A-LevelFurther MathematicsMatrices9231/12 · May/June 2018 · Q1214 Marks

OR It is given that e is an eigenvector of the matrix A, with corresponding eigenvalue λ. Let A = ⎛ 3 0 0 ⎞ ⎜ 2 7 0 ⎟ ⎝ 4 8 1 ⎠

(i)Write down another eigenvector of A corresponding to λ.[1]
(ii)Write down an eigenvector and corresponding eigenvalue of Aⁿ, where n is a positive integer.[2]
(iii)Find a matrix P and a diagonal matrix D such that Aⁿ = PDP⁻¹.[7]
(iv)Determine the set of values of the real constant k such that ∑∞n=1 kⁿ(Aⁿ − kAⁿ⁺¹) = kA.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

B1 (i) Allow any scalar multiple μe where μ≠0,1
B1 (ii) Eigenvector of Aⁿ is e.
B1 (ii) Corresponding eigenvalue of Aⁿ is λⁿ.
B1 (iii) Eigenvalues are λ=3, 7, 1.
M1 (iii) Uses (A-λI)e=0 to find eigenvectors.
A1 (iii) Eigenvector for λ=3 is e₁ = (-2, 1, 0)ᵀ (or proportional to).
A1 (iii) Eigenvector for λ=7 is e₂ = (0, 3, 4)ᵀ (or proportional to).
A1 (iii) Eigenvector for λ=1 is e₃ = (0, 0, 1)ᵀ (or proportional to).
B1 FT (iii) Matrix P formed from eigenvectors. P = (-2 0 0 / 1 3 0 / 0 4 1)
B1 FT (iii) Diagonal matrix D formed from eigenvalues. D = (3 0 0 / 0 7 0 / 0 0 1)
M1 (iv) Method of differences. ∑ kⁿ(Aⁿ - kAⁿ⁺¹) = kA - k^(N+1)A^(N+1).
A1 (iv) Correct application of method of differences and identifying the terms.
M1 (iv) For k^(N+1)A^(N+1) → 0 as N → ∞, eigenvalues of kA must be < 1 in magnitude. Eigenvalues of kA are 3k, 7k, k.
A1 (iv) Range of k: -1/7 < k < 1/7.

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