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Date May Example question Marks available 4 Reference code EXM.2.AHL.TZ0.3
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 3 Adapted from N/A

Question

Let M=(a221)M=(a221), where aZ.

Find M2 in terms of a.

[4]
a.

If M2 is equal to (5445), find the value of a.

[2]
b.

Using this value of a, find M1 and hence solve the system of equations:

x+2y=3

2xy=3

[6]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

M2=(a221)(a221)

=(a2+42a22a25)       (A1)(A1)(A1)(A1)

[4 marks]

a.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

2a2=4

a=1      (A1)

Substituting: a2+4=(1)2+4=5      (A1)

Note: Candidates may solve a2+4=5 to give a=±1, and then show that only a=1 satisfies 2a2=4.

[2 marks]

b.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

M=(1221)

M1=13(1221)       (M1)

=13(1221) or (13232313)      (A1)

x+2y=3

2xy=3

(1221)(xy)=(33)       (M1)(M1)

(13232313)(1221)(xy)=(13232313)(33)      (A1)

(xy)=(11)       (A1)

ie   x=1

     y=1

Note: The solution must use matrices. Award no marks for solutions using other methods.

[6 marks]

c.

Examiners report

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

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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