Question

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

Answer 1

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



$$\begin{gathered} ∆=\left|\begin{array}{ccc}{\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| \\ \mathrm{Applying}{C}_1\to C_1-C_2 \\ =\left|\begin{array}{ccc}{\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}{ccc}\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}{ccc}\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}{ccc}\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| \\ \mathrm{Taking}\mathrm{out}\left(4\right)\mathrm{common}\mathrm{from}{C}_1 \\ =4\left|\begin{array}{ccc}\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}{ccc}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\left(0\right)\left(\because \mathrm{if}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =0 \end{gathered}$$

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