JEE Main 2025 — Matrices Question with 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 ${\left(\frac{1}{\sqrt{2}}\right)}^{10}+1$ $\begin{array}{rl}& =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}}\\ & g\; cd(\mathrm{m},\mathrm{n})=1(\text{ Given })\\ & \Rightarrow \mathrm{m}+\mathrm{n}=65\end{array}$