JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
Let $A=\begin{bmatrix}\dfrac{1}{\sqrt{2}}&-2\\0&1\end{bmatrix}$ and $P=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, $\theta>0$. If $B=PAP^{T}$, $C=P^{T}B^{10}P$ and the sum of the diagonal elements of $C$ is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is :
Matrices
Solution
$P^TP=I$, so $B^{10}=PA^{10}P^T$ and $C=A^{10}=\begin{bmatrix}2^{-5}&*\\0&1\end{bmatrix}$. Its trace is $\frac1{32}+1=\frac{33}{32}$, so $m+n=65$.