A-LevelFurther MathematicsMatrices9231/12 · Oct/Nov 2016 · Q58 Marks
The linear transformation T : R⁴ → R⁴ is represented by the matrix A, where A = (1 3 5 7) (2 8 7 9) (3 13 9 11) (6 24 21 27) Find (i) the rank of A, (ii) a basis for the range space of T, (iii) a basis for the null space of T.
- (i)the rank of A,[3]
- (ii)a basis for the range space of T,[1]
- (iii)a basis for the null space of T.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.
- [M1A1] Row reduction/Echelon form of A yields: (1 3 5 7) (0 2 -3 -5) (0 0 0 0) (0 0 0 0)
- [A1] r(A) = 4 - 2 = 2
- [B1] Basis for range space is ((1), (2), (3), (6)), ((3), (8), (13), (24)) oe
- [B1] From echelon form, obtain equations: x+3y+5z+7t = 0, 2y-3z-5t = 0
- [M1A1A1] Solving the system (e.g., let z=2, t=0 and z=0, t=2) yields a basis for the null space: ((-29), (5), (2), (0)) , ((-19), (3), (0), (2)) oe
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