Let $A=\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \ 0 & 1\end{matrix}\right]&P=\left[\begin{matrix}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{matrix}\right],\theta 0$. If $B=PA{P}^{T},C={P}^{T}{B}^{10}P&$ the sum of the diagonal element of ' C ' is $\frac{m}{n}$ where $gcd(m,n)=1$, then $(m+n)$ is

Question

Let $A=\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \ 0 & 1\end{matrix}\right]&P=\left[\begin{matrix}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{matrix}\right],\theta >0$. If $B=PA{P}^{T},C={P}^{T}{B}^{10}P&$ the sum of the diagonal element of ' C ' is $\frac{m}{n}$ where $gcd(m,n)=1$, then $(m+n)$ is

YESPYQ · Accepted Answer

Correct answer: a) 65. \(P=\left[\begin{matrix}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{matrix}\right]\ P{P}^{T}={P}^{T}P=I\ {B}^{2}=PA{P}^{T}PA{P}^{T}\ {B}^{2}=P{A}^{2}{P}^{T}\ \mathrm{Similarly},{B}^{10}=P{A}^{10}{P}^{T}\ C={P}^{T}{B}^{10}P\ ={P}^{T}P{A}^{10}{P}^{T}P={A}^{10}\ {A}^{2}=\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \ 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \ 0 & 1\end{matrix}\right]=\left[\begin{matrix}{\left(\frac{1}{\sqrt{2}}\right)}^{2} & - \ 0 & 1\end{matrix}\right]\ {A}^{3}=\left[\begin{matrix}{\left(\frac{1}{\sqrt{2}}\right)}^{2} & - \ 0 & 1\end{matrix}\right]\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \ 0 & 1\end{matrix}\right]=\left[\begin{matrix}{\left(\frac{1}{\sqrt{2}}\right)}^{3} & - \ 0 & 1\end{matrix}\right]\ .\ .\ .\ .\ \mathrm{Similarly}\mathrm{for}{A}^{10}=\left[\begin{matrix}{\left(\frac{1}{\sqrt{2}}\right)}^{10} & - \ 0 & 1\end{matrix}\right]\ \mathrm{Sum}\mathrm{of}\mathrm{diagonal}\mathrm{elements}\mathrm{of}C=\frac{1}{32}+1=\frac{33}{32}=\frac{m}{n}\ \mathrm{and}m+n=33+32=65.\)

Let \(A=\left[\begin{matrix}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{matrix}\right]&P=\left[\begin{matrix}\cos \theta & -\si…

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