• # 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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