A) 0, 2/3
B) 2/3, 11/3
C) 1/2, 1
D) 11/3, 1
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \\ \end{matrix}\, \right|=0\] \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\], we get Þ \[(3x-2)\left| \,\begin{matrix} 1 & 3 & 3 \\ 1 & 3x-8 & 3 \\ 1 & 3 & 3x-8 \\ \end{matrix}\, \right|=0\] \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\]and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\], we get Þ \[(3x-2)\left| \,\begin{matrix} 0 & -3x+11 & 0 \\ 0 & 3x-11 & -3x+11 \\ 1 & 3 & 3x-8 \\ \end{matrix}\, \right|=0\] \[(3x-2)\left[ {{(-3x+11)}^{2}} \right]=0\] \[x=\frac{2}{3}\]or \[x=\frac{11}{3}\,\,\,\,\Rightarrow x=\frac{2}{3},\frac{11}{3}\].
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