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Using Cayley-Hamilton for a 2x2 matrix: write A inverse or a power of A as alpha A + beta I

This type of question, asked 4 times: 2021 Paper 16, 27 Jul 2022, 8 Apr 2023, 4 Apr 2024. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 16

    56
    If A=[12−14]A−1=αI+βAA=\begin{bmatrix} 1 & 2 \\ - 1 & 4 \end{bmatrix}A^{- 1}=\alpha I+\beta A, find 5(α−β)=5(\alpha-\beta) = ?

  2. Question 2

    27 Jul 2022

    64
    Let A=(12−2−5)A =\begin{pmatrix} 1 & 2 \\ - 2 & - 5 \end{pmatrix}. Let α,β∈R\alpha,\beta \in R be such that αA2+βA=2I\alpha A^{2}+ \beta A = 2I. Then α+β\alpha + \beta is equal to -

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  3. Question 3

    27 Jul 2022

    154
    Let A=(4−2αβ)A =\begin{pmatrix} 4 & - 2 \\ \alpha & \beta \end{pmatrix} If A2+γA+18I=OA^{2}+ \gamma A + 18I = O, then det(a)det(a) is equal to

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  4. Question 4

    8 Apr 2023

    168
    If A=[15λ10],A−1=αA+βIA =\begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix},A^{- 1}= \alpha A + \beta I and α+β=−2\alpha + \beta = - 2, then 4α2+β2+λ24\alpha^{2}+\beta^{2}+\lambda^{2} is equal to:

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  5. Question 5

    4 Apr 2024

    175
    Let A be a 2×22 \times 2 symmetric matrix such that A[11]=[37]A\begin{bmatrix} 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 \\ 7 \end{bmatrix} and the determinant of AA be 1. If A−1=αA+βIA^{- 1}= \alpha A + \beta I, where II is an identity matrix of order 2×22 \times 2, then α+β\alpha + \beta equals....