If $\alpha,\beta,\gamma \in R$, then the determinant $\mathrm{\Delta} = \left| \begin{matrix} \left( e^{i\alpha} + e^{- i\alpha} ight)^{2} & \left( e^{i\alpha} - e^{- i\alpha} ight)^{2} & 4 \ \left( e^{i\beta} + e^{- i\beta} ight)^{2} & \left( e^{i\beta} - e^{- i\beta} ight)^{2} & 4 \ \left( e^{i\gamma} + e^{- i\gamma} ight)^{2} & \left( e^{i\gamma} - e^{- i\gamma} ight)^{2} & 4 \end{matrix} ight|$ is

Question

If $\alpha,\beta,\gamma \in R$, then the determinant $\mathrm{\Delta} = \left| \begin{matrix} \left( e^{i\alpha} + e^{- i\alpha} ight)^{2} & \left( e^{i\alpha} - e^{- i\alpha} ight)^{2} & 4 \ \left( e^{i\beta} + e^{- i\beta} ight)^{2} & \left( e^{i\beta} - e^{- i\beta} ight)^{2} & 4 \ \left( e^{i\gamma} + e^{- i\gamma} ight)^{2} & \left( e^{i\gamma} - e^{- i\gamma} ight)^{2} & 4 \end{matrix} ight|$ is

Dr. Abubacker · Accepted Answer

(a)
Given, $\mathrm{\Delta} = \left| \begin{matrix}
\left( e^{i\alpha} + e^{- i\alpha} ight)^{2} & \left( e^{i\alpha} - e^{- i\alpha} ight)^{2} & 4 \
\left( e^{i\beta} + e^{- i\beta} ight)^{2} & \left( e^{i\beta} - e^{- i\beta} ight)^{2} & 4 \
\left( e^{i\gamma} + e^{- i\gamma} ight)^{2} & \left( e^{i\gamma} - e^{- i\gamma} ight)^{2} & 4
\end{matrix} ight|$
Applying $C_{1} ightarrow C_{1} - C_{2}$
$= \left| \begin{matrix}
4 & \left( e^{i\alpha} - e^{- i\alpha} ight)^{2} & 4 \
4 & \left( e^{i\beta} - e^{- i\beta} ight)^{2} & 4 \
4 & \left( e^{i\gamma} - e^{- i\gamma} ight)^{2} & 4
\end{matrix} ight|$
$= 0$ $(\because$ two columns are same)
Hence, it is independent of$\ \alpha,\ \beta\ and\ \gamma$.

Explanation

Reviewed by Dr. Abubacker

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