Question

The value of determinant \[\left| {\begin{array}{*{20}{c}}
  1&{{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\
  {{e^{ - i\dfrac{\pi }{3}}}}&1&{{e^{i\dfrac{{2\pi }}{3}}}} \\
  {{e^{ - i\dfrac{\pi }{4}}}}&{{e^{ - i\dfrac{{2\pi }}{3}}}}&1
\end{array}} \right|\] is ___________
A - $2 + \sqrt 2 $
B - $2 - \sqrt 2 $
C - $ - \left( {2 + \sqrt 2 } \right)$
D - $ - \left( {2 - \sqrt 2 } \right)$

Answer 1

Hint: Firstly expand the determinant as general , on solving the determinant we get ${e^{i\theta }}$ form type things in it for this use ${e^{i\theta }} = \cos \theta + i\sin \theta $ apply it in the equation and also remember that $\cos \left( { - \theta } \right) = \cos \theta $ and $\sin \left( { - \theta } \right) = - \sin \theta $ .

Complete step-by-step answer:
As we have to find the value of determinant \[\left| {\begin{array}{*{20}{c}}
  1&{{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\
  {{e^{ - i\dfrac{\pi }{3}}}}&1&{{e^{i\dfrac{{2\pi }}{3}}}} \\
  {{e^{ - i\dfrac{\pi }{4}}}}&{{e^{ - i\dfrac{{2\pi }}{3}}}}&1
\end{array}} \right|\] , So for this now we have to expand the given determinant , with respect to column $1$
So ,
$1\left| {\begin{array}{*{20}{c}}
  1&{{e^{i\dfrac{{2\pi }}{3}}}} \\
  {{e^{ - i\dfrac{{2\pi }}{3}}}}&1
\end{array}} \right|$ $ - $ ${e^{ - i\dfrac{\pi }{3}}}$\[\left| {\begin{array}{*{20}{c}}
  {{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\
  {{e^{ - i\dfrac{{2\pi }}{3}}}}&1
\end{array}} \right|\] $ + $ ${e^{ - i\dfrac{\pi }{4}}}$\[\left| {\begin{array}{*{20}{c}}
  {{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfrac{\pi }{4}}}} \\
  1&{{e^{i\dfrac{{2\pi }}{3}}}}
\end{array}} \right|\]
So on expanding the determinant ,
$ \Rightarrow 1\left( {1 - {e^{i\dfrac{{2\pi }}{3}}}.{e^{ - i\dfrac{{2\pi }}{3}}}} \right) - {e^{ - i\dfrac{\pi }{3}}}\left( {{e^{i\dfrac{\pi }{3}}} - {e^{i\dfrac{\pi }{4}}}.{e^{i\dfrac{{2\pi }}{3}}}} \right) + {e^{ - i\dfrac{\pi }{4}}}\left( {{e^{i\dfrac{\pi }{3}}}.{e^{i\dfrac{{2\pi }}{3}}} - {e^{i\dfrac{\pi }{4}}}} \right)$
On solving further ,
$ \Rightarrow 0 - {e^{ - i\dfrac{\pi }{3}}}\left( {{e^{i\dfrac{\pi }{3}}} - {e^{i\dfrac{{5\pi }}{{12}}}}} \right) + {e^{ - i\dfrac{\pi }{4}}}\left( {{e^{ - \pi }} - {e^{i\dfrac{\pi }{4}}}} \right)$

\[ \Rightarrow - \left( {1 - {e^{i\dfrac{{5\pi }}{{12}}}}.{e^{ - i\dfrac{\pi }{3}}}} \right) + \left( {{e^{ - \pi }}.{e^{ - i\dfrac{\pi }{4}}} - 1} \right)\]
$\Rightarrow - 2 + {e^{ - i\dfrac{\pi }{3} + i\dfrac{{5\pi }}{{12}}}} + {e^{ - i\dfrac{\pi }{4} - i\pi }}$
On solving the power we get the final solution ,
$ \Rightarrow - 2 + {e^{ - i\dfrac{{3\pi }}{4}}} + {e^{i\dfrac{{3\pi }}{4}}}$
 Now we know that from the value of ${e^{i\theta }} = \cos \theta + i\sin \theta $ , apply this on the above equation ,
$ \Rightarrow - 2 + \cos \dfrac{{ - 3\pi }}{4} + i\sin \dfrac{{ - 3\pi }}{4} + \cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}$
We know that from the trigonometry that is $\cos \left( { - \theta } \right) = \cos \theta $ and $\sin \left( { - \theta } \right) = - \sin \theta $
So by using this we get ,
$ \Rightarrow - 2 + 2\cos \dfrac{{3\pi }}{4}$
Hence we know the value of $\cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}$
$ - 2 - \dfrac{2}{{\sqrt 2 }}$ or we write as $ - 2 - \sqrt 2 $
So the option C is correct .

Note: The value of ${e^{i\theta }} = \cos \theta + i\sin \theta $ is known as the Euler's formula . If we have a complex number$z = r\left( {\cos \theta + i\sin \theta } \right)$ written in polar form, we can use Euler's formula to write it even more concisely in exponential form that is \[r.{e^{i\theta }}\]