Question

$A=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]$ and $f\left(x\right)$ is defined as $f\left(x\right)=det.\left(A^T{A}^{-1}\right)$ then find the value of $\underset{\text{n times}}{\underset{}{f\left(f\right(f(f.......f(x\left)\right)\left)\right)}}$ is $(n\ge 2)$.

Answer 1

Given $A=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]$
$|A|=1+{\tan }^{2}x$
$A^T=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$
Here,$c_{11}=1,c_{12}=\tan x{c}_{21}=-\tan x,c_{22}=1$
$adjA={\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]}^T=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$
$A^{-1}=\frac{1}{1+{\tan }^{2}x}\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$
Now, $A^T{A}^{-1}=\frac{1}{1+{\tan }^{2}x}\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$
$A^T{A}^{-1}=\frac{1}{1+{\tan }^{2}x}\left[\begin{array}{cc}1-{\tan }^{2}x& -2\tan x\\ 2\tan x& 1-{\tan }^{2}x\end{array}\right]$
$\Rightarrow A^T{A}^{-1}=\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right]$
$|A^T{A}^{-1}|={\sin }^{2x}+{\cos }^{2x}=1$
Given ,$f\left(x\right)=det.\left(A^T{A}^{-1}\right)$
$\Rightarrow f\left(x\right)=1$
$\Rightarrow \underset{\text{n times}}{\underset{}{f\left(f\right(f(f.......f(x\left)\right)\left)\right)}}=1$