A-LevelFurther MathematicsMatrices9231/12 · May/June 2014 · Q910 Marks

The matrix M, where M = [ [-2, 2, 2], [2, 1, 2], [-3, -6, -7] ], has an eigenvector (0, 1, -1)^T. Find the corresponding eigenvalue. It is given that if the eigenvalues of a general 3 × 3 matrix A, where A = [ [a, b, c], [d, e, f], [g, h, i] ], are λ₁, λ₂ and λ₃ then λ₁ + λ₂ + λ₃ = a + e + i and the determinant of A has the value λ₁λ₂λ₃. Use these results to find the other two eigenvalues of the matrix M, and find corresponding eigenvectors.

The matrix M, where M = [ [-2, 2, 2], [2, 1, 2], [-3, -6, -7] ], has an eigenvector (0, 1, -1)^T. Find the corresponding eigenvalue.[2]
It is given that if the eigenvalues of a general 3 × 3 matrix A, where A = [ [a, b, c], [d, e, f], [g, h, i] ], are λ₁, λ₂ and λ₃ then λ₁ + λ₂ + λ₃ = a + e + i and the determinant of A has the value λ₁λ₂λ₃. Use these results to find the other two eigenvalues of the matrix M, and find corresponding eigenvectors.[8]

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

[M1A1] M [0, 1, -1]^T = [0, -1, 1]^T = -1 * [0, 1, -1]^T ⇒ eigenvalue = -1 (= λ₁)
[M1] λ₁ + λ₂ + λ₃ = -8 ⇒ λ₂ + λ₃ = -7
[M1A1] det M = -2((-7)-(-12)) - 2((-14)-(-6)) + 2((-12)-(-3)) = -10 + 16 - 18 = -12 ⇒ λ₁λ₂λ₃ = -12
[A1A1] λ₂ = -3 and λ₃ = -4 or vice versa. (S.C. Award M1A1 for λ³ + 8λ² + 19λ + 12 = 0 and A1 for both λ₂ = -3 and λ₃ = -4.)
[M1A1] For λ = -3, (M - (-3)I)v = 0: results in [1, 2, 2]^T, [2, 4, 2]^T, [-3, -6, -4]^T. Eigenvector is (-2, 1, 0)^T (OE)
[A1] For λ = -4, (M - (-4)I)v = 0: results in [2, 2, 2]^T, [2, 5, 2]^T, [-3, -6, -3]^T. Eigenvector is (1, 0, -1)^T (OE)

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