Answer only one of the following two alternatives. OR The linear transformation T : R⁴ → R³ is represented by the matrix M, where M = [[2, 1, -1, 4], [3, 4, 6, 1], [-1, 2, 8, -7]] The range space of T is R. In any order,

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.

1. [M1A1] Reduces M to echelon form: [[2, 1, -1, 4], [3, 4, 6, 1], [-1, 2, 8, -7]] → [[2, 1, -1, 4], [0, 1, 3, -2], [0, 0, 0, 0]]
2. [A1] Uses dimension theorem: Dim(M) = 4 - 2 = 2 (number of non-zero rows)
3. [B1] States basis for R: Basis for R is { (2,3,-1)ᵀ, (1,4,2)ᵀ } (OE)
4. [M1A1] Finds cartesian equation for R: If r = x i + y j + z k, and r = λ(2,3,-1)ᵀ + μ(1,4,2)ᵀ, then x = 2λ + μ, y = 3λ + 4μ, z = -λ + 2μ. Eliminating λ and μ gives 2x - y + z = 0.
5. [M1] Finds basis for null space: For Mx=0, 2x+y-z+4t=0 and y+3z-2t=0. Let t=λ and z=μ. Then y=2λ-3μ and x=-3λ+2μ.
6. [A1A1] Basis of null space is { (-3,2,0,1)ᵀ, (2,-3,1,0)ᵀ } (OE)
7. [B1] Evaluates k: (8, 7, k)ᵀ belongs to R means it satisfies 2x - y + z = 0. So 2(8) - 7 + k = 0 ⇒ 16 - 7 + k = 0 ⇒ 9 + k = 0 ⇒ k = -9.
8. [M1A1] Finds a particular solution: Mx = (8, 7, -9)ᵀ. (e.g. from [[2, 1, -1, 4, | 8], [0, 1, 3, -2, | 7], [0, 0, 0, 0, | 0]] a particular solution is (-1, 5, 0, 0)ᵀ (OE))
9. [M1A1] Finds general solution: x = x_p + x_h = (-1, 5, 0, 0)ᵀ + λ(-3,2,0,1)ᵀ + μ(2,-3,1,0)ᵀ