Date | May 2019 | Marks available | 4 | Reference code | 19M.1.AHL.TZ0.F_13 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Prove | Question number | F_13 | Adapted from | N/A |
Question
The function f : M → M where M is the set of 2 × 2 matrices, is given by f(X) = AX where A is a 2 × 2 matrix.
Given that A is non-singular, prove that f is a bijection.
It is now given that A is singular.
By considering appropriate determinants, prove that f is not a bijection.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
suppose f(X) = f(Y) , ie AX = AY (M1)
then A−1AX = A−1AY A1
X = Y A1
since f(X) = f(Y) ⇒ X = Y, f is an injection R1
now suppose C ∈ M and consider f(D) = C , ie AD = C M1
then D = A−1 C (A−1 exists since A is non- singular) A1
since given C ∈ M, there exists D ∈ M such that f(D) = C , f is a surjection R1
therefore f is a bijection AG
[7 marks]
suppose f(X) = Y, ie AX = Y (M1)
then det(A)det(X) = det(Y) A1
since det(A) = 0, it follows that det(Y) = 0 A1
it follows that f is not surjective since the function cannot reach non-singular matrices R1
therefore f is not a bijection AG
[4 marks]