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

Answer 1

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

Given system of equations $AX=O$
where $A=\left[\begin{array}{ccc}\alpha +a& \alpha & \alpha \\ \alpha & \alpha +b& \alpha \\ \alpha & \alpha & \alpha +c\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],O=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
For non-trivial solution,
$D=0$
$\left|\begin{array}{ccc}\alpha +a& \alpha & \alpha \\ \alpha & \alpha +b& \alpha \\ \alpha & \alpha & \alpha +c\end{array}\right|=0$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Rightarrow \left|\begin{array}{ccc}a& -b& 0\\ 0& b& -c\\ \alpha & \alpha & \alpha +c\end{array}\right|=0$
$\Rightarrow ab\alpha +abc+ac\alpha +bc\alpha =0$
$\Rightarrow abc=-\alpha (ab+bc+ca)$
$\Rightarrow \frac{1}{\alpha }=-(\frac{1}{c}+\frac{1}{a}+\frac{1}{b})$

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