A) \[1\]
B) \[-2\]
C) \[2\]
D) \[\omega \]
Correct Answer: B
Solution :
[b] Given matrix is: \[\left| \begin{matrix} {{x}^{2}} & -2x & -2{{\omega }^{2}} \\ 2 & \omega & -\omega \\ 0 & \omega & 1 \\ \end{matrix} \right|=0\] By \[{{C}_{2}}\to {{C}_{2}}+{{C}_{3}},\] we get \[\Rightarrow \left| \begin{matrix} {{x}^{2}} & -2x-2{{\omega }^{2}} & -2{{\omega }^{2}} \\ 2 & 0 & -\omega \\ 0 & 1+\omega & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \left| \begin{matrix} {{x}^{2}} & -2x-2{{\omega }^{2}} & -2{{\omega }^{2}} \\ 2 & 0 & -\omega \\ 0 & -{{\omega }^{2}} & 1 \\ \end{matrix} \right|=0\] \[[\because \,\,1+\omega =-{{\omega }^{2}}]\] \[\Rightarrow {{\omega }^{2}}\left| \begin{matrix} {{x}^{2}} & -2{{\omega }^{2}} \\ 2 & -\omega \\ \end{matrix} \right|+1\left| \begin{matrix} {{x}^{2}} & -2x-2{{\omega }^{2}} \\ 2 & -0 \\ \end{matrix} \right|=0\] \[\Rightarrow {{\omega }^{2}}(-\omega {{x}^{2}}+4{{\omega }^{2}})-(-4x-4{{\omega }^{2}})=0\] \[\Rightarrow -{{x}^{2}}+4\omega +4x+4{{\omega }^{2}}=0\] \[\Rightarrow -{{x}^{2}}+4\omega -4x-4-4\omega =0\] \[\Rightarrow -{{x}^{2}}-4x-4=0\] \[\Rightarrow {{(x+2)}^{2}}=0\Rightarrow x=-2\]
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