Date | May Example question | Marks available | 6 | Reference code | EXM.2.AHL.TZ0.3 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Hence and Find | Question number | 3 | Adapted from | N/A |
Question
Let M=(a22−1), where a∈Z.
Find M2 in terms of a.
If M2 is equal to (5−4−45), find the value of a.
Using this value of a, find M−1 and hence solve the system of equations:
−x+2y=−3
2x−y=3
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M2=(a22−1)(a22−1)
=(a2+42a−22a−25) (A1)(A1)(A1)(A1)
[4 marks]
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
2a−2=−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 2a−2=−4.
[2 marks]
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M=(−122−1)
M−1=−13(−1−2−2−1) (M1)
=13(1221) or (13232313) (A1)
−x+2y=−3
2x−y=3
⇒(−122−1)(xy)=(−33) (M1)(M1)
⇒(13232313)(−122−1)(xy)=(13232313)(−33) (A1)
⇒(xy)=(1−1) (A1)
ie x=1
y=−1
Note: The solution must use matrices. Award no marks for solutions using other methods.
[6 marks]