A-LevelFurther MathematicsIntegration9231/12 · May/June 2016 · Q1114 Marks
ANSWER ONLY ONE OF THE FOLLOWING TWO ALTERNATIVES. OR The linear transformation T : R⁴ → R⁴ is represented by the matrix M, where M = [[1,-2,3,-4],[2,-4,7,-9],[4,-8,14,-18],[5,-10,17,-22]] Find the rank of M. Obtain a basis for the null space K of T. Evaluate M [[1],[-2],[2],[-1]] and hence show that any solution of Mx = [[15],[33],[66],[81]] has the form [[1],[-2],[2],[-1]] + λe₁ + μe₂, where λ and μ are scalars and {e₁, e₂} is a basis for K. Hence obtain a solution x' of (*) such that the sum of the components of x' is 6 and the sum of the squares of the components of x' is 26.
- Find the rank of M.[3]
- Obtain a basis for the null space K of T.[3]
- Evaluate M [[1],[-2],[2],[-1]] and hence show that any solution of Mx = [[15],[33],[66],[81]] has the form [[1],[-2],[2],[-1]] + λe₁ + μe₂, where λ and μ are scalars and {e₁, e₂} is a basis for K.[2]
- Hence obtain a solution x' of (*) such that the sum of the components of x' is 6 and the sum of the squares of the components of x' is 26.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.
- [M1A1] [[1,-2,3,-4],[0,0,1,-1],[0,0,0,0],[0,0,0,0]]
- [A1] r(M) = 2
- [M1] x-2y+3z-4t = 0
- [M1] z-t = 0; t = z = λ and y = μ⇒ x = 2μ + λ
- [A1] Basis for K is [[2],[1],[0],[0]] and [[1],[0],[1],[1]]
- [B1] M [[1],[-2],[2],[-1]] = [[15],[33],[66],[81]]
- [B1^] x = [[1],[-2],[2],[-1]] + λe₁ + μe₂
- [B1] Sum of components = 6 → 3λ + 3μ = 6 (⇒ μ = 2 - λ)
- [M1A1] Sum of squares of components = 26 → 5λ² + 3μ² + 4λμ + 4μ + 10 = 26
- [M1] ⇒ 4λ² −8λ+4=0 ⇒(λ−1)² = 0
- [A1] ⇒ λ = 1, μ = 1
- [A1] x' = [[4],[-1],[3],[0]]
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →