -\alpha & \lambda +\alpha \end{array}\right|$ $[\because C_1\to C_1-C_2,C_2\to C_2-C_3]$ $=(\lambda +1+\alpha +\beta ).1[(\alpha -\lambda -\beta )(1-\lambda -\alpha )$ $-(\beta -1)(\lambda +\beta -1)]$ $=(\lambda +1+\alpha +\beta )(\alpha -{\alpha }^2+{\lambda }^2+\alpha \beta -{\beta }^2+\beta -1)$ $(\lambda +1+\omega +{\omega }^2)(\omega -{\omega }^2+{\lambda }^2+{\omega }^3-{\omega }^4+{\omega }^2-1)$ [putting $\alpha =\omega$ and $\beta ={\omega }^2$ ] $=\lambda (\omega -{\omega }^2+{\lambda }^2+1-\omega +{\omega }^2-1)$ $={\lambda }^3$