JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer Let\[\alpha \]and \[\beta \]be the roots of the equation \[{{x}^{2}}+x+1=0.\] Then for \[y\ne 0\]in R, \[\left| \begin{matrix}    y+1 & \alpha  & \beta   \\    \alpha  & y+\beta  & 1  \\    \beta  & 1 & y+\alpha   \\ \end{matrix} \right|\]is equal to             [JEE Main 9-4-2019 Morning]

    A) \[{{y}^{3}}\]           

    B) \[{{y}^{3}}-1\]

    C) \[y({{y}^{2}}-1)\]                

    D) \[y({{y}^{2}}-3)\]

    Correct Answer: A

    Solution :

    Roots of the equation \[{{x}^{2}}+x+1=0\]are \[\alpha =\omega \] and \[\beta ={{\omega }^{2}}\] where \[\omega ,{{\omega }^{2}}\]are complex cube roots of unity \[\therefore \]\[\Delta =\left| \begin{matrix}    y+1 & \omega  & {{\omega }^{2}}  \\    \omega  & y+{{\omega }^{2}} & 1  \\    {{\omega }^{2}} & 1 & y+\omega   \\ \end{matrix} \right|\] \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] \[\Rightarrow \Delta =y\left| \begin{matrix}    1 & 1 & 1  \\    \omega  & y+{{\omega }^{2}} & 1  \\    {{\omega }^{2}} & 1 & y+\omega   \\ \end{matrix} \right|\] Expanding along \[{{R}_{1}},\]we get\[\Delta =y.{{y}^{2}}\Rightarrow D={{y}^{3}}\]                     


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