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. $\alpha =a+b+c$

Answer 1

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

Homogeneous steamy has non- trivial solution it's mean determinant is zero

So, $\left|\begin{array}{ccc}(\alpha +a)& \alpha & \alpha \\ \alpha & (\alpha +b)& \alpha \\ \alpha & \alpha & \alpha +c\end{array}\right|=0$

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

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

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

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

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

$\Rightarrow \frac{1}{\alpha }=-\frac{-(ab+bc+ac)}{abc}$

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

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