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Powers of a matrix that satisfies A^2 = kA (rank one or idempotent)

This type of question, asked 4 times: 29 Jun 2022, 31 Jan 2023, 29 Jan 2025, 2 Apr 2025. All 4 questions below, oldest first.

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  1. Question 1

    29 Jun 2022

    70
    Let A=[aij]A =\left\lbrack a_{ij} \right\rbrack be a square matrix of order 3 such that aij=2j−ia_{ij}=2^{j - i}, for all i,j=1,2,3i,j = 1,2,3. Then, the matrix A2+A3+…+A10A^{2}+A^{3}+ \ldots +A^{10} is equal to:

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  2. Question 2

    31 Jan 2023

    77
    Let A=(10004−1012−3)A =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{pmatrix}. Then the sum of the diagonal elements of the matrix (A+I)11(A + I)^{11} is equal to

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  3. Question 3

    29 Jan 2025

    137
    Let A=[aij]A =\left\lbrack a_{ij} \right\rbrack be a matrix of order 3×33 \times 3, with aij=(2)i+ja_{ij}= (\sqrt{2})^{i + j}. If the sum of all the elements in the third row of A2A^{2} is α+β2,α,β∈Z\alpha + \beta\sqrt{2},\alpha,\beta \in Z, then α+β\alpha + \beta is equal to

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  4. Question 4

    2 Apr 2025

    55
    Let A=[α−16β],α>0A=\begin{bmatrix} \alpha & - 1 \\ 6 & \beta \end{bmatrix},\alpha> 0, such that det(A)=0det(A) = 0 and α+β=1\alpha+\beta= 1. If I denotes 2×22 \times 2 identity matrix, then the matrix (1+A)8(1 +A)^{8} is:

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