Date | May 2019 | Marks available | 6 | Reference code | 19M.2.AHL.TZ0.F_3 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Question number | F_3 | Adapted from | N/A |
Question
The matrix M is given by M =[122311231].
Given that M3 can be written as a quadratic expression in M in the form aM2 + bM + cI , determine the values of the constants a, b and c.
Show that M4 = 19M2 + 40M + 30I.
Using mathematical induction, prove that Mn can be written as a quadratic expression in M for all positive integers n ≥ 3.
Find a quadratic expression in M for the inverse matrix M–1.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
M2 =[11106810813108]; M3 =[535038545034596044] (A1)(A1)
let [535038545034596044]=a[11106810813108]+b[122311231]+c[100010001] (M1)
then, for example
11a + b + c = 53
10a + 2b = 50 M1A1
6a + 2b = 38
the solution is a = 3, b =10, c =10 (M1)A1
(M3 = 3M2 +10M +10I)
METHOD 2
det(M − λI) = 0 (M1)
⇒(1−λ)((1−λ)2−3)−2(3(1−λ)−2)+2(9−2(1−λ))=0 M1A1
−λ3+3λ2+10λ+10=0 M1A1
applying the Cayley – Hamilton theorem (M1)
M3 = 3M2 +10M +10I and so a = 3, b =10, c =10 A1
[7 marks]
M4 = 3M3 + 10M2 + 10M M1
= 3(3M2 + 10M + 10I) + 10M2 + 10M M1
=19M2 + 40M +30I AG
[2 marks]
the statement is true for n = 3 as shown in part (a) A1
assume true for n = k, ie Mk = pM2 + qM + rI M1
Note: Subsequent marks after this M1 are independent and can be awarded.
Mk+1 = pM3 + qM2 + rM M1
= p(3M2 +10M +10I) + qM2 + rM M1
= (3p +q)M2 + (10p + r)M + 10pI A1
hence true for n = k ⇒ true for n = k +1 and since true for n = 3, the statement is proved by induction R1
Note: Award R1 provided at least four of the previous marks are gained.
[6 marks]
M2 = 3M + 10I + 10M–1 M1
M–1 = 110(M2 − 3M − 10I) A1
[2 marks]