Determinants Matrices Question 179
Question: If $ a _{r}={{(\cos 2r\pi +i\sin 2r\pi )}^{\frac{1}{9}}}, $ then the value of $ \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \\ \end{vmatrix} $ is
Options:
A) $ 1 $
B) $ -1 $
C) $ 0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ a _{r}={{(cos2r\pi +isin2r\pi )}^{\frac{1}{9}}}=a _{r} $
$ ={{({e^{i2r\pi }})}^{\frac{1}{9}}}={e^{\frac{i2r\pi }{9}}} $
$ \Rightarrow \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \\ \end{vmatrix}= \begin{vmatrix} {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{vmatrix} $
$ ={e^{\frac{i8\pi }{9}}} \begin{vmatrix} {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{vmatrix} $ = (taking $ {e^{\frac{i8\pi }{9}}} $ common from second row) = 0
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