A) \[1\]
B) \[-1\]
C) \[0\]
D) None of these
Correct Answer: C
Solution :
[c] \[{{a}_{r}}={{(cos2r\pi +i\,\,sin\,\,2r\pi )}^{\frac{1}{9}}}={{a}_{r}}\] \[={{({{e}^{i2r\pi }})}^{\frac{1}{9}}}={{e}^{\frac{i2r\pi }{9}}}\] \[\Rightarrow \left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\ {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\ \end{matrix} \right|=\left| \begin{matrix} {{e}^{\frac{i2\pi }{9}}} & {{e}^{\frac{i4\pi }{9}}} & {{e}^{\frac{i6\pi }{9}}} \\ {{e}^{\frac{i2\pi }{9}}} & {{e}^{\frac{i10\pi }{9}}} & {{e}^{\frac{i12\pi }{9}}} \\ {{e}^{\frac{i14\pi }{9}}} & {{e}^{\frac{i16\pi }{9}}} & {{e}^{\frac{i18\pi }{9}}} \\ \end{matrix} \right|\] \[={{e}^{\frac{i8\pi }{9}}}\left| \begin{matrix} {{e}^{\frac{i2\pi }{9}}} & {{e}^{\frac{i4\pi }{9}}} & {{e}^{\frac{i6\pi }{9}}} \\ {{e}^{\frac{i2\pi }{9}}} & {{e}^{\frac{i4\pi }{9}}} & {{e}^{\frac{i6\pi }{9}}} \\ {{e}^{\frac{i14\pi }{9}}} & {{e}^{\frac{i16\pi }{9}}} & {{e}^{\frac{i18\pi }{9}}} \\ \end{matrix} \right|\] = (taking \[{{e}^{\frac{i8\pi }{9}}}\] common from second row) = 0
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