Date | May Example question | Marks available | 2 | Reference code | EXM.1.AHL.TZ0.4 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Consider the matrix A =(5−271)=(5−271).
B, C and X are also 2 × 2 matrices.
Write down the inverse, A–1.
Given that XA + B = C, express X in terms of A–1, B and C.
Given that B =(675−2)=(675−2), and C =(−50−87)=(−50−87), find X.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
det A = 5(1) − 7(−2) = 19
A–1 =119(12−75)=(119219−719519)=119(12−75)=(119219−719519) (A2)
Note: Award (A1) for (12−75)(12−75), (A1) for dividing by 19.
OR
A–1 =(0.05260.105−0.3680.263)=(0.05260.105−0.3680.263) (G2)
[2 marks]
XA + B = C ⇒ XA = C – Β (M1)
X = (C – Β)Α–1 (A1)
OR
X = (C – B)A–1 (A2)
[2 marks]
(C – Β)Α–1 = (−11−7−139)(119219−719519)(−11−7−139)(119219−719519) (A1)
⇒ X = (3819−5719−76191919)=(2−3−41)(3819−5719−76191919)=(2−3−41) (A1)
OR
X = (2−3−41)(2−3−41) (G2)
Note: If premultiplication by A–1 is used, award (M1)(M0) in part (i) but award (A2) for (−3719111912199419)(−3719111912199419) in part (ii).
[2 marks]