The value of determinant $\left| \begin{matrix} \left( a^{x} + a^{- x} ight)^{2} & \left( a^{x} - a^{- x} ight)^{2} & 1 \ \left( b^{x} + b^{- x} ight)^{2} & \left( b^{x} - b^{- x} ight)^{2} & 1 \ \left( c^{x} + c^{- x} ight)^{2} & \left( c^{x} - c^{- x} ight)^{2} & 1 \end{matrix} ight|$ is

Question

The value of determinant $\left| \begin{matrix} \left( a^{x} + a^{- x} ight)^{2} & \left( a^{x} - a^{- x} ight)^{2} & 1 \ \left( b^{x} + b^{- x} ight)^{2} & \left( b^{x} - b^{- x} ight)^{2} & 1 \ \left( c^{x} + c^{- x} ight)^{2} & \left( c^{x} - c^{- x} ight)^{2} & 1 \end{matrix} ight|$ is

Dr. Abubacker · Accepted Answer

(a)
$\left| \begin{matrix}
\left( a^{x} + a^{- x} ight)^{2} & \left( a^{x} - a^{- x} ight)^{2} & 1 \
\left( b^{x} + b^{- x} ight)^{2} & \left( b^{x} - b^{- x} ight)^{2} & 1 \
\left( c^{x} + c^{- x} ight)^{2} & \left( c^{x} - c^{- x} ight)^{2} & 1
\end{matrix} ight|$
Applying $C_{1} ightarrow C_{1} - C_{2}$
$= \left| \begin{matrix}
4 & \left( a^{x} - a^{- x} ight)^{2} & 1 \
4 & \left( b^{x} - b^{- x} ight)^{2} & 1 \
4 & \left( c^{x} - c^{- x} ight)^{2} & 1
\end{matrix} ight| = 4\left| \begin{matrix}
1 & \left( a^{x} - a^{- x} ight)^{2} & 1 \
1 & \left( b^{x} - b^{- x} ight)^{2} & 1 \
1 & \left( c^{x} - c^{- x} ight)^{2} & 1
\end{matrix} ight| = 0$ $\ (\because$two columns are identical)

Explanation

Reviewed by Dr. Abubacker

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