A) 0
B) \[{{\lambda }^{3}}\]
C) \[{{x}^{2}}+2bx+4=0\]
D) \[{{x}^{2}}-bx+1=0\]
Correct Answer: B
Solution :
It is given that \[\alpha \] and \[\beta \] are the non-real roots of the equation\[{{x}^{3}}-1=0\]. We have, \[{{x}^{3}}-1=0\] \[\Rightarrow \]\[{{x}^{3}}=1\] \[\Rightarrow \]\[x=1,\omega ,{{\omega }^{2}}\] Hence, \[\alpha =\omega \] and \[\beta ={{\omega }^{2}}\] Now, \[\left| \begin{matrix} \lambda +1 & \alpha & \beta \\ \alpha & \lambda +\beta & 1 \\ \beta & 1 & \lambda +\alpha \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \lambda +1+\alpha +\beta & \lambda +1+\alpha +\beta & \lambda +1+a+\beta \\ \alpha & \alpha +\beta & 1 \\ \beta & 1 & \lambda +\alpha \\ \end{matrix} \right|\] \[[\because {{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}]\] \[=(\lambda +1+\alpha +\beta )\left| \begin{matrix} 0 & 0 & 1 \\ \alpha -\lambda -\beta & \lambda +\beta -1 & 1 \\ \beta -1 & 1-\lambda -\alpha & \lambda +\alpha \\ \end{matrix} \right|\] \[[\because {{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}]\] \[=(\lambda +1+\alpha +\beta ).1[(\alpha -\lambda -\beta )(1-\lambda -\alpha )\] \[-(\beta -1)(\lambda +\beta -1)]\] \[=(\lambda +1+\alpha +\beta )(\alpha -{{\alpha }^{2}}+{{\lambda }^{2}}+\alpha \beta -{{\beta }^{2}}+\beta -1)\] \[(\lambda +1+\omega +{{\omega }^{2}})(\omega -{{\omega }^{2}}+{{\lambda }^{2}}+{{\omega }^{3}}-{{\omega }^{4}}+{{\omega }^{2}}-1)\] [putting \[\alpha =\omega \]and \[\beta ={{\omega }^{2}}\]] \[=\lambda (\omega -{{\omega }^{2}}+{{\lambda }^{2}}+1-\omega +{{\omega }^{2}}-1)\] \[={{\lambda }^{3}}\]
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