If P = beginbmatrix √(3)/2 & 1/2 \ -1/2 & √(3)/2… — JEE Main 2005 Mathematics PYQ

AIEEE 2005

P = \begin{bmatrix} \sqrt{3}/2 & 1/2 \\ -1/2 & \sqrt{3}/2 \end{bmatrix}

A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

and

Q = PAP^T

, then

P^TQ^{2005}P

A.
$\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}$
B.
$\begin{bmatrix} 1 & 2005 \\ 2005 & 1 \end{bmatrix}$
C.
$\begin{bmatrix} 1 & 0 \\ 2005 & 1 \end{bmatrix}$
D.
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Show answer and solution

Correct option: A

Solution

P is a rotation matrix, so PP^T = I, meaning P^T = P^(-1).

Since Q = PAP^T, we get Q²⁰⁰5 = (PAP^T)²⁰⁰5 = PA²⁰⁰5P^T (since intermediate P^T·P = I terms cancel).

Then P^T·Q²⁰⁰5·P = P^T·(PA²⁰⁰5P^T)·P = A²⁰⁰5.

For A = [[1,1],[0,1]], Aⁿ = [[1,n],[0,1]], so A²⁰⁰5 = [[1,2005],[0,1]].

[AIEEE 2005]

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