Question

If the value of determinant $\left|\begin{array}{ccc}x+1& \alpha & \beta \\ \alpha & x+\beta & 1\\ \beta & 1& x+\alpha \end{array}\right|$is equal to -8, then the value of x, is (where $\alpha ,\beta$ are non real cube roots of unity)

• A. $\pm 2$
  
• B. -2
  
• C. 0
  
• D. 1

Answer 1

The correct option is B

-2

We have
$\alpha =\omega and\beta ={\omega }^2△=\left|\begin{array}{ccc}x& \alpha & \beta \\ x& x+\beta & 1\\ x& 1& x+\alpha \end{array}\right|[C_1\to C_1+C_2+C_{3}andusing1+\omega +{\omega }^2=0]=\left|\begin{array}{ccc}x& \alpha & \beta \\ 0& x+\beta -\alpha & 1-\beta \\ 0& 1-\alpha & x+\alpha -\beta \end{array}\right|[R_2\to R_2-R_{1}and{R}_3\to R_3-R_1]=x[x^2-(\alpha -\beta {)}^2-(1-\alpha )(1-\beta )]=x(x^2-{\alpha }^2-{\beta }^2+2\alpha \beta -1+\alpha +\beta -\alpha \beta )=x[x^2-{\omega }^2-\omega +2-1+\omega +{\omega }^2-1\left]\right[using{\omega }^2=1]=x^3$
According to the given condition, we have
$x^3=-8givesx=-2$