A-LevelFurther MathematicsMatrices9231/12 · Oct/Nov 2013 · Q69 Marks

The linear transformation T : R⁴ → R⁴ is represented by the matrix M, where M = [[1, -3, -1, 2], [4, -10, 0, 2], [1, -1, 3, -4], [5, -12, 1, 1]]. Find, in either order, the rank of M and a basis for the null space K of T. Evaluate M [[1], [-2], [-3], [-4]] and hence show that every solution of Mx = [[2], [16], [10], [22]] has the form x = [[1], [-2], [-3], [-4]] + λe₁ + µe₂, where λ and µ are real numbers and {e₁, e₂} is a basis for K.

(a)Find, in either order, the rank of M and a basis for the null space K of T.[6]
(b)Evaluate M [[1], [-2], [-3], [-4]] and hence show that every solution of Mx = [[2], [16], [10], [22]] has the form x = [[1], [-2], [-3], [-4]] + λe₁ + µe₂, where λ and µ are real numbers and {e₁, e₂} is a basis for K.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2013). The mark or mark type is shown in brackets.

[M1A1] (a) Reduces to echelon form. [[1, -3, -1, 2], [0, 1, 2, -3], [0, 0, 0, 0], [0, 0, 0, 0]]
[A1] Obtains rank r(M) = 4 − 2 = 2
[M1] x - 3y - z + 2t = 0; y + 2z - 3t = 0; ⇒ t = μ, z = λ, y = 3μ – 2λ, x = 7μ – 5λ
[A1] and basis for null space. Basis is: [[5], [-2], [1], [0]], [[7], [3], [0], [1]] (OE) e.g. or [[1], [0], [-3], [2]], [[0], [1], [7], [5]]
[B1] (b) Evaluates matrix product. M [[1], [-2], [-3], [-4]] = [[2], [16], [10], [22]]
[M1A1] Finds general solution of equations. x - [[1], [-2], [-3], [-4]] ∈ K
[A1] x = [[1], [-2], [-3], [-4]] + λe₁ + µe₂ (AG)

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