Question

If $\alpha ,\beta$ 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 equal to

• A. \N
• B. ${\lambda }^3$
• C. ${\lambda }^3+1$
• D. ${\lambda }^3-1$

Answer 1

The correct option is B ${\lambda }^3$

$x^3-1=0\therefore x=1,\omega ,{\omega }^2$
Here, $\alpha =\omega ,\beta ={\omega }^2\left|\begin{array}{ccc}\lambda +1& \omega & {\omega }^2\\ \omega & \lambda +{\omega }^2& 1\\ {\omega }^2& 1& \lambda +\omega \end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$, then
$\left|\begin{array}{ccc}\lambda & \omega & {\omega }^2\\ \lambda & \lambda +{\omega }^2& 1\\ \lambda & 1& \lambda +\omega \end{array}\right|$
Applying $R-2\to R_2-R_1$ and $R_3\to R_3-R_1$, then we get
$\left|\begin{array}{ccc}\lambda & \omega & {\omega }^2\\ 0& \lambda +{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & \lambda +\omega -{\omega }^2\end{array}\right|=\lambda \left(\right(\lambda +{\omega }^2-\omega \left)\right(\lambda +\omega -{\omega }^2)-(1-\omega \left)\right(1-{\omega }^2\left)\right)=\lambda \left({\lambda }^2\right)={\lambda }^3$