Solveeit Logo
Login

Question

Question: The value of determinant \[\left| {\begin{array}{*{20}{c}} 1&{{e^{i\dfrac{\pi }{3}}}}&{{e^{i\dfr...

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+22 + \sqrt 2
B - 2−22 - \sqrt 2
C - −(2+2) - \left( {2 + \sqrt 2 } \right)
D - −(2−2) - \left( {2 - \sqrt 2 } \right)

Explanation

Solution

Firstly expand the determinant as general , on solving the determinant we get eiθ{e^{i\theta }} form type things in it for this use eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta apply it in the equation and also remember that cos⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta and sin⁡(−θ)=−sin⁡θ\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 11
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 ,
⇒1(1−ei2π3.e−i2π3)−e−iπ3(eiπ3−eiπ4.ei2π3)+e−iπ4(eiπ3.ei2π3−eiπ4)\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 ,
⇒0−e−iπ3(eiπ3−ei5π12)+e−iπ4(e−π−eiπ4)\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)

⇒−(1−ei5π12.e−iπ3)+(e−π.e−iπ4−1) \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)
⇒−2+e−iπ3+i5π12+e−iπ4−iπ\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 ,
⇒−2+e−i3π4+ei3π4\Rightarrow - 2 + {e^{ - i\dfrac{{3\pi }}{4}}} + {e^{i\dfrac{{3\pi }}{4}}}
Now we know that from the value of eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta , apply this on the above equation ,
⇒−2+cos⁡−3π4+isin⁡−3π4+cos⁡3π4+isin⁡3π4\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⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta and sin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta
So by using this we get ,
⇒−2+2cos⁡3π4\Rightarrow - 2 + 2\cos \dfrac{{3\pi }}{4}
Hence we know the value of cos⁡3π4=−12\cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}
−2−22- 2 - \dfrac{2}{{\sqrt 2 }} or we write as −2−2- 2 - \sqrt 2
So the option C is correct .

Note: The value of eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta is known as the Euler's formula . If we have a complex numberz=r(cos⁡θ+isin⁡θ)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.eiθr.{e^{i\theta }}