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Powers of P A P^T (or A^T B A) with P orthogonal: the middle power survives

This type of question, asked 4 times: 2021 Paper 18, 25 Jan 2023, 8 Apr 2023, 28 Jan 2025. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 18

    76
    If A=(1525−2515),B=(10i1),i=−1A =\begin{pmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix},B =\begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix},i=\sqrt{- 1}, and Q=ATBAQ=A^{T}BA, then the inverse of the matrix AQ2021 ATAQ^{2021}{\text{ }A}^{T} is equal to:

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  2. Question 2

    25 Jan 2023

    158
    Let A=[110310−310110]A =\begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} and B=[1−i01]B =\begin{bmatrix} 1 & - i \\ 0 & 1 \end{bmatrix}, where i=−1i =\sqrt{- 1}. If M=ATBAM =A^{T}BA, then the inverse of the matrix A2023ATA^{2023}A^{T} is

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  3. Question 3

    8 Apr 2023

    65
    Let P=[3212−1232],A=[1101]P =\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix},A =\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} And Q=PQPTQ =PQP^{T}. If  PTQ2007P=[abcd]\ P^{T}Q^{2007}P =\begin{bmatrix} a & b \\ c & d \end{bmatrix}, then 2a+b−3c−4d2a + b - 3c - 4d equal to

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  4. Question 4

    28 Jan 2025

    135
    Let A=[12−201]A =\begin{bmatrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{bmatrix} and P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ],θ>0P =\begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix},\theta > 0. If B=PAPT,C=PTB10PB = PAP^{T},C =P^{T}B^{10}P and the sum of the diagonal elements of C is mn\frac{m}{n}, where gcd(m,n)=gcd(m,n) = 1 , then m+nm + n is :

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