A) \[-2\]
B) \[-1\]
C) 0
D) 1
E) 2
Correct Answer: C
Solution :
Given,\[\alpha ,\beta \]and\[\gamma \]are the cube roots of unity, then assume\[\alpha =1,\text{ }\beta =\omega \]and\[\gamma ={{\omega }^{2}}\]. \[\therefore \]\[\left| \begin{matrix} {{e}^{\alpha }} & {{e}^{2\alpha }} & {{e}^{3\alpha }}-1 \\ {{e}^{\beta }} & {{e}^{2\beta }} & {{e}^{3\beta }}-1 \\ {{e}^{\gamma }} & {{e}^{2\gamma }} & {{e}^{3\gamma }}-1 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{e}^{\alpha }} & {{e}^{2\alpha }} & {{e}^{3\alpha }} \\ {{e}^{\beta }} & {{e}^{2\beta }} & {{e}^{3\beta }} \\ {{e}^{\gamma }} & {{e}^{2\gamma }} & {{e}^{3\gamma }} \\ \end{matrix} \right|+\left| \begin{matrix} {{e}^{\alpha }} & {{e}^{2\alpha }} & -1 \\ {{e}^{\beta }} & {{e}^{2\beta }} & -1 \\ {{e}^{\gamma }} & {{e}^{2\gamma }} & -1 \\ \end{matrix} \right|\] \[={{e}^{\alpha }}{{e}^{\beta }}{{e}^{\gamma }}\left| \begin{matrix} 1 & {{e}^{\alpha }} & {{e}^{2\alpha }} \\ 1 & {{e}^{\beta }} & {{e}^{2\beta }} \\ 1 & {{e}^{\gamma }} & {{e}^{2\gamma }} \\ \end{matrix} \right|-\left| \begin{matrix} 1 & {{e}^{\alpha }} & {{e}^{2\alpha }} \\ 1 & {{e}^{\beta }} & {{e}^{2\beta }} \\ 1 & {{e}^{\gamma }} & {{e}^{2\gamma }} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & {{e}^{\alpha }} & {{e}^{2\alpha }} \\ 1 & {{e}^{\beta }} & {{e}^{2\beta }} \\ 1 & {{e}^{\gamma }} & {{e}^{2\gamma }} \\ \end{matrix} \right|-[{{e}^{\alpha }}{{e}^{\beta }}{{e}^{\gamma }}-1]=0\] \[(\because {{e}^{\alpha }}{{e}^{\beta }}{{e}^{\gamma }}={{e}^{1+\omega +{{\omega }^{2}}}}={{e}^{0}}=1)\]
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