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Without expanding, show that the value of the following determinant is zero: ∣ ∣ ∣ ∣ ∣ ∣ ( 2 x + 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 - Mathematics

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Question

Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]

Solution

\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]
\[ = \begin{vmatrix}\left( 2^{2x} + 2^{- 2x} + 2 \right) & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ \left( 3^{2x} + 3^{- 2x} + 2 \right) & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ \left( 4^{2x} + 4^{- 2x} + 2 \right) & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix}\]
\[ = \begin{vmatrix}4 & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ 4 & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ 4 & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix} \left[ \text{ Applying }C_1 \to C_1 - C_2 \right]\]
\[ = 4\begin{vmatrix}1 & \left( 2^{2x} + 2^{- 2x} - 2 \right) & 1 \\ 1 & \left( 3^{2x} + 3^{- 2x} - 2 \right) & 1 \\ 1 & \left( 4^{2x} + 4^{- 2x} - 2 \right) & 1\end{vmatrix}\]
\[ = 0\]

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Chapter 6: Determinants - Exercise 6.2 [Page 57]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 2.12 | Page 57

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