JEE Main 2025 — 28 January, Evening Shift — Question 10

$$P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ $$\therefore P^T P = I$$ $$B = PAP^T$$ $$\text{Pre\; multiply\; by } P^T\; \text{ (Given)}$$ $$P^T B = P^T P A P^T = A P^T$$ $$\text{Now\; post\; multiply\; by\; } P$$ $$P^T B P = A P^T P = A$$ $$\text{So\; } A^2 = P^T B P \ P^T B P$$ $$A^2 = P^T B^2 P$$ $$\text{Similarly\; } A^{10} = P^T B^{10} P = C$$ $$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \text{ (Given)}$$ $$\Rightarrow A^2 = \begin{bmatrix} \frac{1}{2} & -\sqrt{2} - 2 \\ 0 & 1 \end{bmatrix}$$ $$\text{Similarly\; check\; } A^3 \text{ and\; so\; on\; since } C = A^{10}$$ $$\Rightarrow \text{ Sum\; of\; diagonal\; elements\; of\; } C\; \text{ is } \left( \frac{1}{\sqrt{2}} \right)^{10} + 1$$

$=\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}}$

$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$ (Given)

$\Rightarrow \mathrm{m}+\mathrm{n}=65$