JEE Main 2025 — Mathematics Algebra

$\begin{aligned}
& \mathrm{P}=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \
\sin \theta & \cos \theta
\end{array}\right] \
& \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \
& \mathrm{B}=\mathrm{PAPT}
\end{aligned}$Pre multiply by$Premultiplyby\mathrm{P}^{\mathrm{T}}$( Given)$(Given)\mathrm{P}^{\mathrm{T}} \mathrm{B}=\mathrm{P}^{\mathrm{T}} \mathrm{P} \mathrm{AP}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$Now post multiply by P$NowpostmultiplybyP\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}$

$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$A2=PT B2P Similarly $A^{10}=P^T B^{10} P=C$A10=PTB10P=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$SimilarlycheckA^3$and so on since$andsoonsinceC=A^{10}$$\Rightarrow$Sum of diagonal elements of C is$SumofdiagonalelementsofCis\left(\frac{1}{\sqrt{2}}\right)^{10}+1$$\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}$