A) \[\alpha \]
B) \[\beta \]
C) \[\gamma \]
D) All of these
Correct Answer: A
Solution :
\[\left| \begin{matrix} x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1 \\ \end{matrix} \right|=0\] Applying, \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}-{{R}_{1}},\]we get \[\left| \begin{matrix} x & \alpha & 1 \\ \beta -x & x-\alpha & 0 \\ \beta -x & \gamma -a & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(\beta -x)(\gamma -\alpha )-(\beta -x)(x-\alpha )=0\] \[\Rightarrow \] \[(\beta -x)(\gamma -\alpha -x+\alpha )=0\] \[\Rightarrow \] \[(\beta -x)(\gamma -x)=0\] \[\Rightarrow \] \[x=\beta ,\gamma \] Hence, roots of the given equations are independent of a.
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