A-LevelFurther MathematicsMatrices9231/12 · May/June 2012 · Q710 Marks

The linear transformations T₁ : R⁴ → R⁴ and T₂ : R⁴ → R⁴ are represented by the matrices M₁ = [[1, 1, 1, 4], [2, 1, 4, 11], [3, 4, 1, 9], [4, -3, 18, 37]] and M₂ = [[1, 1, 1, -1], [2, 3, 0, 1], [3, 4, 1, 0], [4, 5, 2, 0]] respectively. The null space of T₁ is denoted by K₁ and the null space of T₂ is denoted by K₂. Show that the dimension of K₁ is 2 and that the dimension of K₂ is 1. Find the basis of K₁ which has the form [[p], [q], [r], [s]] such that [[r], [s]] = [[1], [0]] and show that K₂ is a subspace of K₁.

Show that the dimension of K₁ is 2 and that the dimension of K₂ is 1.[5]
Find the basis of K₁ which has the form [[p], [q], [r], [s]] such that [[r], [s]] = [[1], [0]] and show that K₂ is a subspace of K₁.[5]

Mark scheme

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

[M1A1] Reduces M₁ to echelon form. [[1,1,1,4],[0,-1,2,3],[0,0,0,0],[0,0,0,0]]
[A1] Finds Dim(K₁). Dim(K₁) = 4 – 2 = 2 (AG)
[A1] Reduces M₂ to echelon form. [[1,1,1,-1],[0,1,2,3],[0,0,0,1],[0,0,0,0]] (aef)
[A1] Finds Dim(K₂). Dim(K₂) = 4 - 3 = 1 (AG)
[M1] Obtains basis for K₁. x + y + z + 4t = 0, - y + 2z + 3t = 0
[A1A1] Basis for K₁ is [[-3],[2],[1],[0]], [[-7],[3],[0],[1]] (OE)
[M1] Obtains basis for K₂, x + y + z - t = 0, y - 2z + 3t = 0, t = 0
[A1] Basis for K₂ is [[-3],[2],[1],[0]] (OE) ⇒ K₂ ⊂ K₁

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