Question

If the system of equations
$\begin{array}{ccc}x+\alpha y+\alpha z=0& & \\ bx+y+bz=0& & \\ cx+cy+z=0& & \end{array}$
where $\alpha ,b,c$ are non-zero non-unity,has a non-trivial solution,then the value of $\frac{\alpha }{1-\alpha }+\frac{b}{1-b}+\frac{c}{1-c}$ is

• A. 0
• B. 1
• C. -1
• D. $\frac{\alpha bc}{{\alpha }^2+b^2+c^2}$

Answer 1

The correct option is C -1

Here, $D=\left|\begin{array}{ccc}1& \alpha & \alpha \\ b& 1& b\\ c& c& 1\end{array}\right|$
For non-trivial solution,$D=0$
$\left|\begin{array}{ccc}1& \alpha & \alpha \\ b& 1& b\\ c& c& 1\end{array}\right|=0$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}1-\alpha & 0& \alpha \\ -(1-b)& 1-b& b\\ 0& -(1-c)& 1\end{array}\right|=0$
$\Rightarrow (1-\alpha )(1-b)+b(1-c)(1-\alpha )+\alpha (1-b)(1-c)=0$
$\Rightarrow b(1-c)(1-\alpha )+\alpha (1-b)(1-c)=-(1-\alpha )(1-b)$ ...(i)
Now consider,
$\frac{\alpha }{1-\alpha }+\frac{b}{1-b}+\frac{c}{1-c}$
$=\frac{\alpha (1-b)(1-c)+b(1-\alpha )(1-c)+c(1-\alpha )(1-b)}{(1-\alpha )(1-b)(1-c)}$
$=\frac{-(1-\alpha )(1-b)+c(1-\alpha )(1-b)}{(1-\alpha )(1-b)(1-c)}$ (by (i))
$=\frac{-(1-c)(1-\alpha )(1-b)}{(1-\alpha )(1-b)(1-c)}$
$=-1$