Answer only one of the following two alternatives. EITHER The roots of the quartic equation x⁴ + 4x³ + 2x² – 4x + 1 = 0 are α, β, γ and δ. Find the values of (i) α + β + γ + δ, (ii) α² + β² + γ² + δ², (iii) 1/α + 1/β + 1/γ + 1/δ, (iv) α/(βγδ) + β/(αγδ) + γ/(αβδ) + δ/(αβγ). Using the substitution y = x + 1, find a quartic equation in y. Solve this quartic equation and hence find the roots of the equation x⁴ + 4x³ + 2x² − 4x + 1 = 0. OR The square matrix A has λ as an eigenvalue with e as a corresponding eigenvector. Show that if A is non-singular then (i) λ ≠ 0, (ii) the matrix A⁻¹ has λ⁻¹ as an eigenvalue with e as a corresponding eigenvector. The 3 × 3 matrices A and B are given by A = ((-2, 2, -4), (0, -1, 5), (0, 0, 3)) and B = (A + 3I)⁻¹, where I is the 3 × 3 identity matrix. Find a non-singular matrix P, and a diagonal matrix D, such that B = PDP⁻¹.

Mark scheme

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

1. [B1] EITHER (i) From the coefficients, for x⁴ + 4x³ + 2x² – 4x + 1 = 0, Σα = -4.
2. [M1A1] EITHER (ii) Σα² = (Σα)² - 2Σαβ = (-4)² - 2(2) = 16 - 4 = 12.
3. [M1A1] EITHER (iii) Σ(1/α) = (Σαβγ) / (αβγδ) = 4/1 = 4.
4. [M1A1] EITHER (iv) Σ(α/βγδ) = (α² + β² + γ² + δ²) / (αβγδ) = (Σα²) / (αβγδ) = 12/1 = 12.
5. [M1A1] EITHER (v) Let y = x+1, so x = y-1. Substitute into the equation: (y-1)⁴ + 4(y-1)³ + 2(y-1)² - 4(y-1) + 1 = 0. Expanding and collecting terms yields y⁴ - 4y² + 4 = 0.
6. [A1] y⁴ - 4y² + 4 = 0.
7. [A1] This equation can be written as (y² - 2)² = 0.
8. [M1A1] Solving for y: y² - 2 = 0 ⇒ y² = 2 ⇒ y = ±√2 (twice).
9. [M1A1] Since x = y-1, the roots of the original equation are x = ±√2 - 1 (twice).
10. [M1A1] OR (i) Given Ae = λe. If A is non-singular, then A⁻¹ exists. If λ = 0, then Ae = 0. Since A is non-singular, its determinant is non-zero, meaning its null space contains only the zero vector. Thus, e must be the zero vector, which contradicts e being an eigenvector. Therefore, λ ≠ 0 (since e ≠ 0).
11. [M1A1] OR (ii) Given Ae = λe. Multiply by A⁻¹ from the left: A⁻¹Ae = A⁻¹λe. This gives e = λA⁻¹e. Since λ ≠ 0, we can divide by λ: (1/λ)e = A⁻¹e. Therefore, A⁻¹ has λ⁻¹ as an eigenvalue with e as its corresponding eigenvector.
12. [B2,1,0] OR (iii) The eigenvalues of A are the diagonal entries of the upper triangular matrix: λ_A = -2, -1, 3. (1 mark for any one, 2 marks for all three).
13. [M1A1 A1A1] For λ = -2: A - (-2)I = ((0, 2, -4), (0, 1, 5), (0, 0, 5)). Solving (A+2I)e₁ = 0 gives e₁ = (1, 0, 0)ᵀ. For λ = -1: A - (-1)I = (( -1, 2, -4), (0, 0, 5), (0, 0, 4)). Solving (A+I)e₂ = 0 gives e₂ = (2, 1, 0)ᵀ. For λ = 3: A - 3I = ((-5, 2, -4), (0, -4, 5), (0, 0, 0)). Solving (A-3I)e₃ = 0 gives e₃ = (-6, 25, 20)ᵀ. So, the eigenvectors are (1, 0, 0)ᵀ, (2, 1, 0)ᵀ, (-6, 25, 20)ᵀ.
14. [B1] The eigenvalues of B = (A + 3I)⁻¹. The eigenvalues of A+3I are λ_A+3I = λ_A + 3. So, (-2+3), (-1+3), (3+3) = 1, 2, 6.
15. [B1] The eigenvalues of B = (A+3I)⁻¹ are 1/λ_(A+3I) = 1/1, 1/2, 1/6.
16. [B1] The non-singular matrix P consists of the eigenvectors of A (which are also the eigenvectors of A+3I and (A+3I)⁻¹): P = ((1, 2, -6), (0, 1, 25), (0, 0, 20)).
17. [B1] The diagonal matrix D consists of the eigenvalues of B: D = ((1, 0, 0), (0, 1/2, 0), (0, 0, 1/6)).