A) \[{{b}^{2}}pr={{q}^{2}}ac\]
B) \[bp{{r}^{2}}=qa{{c}^{2}}\]
C) \[b{{p}^{2}}r=q{{a}^{2}}c\]
D) None of these
Correct Answer: A
Solution :
Given system has non-trivial solution then \[\left| \begin{matrix} {{\alpha }_{1}} & {{\alpha }_{2}} \\ {{\beta }_{1}} & {{\beta }_{2}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[{{\alpha }_{1}}{{\beta }_{2}}={{\alpha }_{2}}{{\beta }_{1}}\] \[\Rightarrow \] \[\frac{{{\alpha }_{1}}}{{{\beta }_{1}}}=\frac{{{\alpha }_{2}}}{{{\beta }_{2}}}=\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{{{\beta }_{1}}+{{\beta }_{2}}}=\sqrt{\frac{{{\alpha }_{1}}{{\alpha }_{2}}}{{{\beta }_{1}}{{\beta }_{2}}}}\] \[\Rightarrow \] \[\frac{-b/a}{-q/p}=\sqrt{\frac{c/a}{r/p}}\] \[\Rightarrow \] \[\frac{bp}{aq}=\sqrt{\frac{cp}{ar}}\] \[\Rightarrow \] \[{{b}^{2}}{{p}^{2}}ar={{a}^{2}}{{q}^{2}}cp\] \[\Rightarrow \] \[{{b}^{2}}pr={{q}^{2}}ac\]
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