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.
You will be redirected in
3 sec
