A-LevelFurther MathematicsMatrices9231/12 · May/June 2013 · Q810 Marks
The linear transformations T₁ : R⁴ → R⁴ and T₂ : R⁴ → R⁴ are represented by the matrices M₁ and M₂ respectively, where M₁ = [[1, -2, 3, 5], [3, -4, 17, 33], [5, -9, 20, 36], [4, -7, 16, 29]] and M₂ = [[1, -2, 0, -3], [2, -1, 0, 0], [4, -7, 1, -9], [6, -10, 0, -14]]. The null spaces of T₁ and T₂ are denoted by K₁ and K₂ respectively. Find a basis for K₁ and a basis for K₂. It is given that a = [[1], [2], [3], [4]]. The vectors x₁ and x₂ are such that M₁x₁ = M₁a and M₂x₂ = M₂a. Given that x₁ - x₂ = [[p], [5], [7], [q]], find p and q.
- (a)The null spaces of T₁ and T₂ are denoted by K₁ and K₂ respectively. Find a basis for K₁ and a basis for K₂.[6]
- (b)It is given that a = [[1], [2], [3], [4]]. The vectors x₁ and x₂ are such that M₁x₁ = M₁a and M₂x₂ = M₂a. Given that x₁ - x₂ = [[p], [5], [7], [q]], find p and q.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.
- M1A1 Reduces matrices to echelon form. (M1 for attempt, A1 for M1 echelon form)
- A1 (A1 for M2 echelon form)
- M1A1 Finds basis for each null space. (M1 for attempt, A1 for K1 basis)
- A1 (A1 for K2 basis)
- B1√ Writes x₁ and x₂ appropriately. x₁ = a + λv₁ and x₂ = a + μv₂
- M1 Finds difference x₁ - x₂ = [[μ-λ], [2μ-λ], [μ-2λ], [λ-μ]]. ⇒ λ-2μ=5 and -λ+μ=7
- A1 and solves. ⇒ λ = 3 and μ = -1
- A1 x₁-x₂ = (4 5 7 -4)ᵀ ⇒ p=4 and q=-4
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 →