A) 0, 2a
B) a, 2a
C) 0, 3a
D) None of these
Correct Answer: C
Solution :
\[\Delta =\left| \begin{matrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \\ \end{matrix} \right|=0\] \[\Rightarrow \Delta =\left| \begin{matrix} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \\ \end{matrix} \right|,{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \\ \end{matrix} \right|=0\] Using \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\Rightarrow \] \[\Delta =(3a-x)\left| \begin{matrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \\ \end{matrix} \right|=0\] Or, \[4{{x}^{2}}(3a-x)=0\Rightarrow x=0\]
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