Question

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

Answer 1

Evaluating the given determinant by using the column operation of determinant.

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

Applying $C_1\to C_1-C_2$

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

$=\left|\begin{array}{ccc}{4.2}^x{.2}^{-x}& (2^x-2^{-x}{)}^2& 1\\ {4.3}^x{.3}^{-x}& (3^x-3^{-x}{)}^2& 1\\ {4.4}^x{.4}^{-x}& (4^x-4^{-x}{)}^2& 1\end{array}\right|$

$[\because (a+b{)}^2-(a-b{)}^2=4ab]$

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

Taking 4 common from $C_1$

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

Here, $C_1$ and $C_3$ are identical columns. If any two rows or columns are identical then value of determinant is zero.

$\Rightarrow \Delta =4\times 0=0$

Hence, the value of determinant is 0.