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

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Matrices

Let

A = [\frac{1}{2} 0 - 2 1]

and

P = [cos θ sin θ - sin θ cos θ], θ > 0

. If

B = PAP^{T}, C = P^{T} B^{10} P

and the sum of the diagonal elements of

C

\frac{m}{n}

, where

gcd (m, n) =

1 , then

m + n

is :

Answer: (A)

Since

P

is a rotation matrix,

P^{T} P = I

, so from

B = P A P^{T}

P^{T} B = A P^{T} ⟹ P^{T} B P = A

Squaring the relation

B = P A P^{T}

repeatedly gives

B^{10} = P A^{10} P^{T}

, so

C = P^{T} B^{10} P = A^{10}

Since

A = (\frac{1}{2} 0 - 2 1)

is upper triangular,

A^{10}

is upper triangular with diagonal entries

(\frac{1}{2})^{10}

and

1

. The trace of

C

(\frac{1}{2})^{10} + 1 = \frac{1}{32} + 1 = \frac{33}{32}

Since

g cd (33, 32) = 1

m = 33

n = 32

, so

m + n = 65

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option