A-LevelFurther MathematicsMatrices9231/12 · Oct/Nov 2011 · Q47 Marks
The linear transformation T : R⁴ → R⁴ is represented by the matrix M, where M = ( 3 4 2 5 ) ( 6 7 5 8 ) ( 9 9 9 9 ) ( 15 16 14 17 )
- (i)the rank of M and a basis for the range space of T,[4]
- (ii)a basis for the null space of T.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.
- [M1A1] Reduces matrix to echelon form. (3 4 2 5; 0 -1 1 -2; 0 0 0 0; 0 0 0 0)
- [A1] States rank, ⇒ R(M) = 2.
- [A1] Basis for range space is: (3 4; 6 7; 9 9; 15 16) (OE)
- [M1] Forms equations. 3x + 4y + 2z + 5t = 0 -y + z - 2t = 0
- [A1A1] States basis of null space. Basis of null space is (1 -2 0 1)ᵀ, (2 1 1 0)ᵀ or (-3 0 2 1)ᵀ, (0 -3 0 2)ᵀ (Or by reducing transpose to echelon form, or by any other valid method.).
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