Question

If a,b,c are non-zeros, then the system of equations
$(\alpha +a)x+\alpha y+\alpha z=0$
$\alpha x+(\alpha +b)y+\alpha z=0$
$\alpha x+\alpha y+(\alpha +c)z=0$
has a non-trivial solution if

• A. ${\alpha }^{-1}=-(a^{-1}+b^{-1}+c^{-1})$
• B. $a^{-1}=a+b+c$
• C. $\alpha +a+b+c=1$
• D. none of these

Answer 1

The correct option is A ${\alpha }^{-1}=-(a^{-1}+b^{-1}+c^{-1})$

Non trivial solutions

$\left[\begin{array}{ccc}\alpha +a& \alpha & \alpha \\ \alpha & \alpha +b& \alpha \\ \alpha & \alpha & \alpha +c\end{array}\right]=0$

$\Rightarrow (\alpha +a)\left[\right(\alpha +b\left)\right(\alpha +c)-{\alpha }^2)-\alpha \left(\alpha \right(\alpha +c)-{\alpha }^2)+\alpha ({\alpha }^2-\alpha (\alpha +b\left)\right)=0$

$\Rightarrow (\alpha +a)({\alpha }^2+c\alpha +b\alpha +bc-{\alpha }^2)-\alpha ({\alpha }^2+\alpha c-{\alpha }^2)+\alpha ({\alpha }^2-{\alpha }^2-b\alpha )=0$

$\Rightarrow (\alpha +a)(b\alpha +c\alpha +bc){\alpha }^{2}c-{\alpha }^{2}b=0$

$\Rightarrow b{\alpha }^2+c{\alpha }^2+bc\alpha +ab\alpha +ca\alpha +abc-{\alpha }^{2}c-{\alpha }^{2}b=0$

$\Rightarrow \alpha (bc+ab+ca)=-abc$

$\Rightarrow \frac{1}{a}+\frac{1}{c}+\frac{1}{b}=-\frac{1}{\alpha }$