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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 = ( a 2 2 1 ) , where  a Z .

Find  M 2 in terms of a .

[4]
a.

If  M 2  is equal to ( 5 4 4 5 ) , find the value of a .

[2]
b.

Using this value of a , find M 1 and hence solve the system of equations:

x + 2 y = 3

2 x y = 3

[6]
c.

Markscheme

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

M 2 = ( a 2 2 1 ) ( a 2 2 1 )

= ( a 2 + 4 2 a 2 2 a 2 5 )        (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.

2 a 2 = 4

a = 1       (A1)

Substituting:  a 2 + 4 = ( 1 ) 2 + 4 = 5       (A1)

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

[2 marks]

b.

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

M = ( 1 2 2 1 )

M 1 = 1 3 ( 1 2 2 1 )        (M1)

= 1 3 ( 1 2 2 1 ) or ( 1 3 2 3 2 3 1 3 )       (A1)

x + 2 y = 3

2 x y = 3

( 1 2 2 1 ) ( x y ) = ( 3 3 )        (M1)(M1)

( 1 3 2 3 2 3 1 3 ) ( 1 2 2 1 ) ( x y ) = ( 1 3 2 3 2 3 1 3 ) ( 3 3 )       (A1)

( x y ) = ( 1 1 )        (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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