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Question
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]
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Solution
\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]
\[ = \begin{vmatrix}\left( 2^{2x} + 2^{- 2x} + 2 \right) & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ \left( 3^{2x} + 3^{- 2x} + 2 \right) & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ \left( 4^{2x} + 4^{- 2x} + 2 \right) & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix}\]
\[ = \begin{vmatrix}4 & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ 4 & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ 4 & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix} \left[ \text{ Applying }C_1 \to C_1 - C_2 \right]\]
\[ = 4\begin{vmatrix}1 & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ 1 & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ 1 & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix}\]
\[ = 0\]
