Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}\right]$ and…

Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}\right]$ and $P=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], \theta \gt 0$. 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 $\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is :
  1. $127$
  2. $258$
  3. $65$
  4. $2049$

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 $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$
$\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \operatorname{g~cd}(\mathrm{m}, \mathrm{n})=1(\text { Given }) \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65\end{aligned}$ ,

Asked in: JEE Main 2025 (28 Jan Shift 2)

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