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. ${\lambda }^3+1$
• D. ${\lambda }^3-1$

Answer 1

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

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

$=\left|\begin{array}{ccc}\lambda +1+\alpha +\beta & \lambda +1+\alpha +\beta & \lambda +\alpha +\beta +1\\ \alpha & \lambda +\beta & 1\\ \beta & 1& \lambda +\alpha \end{array}\right|$ $[\because R_1\to R_1+R_2+R_3]$

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

$[\because C_1\to C_1-C_2,C_2\to C_2-C_3]$

$=(\lambda +1+\alpha +\beta )\left|\begin{array}{ccc}0& 0& 1\\ \alpha -\lambda -\beta & \lambda +\beta -1& 1\\ \beta -1& 1-\lambda -\alpha & \lambda +\alpha \end{array}\right|$

$=(\lambda +1+\alpha +\beta )\cdot 1\left[\right(\alpha -\lambda -\beta \left)\right(1-\lambda -\alpha )-(\beta -1\left)\right(\lambda +\beta -1\left)\right]$
$=(\lambda +1+\alpha +\beta )(\alpha -{\alpha }^2+{\lambda }^2+\alpha \beta -{\beta }^2+\beta -1)$
$=(\lambda +1+\omega +{\omega }^2)(\omega -{\omega }^2+{\lambda }^2+{\omega }^3-{\omega }^4+{\omega }^2-1)$ (putting $\alpha =\omega$ and $\beta ={\omega }^2$)
$=\lambda (\omega -{\omega }^2+{\lambda }^2+1-\omega +{\omega }^2-1)$ ..... $[\because {\omega }^3=1⟹1+\omega +{\omega }^2=0]$
$={\lambda }^3$.