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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 : MM 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.

[7]
a.

It is now given that A is singular.

By considering appropriate determinants, prove that f is not a bijection.

[4]
b.

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 CM 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]

a.

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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » AHL 1.14—Introduction to matrices
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Topic 1—Number and algebra

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