A) \[-2+\sqrt{2}\]
B) \[2-\sqrt{2}\]
C) \[-2-\sqrt{2}\]
D) 1
Correct Answer: C
Solution :
[c] Expanding by Sarrus rule, \[\left| \begin{matrix} 1 & {{e}^{i\pi /3}} & {{e}^{i\pi /4}} \\ {{e}^{-i\pi /3}} & 1 & {{e}^{i2\pi /3}} \\ {{e}^{-i\pi /4}} & {{e}^{-i2\pi /3}} & 1 \\ \end{matrix} \right|=1+{{e}^{i\pi /3}}\times {{e}^{i2\pi /3}}\times {{e}^{-i\pi /4}}+\] \[{{e}^{-i\pi /3}}\times {{e}^{-i2\pi /3}}\times {{e}^{-i\pi /4}}-({{e}^{i2\pi /4}}\times {{e}^{-i\pi /4}}+{{e}^{-i\pi /3}}\times \] \[{{e}^{-i\pi /3}}+{{e}^{-i\pi /3}}\times {{e}^{-i2\pi /3}})\] \[=1+{{e}^{i\,3\pi /4}}+{{e}^{-i3\pi /4}}-(1+1+1)\] \[=-2+2\cos (3\pi /4)=-2-\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
