# If $x,y, z\in R,$ then the value of determinant $\small\begin{vmatrix}(2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1\end{vmatrix}$ is

$\begin{array}{1 1} 1 \\ -1 \\ 0 \\ \pm 1 \end{array}$

Toolbox:
• Elementary transformation of a matrix can be done by the addition (or subtraction)of the elements of two columns(or rows).
• If each element of a row (or column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
$\Delta=\begin{vmatrix}(2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1\end{vmatrix}$
Apply $C_1\rightarrow C_1-C_2$
$\Delta=\begin{vmatrix}(2^x + 2^{-x})^2 -(2^x - 2^{-x})^2&(2^x - 2^{-x})^2 & 1\\ (3^x + 3^{-x})^2 - (3^x - 3^{-x})^2&(3^x - 3^{-x})^2 & 1\\(4^x + 4^{-x})^2 - (4^x - 4^{-x})^2&4^x - 4^{-x})^2 & 1\end{vmatrix}$
But we know $(a+b)^2-(a-b)^2=4ab$
Similarly $(2^x+2^{-x})^2-(2^x-2^{-x})^2=4.2^x.2^{-x}=4$
$\quad(3^x+3^{-x})^2-(3^x-3^{-x})^2=4.3^x.3^{-x}=4$
$\quad\quad(4^x+4^{-x})^2-(4^x-4^{-x})^2=4.4^x.4^{-x}=4$
$\Delta=\begin{vmatrix}4 & (2^x-2^{-x})^2& 1\\4 & (3^x-3^{-x})^2& 1\\4 & (4^x-4^{-x})^2& 1\end{vmatrix}$
Take 4 as the common factor from $C_1$,
$\Delta=\begin{vmatrix}1 & (2^x-2^{-x})^2& 1\\1 & (3^x-3^{-x})^2& 1\\1 & (4^x-4^{-x})^2& 1\end{vmatrix}$=0 (since two rows are identical)
Hence the value of the determinant=0.
answered Mar 26, 2013