If x, y, z ∈ R, the value of the determinant ∣∣∣∣

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Question:

If $x, y, z \in R$, the value of the determinant$\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$is equal to_____________

Solution:

Let $\Delta=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$\Delta=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

Applying $C_{1} \rightarrow C_{1}-C_{2}$

$=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$=\left|\begin{array}{lll}\left(2^{x}+2^{-x}+2^{x}-2^{-x}\right)\left(2^{x}+2^{-x}-2^{x}+2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}+3^{x}-3^{-x}\right)\left(3^{x}+3^{-x}-3^{x}+3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}+4^{x}-4^{-x}\right)\left(4^{x}+4^{-x}-4^{x}+4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$=\left|\begin{array}{lll}\left(2^{x}+2^{x}\right)\left(2^{-x}+2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{x}\right)\left(3^{-x}+3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{x}\right)\left(4^{-x}+4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$=\left|\begin{array}{lll}\left(2.2^{x}\right)\left(2.2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(2.3^{x}\right)\left(2.3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(2.4^{x}\right)\left(2.4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

Taking out $(4)$ common from $C_{1}$

$=4\left|\begin{array}{lll}\left(2^{x}\right)\left(2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}\right)\left(3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}\right)\left(4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$=4\left|\begin{array}{lll}1 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 1 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 1 & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$

$=4(0)$   ( $\because$ if two columns are identical then the value of determinant is zero)

$=0$

Hence, the value of the determinant $\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$ is equal to $\underline{0}$.

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