Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}\right]$ and $P=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], \theta\gt0$. If $\mathrm{B}=\mathrm{PAP}^{\mathrm{T}}, \mathrm{C}=\mathrm{P}^{\mathrm{T}} \mathrm{B}^{10} \mathrm{P}$ and the sum of the diagonal elements of $C$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is :

• A

  $127$
• B

  $258$
• Correct answer
  C

  $65$
• D

  $2049$

Solution

$\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}$ Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given) $\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{P} \mathrm{AP}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$ Now post multiply by P $\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}$
$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$ Similarly $A^{10}=P^T B^{10} P=C$ $\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text { (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}$ Similarly check $A^3$ and so on since $C=A^{10}$ $\Rightarrow$ Sum of diagonal elements of C is $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$ $\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \operatorname{g~cd}(\mathrm{m}, \mathrm{n})=1(\text { Given }) \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65\end{aligned}$

Hence, the correct answer is option: 'C'.