Question

The system of equations
$(a\alpha +b)x+ay+bz=0$
$(b\alpha +c)x+by+cz=0$
$(a\alpha +b)y+(b\alpha +c)z=0$ has a non-trivial solution, if

• A. a, b, c are in AP.
• B. a, b, c are in G.P.
• C. a, b, c are in H.P.
• D. $\alpha$ is a root of $a{x}^2+2bx+c=0$

Answer 1

Answer: (B), (D)

a, b, c are in G.P.
$\alpha$ is a root of $a{x}^2+2bx+c=0$

The given system of equations will have a non-trivial solution if
$\Delta =\left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ 0& a\alpha +b& b\alpha +c\end{array}\right|=0$
Applying $R_3\to R_3-\alpha R_1-R_2$, we get
$\Delta =\left|\begin{array}{ccc}a\alpha +b& a& b\\ b\alpha +c& b& c\\ -(a{\alpha }^2+2b\alpha +c)& 0& 0\end{array}\right|=0$
$\Rightarrow -(a{\alpha }^2+2b\alpha +c)(ac-b^2)=0$
$\Rightarrow a{\alpha }^2+2b\alpha +c=0orac-b^2=0$
$\Rightarrow \alpha$ is a root of $a{x}^2+2bx+c=0$ or $a,b,c$ are in G.P.
Hence, option B and D.