A) 1, 2
B) - 1, 2
C) 1, - 2
D) -1, - 2
Correct Answer: B
Solution :
We have \[\left| \,\begin{matrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \\ \end{matrix}\, \right|\,=0\] \[\Rightarrow \]\[\left| \,\begin{matrix} x+1 & 1 & 1 \\ x+1 & x-1 & 1 \\ x+1 & 1 & x-1 \\ \end{matrix}\, \right|\,=0\], {Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]} \[\Rightarrow \]\[(x+1)\,\left| \,\begin{matrix} 1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \\ \end{matrix}\, \right|\]= 0 \[\Rightarrow \] \[(x+1)\,\left| \,\begin{matrix} 1 & 1 & 1 \\ 0 & x-2 & 0 \\ 0 & 0 & x-2 \\ \end{matrix}\, \right|=0\] {Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]} \[\Rightarrow \] \[N=\left[ \begin{matrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{matrix} \right]\].
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