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 $x\subseteq N$, then the minimum value of $\sum (co{s}^x\theta +si{n}^x\theta ),\theta \in R$.

Answer 1

Given $A^x=I$, where $A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$
Now, $A^2=\left[\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right]\left[\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow x=2$
Again $A^3=A^{2}A=A\ne I$
So, $A^3\ne I$
Hence, $x=3\notin X$
Again $A^4=A^2{A}^2=I.I=I$
So, $X=2,4,\mathrm{6....}$
Now, $\sum (co{s}^x\theta +si{n}^x\theta )$
$={\cos }^2\theta +{\cos }^4\theta +.....)+({\sin }^2\theta +{\sin }^4\theta +....)$
$={\cot }^2\theta +{\tan }^2\theta$
$={\sec }^2\theta +cose{c}^2\theta -2$
$=\frac{1}{{\sin }^2\theta {\cos }^2\theta }-2$
Minimum value of $(\sin \theta \cos \theta {)}^n=\frac{1}{2^n}$
So, minimum value of $\frac{1}{{\sin }^2\theta {\cos }^2\theta }-2$ =2