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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  = [ 1 2 2 3 1 1 2 3 1 ] .

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.

[7]
a.

Show that M4 = 19M2 + 40M + 30I.

[2]
b.

Using mathematical induction, prove that Mn can be written as a quadratic expression in M for all positive integers ≥ 3.

[6]
c.

Find a quadratic expression in M for the inverse matrix M–1.

[2]
d.

Markscheme

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

METHOD 1

M2  = [ 11 10 6 8 10 8 13 10 8 ] M3  = [ 53 50 38 54 50 34 59 60 44 ]      (A1)(A1)

let  [ 53 50 38 54 50 34 59 60 44 ] = a [ 11 10 6 8 10 8 13 10 8 ] + b [ 1 2 2 3 1 1 2 3 1 ] + c [ 1 0 0 0 1 0 0 0 1 ]        (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 λ I) = 0       (M1)

( 1 λ ) ( ( 1 λ ) 2 3 ) 2 ( 3 ( 1 λ ) 2 ) + 2 ( 9 2 ( 1 λ ) ) = 0         M1A1

λ 3 + 3 λ 2 + 10 λ + 10 = 0         M1A1

applying the Cayley – Hamilton theorem       (M1)

M3 = 3M2 +10M +10I and so a = 3, b =10, c =10        A1

 

[7 marks]

a.

M4 = 3M3 + 10M2 + 10M     M1

= 3(3M2 + 10M + 10I) + 10M2 + 10M      M1

=19M2 + 40M +30I       AG 

[2 marks]

b.

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]

c.

M2 = 3M + 10I + 10M–1     M1

M–1 = 1 10 (M2 − 3M − 10I)     A1

[2 marks]

d.

Examiners report

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

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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