Question

Let $X$ be the solution set of the equation $A^x=I,$ where $A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$ and $I$ is the corresponding unit matrix and $x\in N,$ then the minimum value of $\sum ({\cos }^x\theta +{\sin }^x\theta ),\theta \in R$ is

Answer 1

$A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$
$\Rightarrow A^2=A.A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right].\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$

$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow A^2=I\Rightarrow A^4=A^6=A^8=...=I$
Now $A^x=I$
$\Rightarrow x=2,4,6,8,...$
$\therefore \sum ({\cos }^x\theta +{\sin }^x\theta )$
$=({\cos }^2\theta +{\sin }^2\theta )+({\cos }^4\theta +{\sin }^4\theta )+({\cos }^6\theta +{\sin }^6\theta )+....$
$=({\cos }^2\theta +{\cos }^4\theta +{\cos }^6\theta +...)+({\sin }^4\theta +{\sin }^4\theta +{\sin }^6\theta +...)$
$=\frac{{\cos }^2\theta }{1-{\cos }^2\theta }+\frac{{\sin }^2\theta }{1-{\sin }^2\theta }$
$={\cot }^2\theta +{\tan }^2\theta$
which has minimum value $2.$