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. $0$
• B. ${\lambda }^3$
• C. ${\lambda }^3+1$
• D. none of these

Answer 1

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

Since, $\alpha ,\beta$ are non-real numbers satisfying $x^3-1=0$

$(x-1)(x^2+x+1)=0$

So, $x=1,\alpha =\omega ,\beta ={\omega }^2$

$\Rightarrow 1+\alpha +\beta =0$
Now,$\left|\begin{array}{ccc}\lambda +1& \alpha & \beta \\ \alpha & \lambda +\beta & 1\\ \beta & 1& \lambda +\alpha \end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}\lambda +1+\alpha +\beta & \alpha & \beta \\ \lambda +1+\alpha +\beta & \lambda +\beta & 1\\ \lambda +1+\alpha +\beta & 1& \lambda +\alpha \end{array}\right|$
$\lambda \left|\begin{array}{ccc}1& \alpha & \beta \\ 1& \lambda +\beta & 1\\ 1& 1& \lambda +\alpha \end{array}\right|$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$\lambda \left|\begin{array}{ccc}1& \alpha & \beta \\ 0& \lambda +\beta -\alpha & 1-\beta \\ 0& 1-\alpha & \lambda +\alpha -\beta \end{array}\right|$
$=\lambda \left[\right(\lambda +\beta -\alpha \left)\right(\lambda -(\beta -\alpha ))-(1-\alpha \left)\right(1-\beta \left)\right]$
$=\lambda [{\lambda }^2-{(\beta -\alpha )}^2-(1-\alpha -\beta +\alpha \beta \left)\right]$
$=\lambda [{\lambda }^2-({\omega }^2+{\omega }^4-2)-(1+1+1\left)\right]$
$={\lambda }^3$