The linear transformation T : R⁴ → R⁴ is represented by the matrix M, where M = (1 -2 -3 1 3 -5 -7 7 5 -9 -13 9 7 -13 -19 11) Find the rank of M and a basis for the null space of T. The vector e = (1 2 3 4)ᵀ is denoted by e. Show that there is a solution of the equation Mx = Me of the form X = (a b -1 -1)ᵀ where the constants a and b are to be found.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.

1. [M1] (a) Attempt row reduction of M
2. [A1] (a) Correct first step of row reduction (e.g., R2-3R1, R3-5R1, R4-7R1)
3. [A1] (a) Complete row reduction to echelon form
4. [M1] (a) State rank = 2 (number of non-zero rows in echelon form)
5. [M1] (a) Set up equations from echelon form and introduce parameters for free variables
6. [A1] (a) Correct basis vectors for the null space (e.g., (-9 -2 1 0)ᵀ and (-4 -4 0 1)ᵀ)
7. [M1] (b) Recognise that x = e + v where Mv = 0 (v is in null space)
8. [M1] (b) Set up equations by equating components of x and given form
9. [A1] (b) Solve for λ and μ (parameters for null space basis)
10. [A1] (b) Find correct values for a and b (a=50, b=30)