A) 0, - 3
B) 0, 0, - 3
C) 0, 0, 0, - 3
D) None of these
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} 1+x & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \\ \end{matrix}\, \right|\,=\,0\] \[\Rightarrow \] \[4\,\left| \,\begin{matrix} 1 & 21 & 9 \\ 0 & 90 & -45 \\ 0 & -4 & 2 \\ \end{matrix}\, \right|\], \[\left( \begin{align} & {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \\ & {{C}_{2}}\to {{C}_{2}}-{{C}_{3}} \\ \end{align} \right)\] \[\Rightarrow \] \[(x+3)\,\left| \,\begin{matrix} 1 & 0 & 1 \\ 1 & x & 1 \\ 1 & -x & 1+x \\ \end{matrix}\, \right|\,=0\] \[\Rightarrow \] \[(x+3)\,\left| \,\begin{matrix} 1 & 0 & 1 \\ 0 & x & 0 \\ 0 & -x & x \\ \end{matrix}\, \right|\,=0\], \[\left( \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align} \right)\] \[\Rightarrow \] \[(x+3){{x}^{2}}=0\Rightarrow x=0,\,0,\,-3\]. Trick: Obviously the equation is of degree three, therefore it must have three solutions. So check for option (b).You need to login to perform this action.
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