Question

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

• A. $\frac{1}{\alpha }=-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
• B. ${\alpha }^{-1}=a+b+c$
• C. $\alpha +a+b+c=1$
• D. None of the above

Answer 1

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

Given that

$\Rightarrow \left|\begin{array}{ccc}\alpha +a& \alpha & \alpha \\ \alpha & \alpha +b& \alpha \\ \alpha & \alpha & \alpha +c\end{array}\right|0$

$\Rightarrow R_1\to R_1-R_3$

$\Rightarrow \left|\begin{array}{ccc}a& 0& -c\\ 0& b& -c\\ \alpha & \alpha & \alpha +c\end{array}\right|=0$

$\Rightarrow a(b\alpha +bc+c\alpha )+\alpha bc=0$

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

$\Rightarrow \frac{1}{\alpha }=-\frac{ab+bc+ca}{abc}$

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