Determinants Matrices Question 180
Question: If $ {e^{i\theta }}=\cos \theta +i\sin \theta , $ then the value of $ \begin{vmatrix} 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{vmatrix} $ is
Options:
A) $ -2+\sqrt{2} $
B) $ 2-\sqrt{2} $
C) $ -2-\sqrt{2} $
D) 1
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Answer:
Correct Answer: C
Solution:
- [c] Expanding by Sarrus rule, $ \begin{vmatrix} 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{vmatrix}=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^{i3\pi /4}}+{e^{-i3\pi /4}}-(1+1+1) $
$ =-2+2\cos (3\pi /4)=-2-\sqrt{2} $
BETA
