Let $\mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]\text{ and }\mathrm{P}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right],\theta >0.$ 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 $\frac{\mathrm{m}}{\mathrm{n}}$ gcd(m, n) = 1, then m + n is :

$\begin{array}{l}\mathrm{P}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\\ \because {\mathrm{P}}^{\mathrm{T}}\mathrm{P}=\mathrm{I}\\ \mathrm{B}=\mathrm{PAPT}\end{array}$

Pre multiply by P^T ( Given)
P^TB = P^TP AP^T = AP^T
Now post multiply by P
P^TBP = AP^TP = A

So ${\mathrm{A}}^2=\underset{\mathrm{I}}{\underbrace{{\mathrm{P}}^{\mathrm{T}}{\mathrm{BPP}}^{\mathrm{T}}}}\mathrm{BP}$

A² = P^TB²P
Similarly A¹⁰ = P^TB¹⁰ P = C

$\begin{array}{r}A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]\text{ (Given) }\\ \Rightarrow A^2=\left[\begin{array}{cc}\frac{1}{2}& -\sqrt{2}-2\\ 0& 1\end{array}\right]\end{array}$

Similarly check A3 and so on since C = A¹⁰
$\Rightarrow$ Sum of diagonal elements of C is ${\left(\frac{1}{\sqrt{2}}\right)}^{10}+1$

$\begin{array}{l}=\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}}\\ \gcd (\mathrm{m},\mathrm{n})=1\text{ (Given) }\\ \Rightarrow \mathrm{m}+\mathrm{n}=65\end{array}$