A-LevelFurther MathematicsMatrices9231/12 · Oct/Nov 2011 · Q811 Marks

The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue μ. Show that e is an eigenvector of the matrix AB with corresponding eigenvalue λμ. [2] State the eigenvalues of the matrix C, where C = ( -1 -1 3 ) ( 0 1 2 ) ( 0 0 2 ) and find corresponding eigenvectors. [4] Show that `(1 6 3)ᵀ` is an eigenvector of the matrix D, where D = ( 1 -1 1 ) ( -6 -3 4 ) ( -9 -3 7 ) and state the corresponding eigenvalue. [3] Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue. [2]

(a)The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue μ. Show that e is an eigenvector of the matrix AB with corresponding eigenvalue λμ.[2]
(b)State the eigenvalues of the matrix C, where C = ( -1 -1 3 ) ( 0 1 2 ) ( 0 0 2 ) and find corresponding eigenvectors.[4]
(c)Show that `(1 6 3)ᵀ` is an eigenvector of the matrix D, where D = ( 1 -1 1 ) ( -6 -3 4 ) ( -9 -3 7 ) and state the corresponding eigenvalue.[3]
(d)Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.

[M1A1] Shows required result, using Ae = λe. ABe = A (μe) = μ(Ae) = μ(λe) = λμe
[B1] States eigenvalues from leading diagonal. Eigenvalues of C are -1, 1 and 2
[M1A1] Finds eigenvectors using cross-product or equations. λ = -1: e₁ = (0 0 1)ᵀ (or proportional)
[A1] λ = 1: e₂ = (-2 4 0)ᵀ (or proportional)
[A1] λ = 2: e₃ = (1 6 3)ᵀ (or proportional)
[M1A1] Uses De = μe. D(1 6 3)ᵀ = (-2 -12 -6)ᵀ
[A1] States eigenvalue. Eigenvalue is –2.
[B1] Recognises that CD has an eigenvector common to C and D and (1 6 3)ᵀ
[B1√] states the corresponding eigenvalue. and the corresponding eigenvalue is -2 × 2 = -4 .

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