Worked solutions
3 Matrices
These are worked solutions to the colour-coded problem-solving questions from the exercises in
the Student’s Book. This excludes the drill questions.
Exercise 3A
23 The matrix for overall results is given by
5
𝐇 + 𝐀 = (4
6
2 3
4 1
2 4) + (4 0
3 1
5 2
9 3 8
5
6) = ( 8 2 10)
11 5 4
3
24 The matrix for next week’s projected takings is given by
864.00 604.80
800 560
1.08 (640 320) = (691.20 345.60)
734.40 324.00
680 300
2𝑥 − 𝑦
25 (
7
3
12
)=(
𝑦 + 3𝑥
7
3
)
13
So
{
2𝑥 − 𝑦 = 12
𝑦 + 3𝑥 = 13
Solving simultaneously: 𝑥 = 5, 𝑦 = −
26
1 2𝑝 0
1
(
)=(
𝑝 −2 3𝑞
𝑝
5−𝑞
−2
0
)
𝑝 − 13
So
{
2𝑝 = 5 − 𝑞
3𝑞 = 𝑝 − 13
{
2𝑝 + 𝑞 = 5
𝑝 − 3𝑞 = 13
Solving simultaneously: 𝑝 = 4, 𝑞 = −3
27 a
−3 0
−1 5
𝐀 + 3𝐁 = (
) +3(
)
5 4
𝑘 −𝑘
−3 0
−3 15
=(
)+(
)
5 4
3𝑘 −3𝑘
−6
15
=(
)
5 + 3𝑘 4 − 3𝑘
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
1
Worked solutions
b
−3 0
1 0
−1 5
)−(
) + 4(
)
5 4
0 1
𝑘 −𝑘
−6 0
4 0
−1 5
=(
)−(
)+(
)
10 8
0 4
𝑘 −𝑘
−1
−5
=(
)
10 − 𝑘 12 + 𝑘
2𝐀 − 𝐁 + 4𝐈 = 2 (
28 a
1 𝑘
−𝑘
𝐀−𝐁 = (
)−(
2 3
1
1 + 𝑘 −𝑘
=(
)
1
0
2𝑘
)
3
b
1 𝑘 −𝑘 2𝑘
𝐀𝐁 = (
)(
)
2 3
1
3
−𝑘 + 𝑘 2𝑘 + 3𝑘
=(
)
−2𝑘 + 3 4𝑘 + 9
0
5𝑘
=(
)
3 − 2𝑘 4𝑘 + 9
29 a
1
𝐏 + 𝐐 = (2
1
7
= (1
3
6 1
𝑘 −3
4 𝑘 ) + (−1 𝑘
2 0
0 5
𝑘+1
1
𝑘 + 4 𝑘 + 3)
0
𝑘+5
4
3)
𝑘
b
6 1 4
1 𝑘 −3
𝐏𝐐 = (2 4 𝑘 ) (−1 𝑘 3)
2 0 𝑘
1 0 5
6−𝑘−6
1 + 𝑘 2 + 0 4 + 3𝑘 − 3𝑘
= (12 − 4 + 2𝑘 2 + 4𝑘 + 0 8 + 12 + 𝑘 2 )
6 + 0 + 10
1+0+0
4 + 0 + 5𝑘
−𝑘
𝑘2 + 1
4
= (2𝑘 + 8 4𝑘 + 2 𝑘 2 + 20)
16
1
5𝑘 + 4
30 a
50 38
𝐂=(
55 30
12300
=(
)
12500
100
24 80
) ( 50 )
26 75 125
30
b Cost from France = £12 300
Cost from Germany = £12 500
Mathematics for the IB Diploma: Applications & Interpretation
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2
𝑥 𝑦
3 1
7
(
)(
)=(
2 −1 1 2
5
3𝑥 + 𝑦 𝑥 + 2𝑦
7
(
)=(
5
5
0
Worked solutions
31
9
)
0
9
)
0
So
{
3𝑥 + 𝑦 = 7
𝑥 + 2𝑦 = 9
Solving simultaneously: 𝑥 = 1, 𝑦 = 4
2 1 1
32 (
)(
−1 4 3
(
𝑝
5
)=(
𝑞
11
−4
)
11
5 2𝑝 + 𝑞
5 −4
)=(
)
11 4𝑞 − 𝑝
11 11
So
{
2𝑝 + 𝑞 = −4
4𝑞 − 𝑝 = 11
Solving simultaneously: 𝑝 = −3, 𝑞 = 2
33 a
55
210 180 320 400 80
𝐌=(
)(
250 150 200 450 48
20
49310 47400
=(
)
44350 43150
50
85
)
40
22
b Engineer Right should use factory 𝐻 as total cost is £47 400 as opposed to £49 310 at
factory 𝐺
Forge Well should also use factory 𝐻 as total cost is £43 150 as opposed to £44 350 at
factory 𝐺
0.28 0.34
17557 18264
26124 28987
34 a (0.42 0.31) (
) = (20311 21427)
30125 29846
0.30 0.35
18381 19142
b Projected total for Yellow Part = 18 381 + 19 142 = 37 523
35
𝐀 + 𝑠𝐁 = 𝑡𝐈
𝑎 3
1 0
2 𝑏
(
)+𝑠(
) = 𝑡(
)
−2 1
0 1
1 3
𝑎 + 2𝑠 3 + 𝑏𝑠
𝑡 0
(
)=(
)
−2 + 𝑠 1 + 3𝑠
0 𝑡
So
𝑎 + 2𝑠 = 𝑡
3 + 𝑏𝑠 = 0
{
−2 + 𝑠 = 0
1 + 3𝑠 = 𝑡
(1)
(2)
(3)
(4)
Mathematics for the IB Diploma: Applications & Interpretation
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Worked solutions
From (3): 𝑠 = 2
From (4): 1 + 6 = 𝑡 ⇒ 𝑡 = 7
From (1): 𝑎 + 4 = 7 ⇒ 𝑎 = 3
From (2): 3 + 2𝑏 = 0 ⇒ 𝑏 = −1.5
36
1 2
(
2 4
1
(
2
1
𝑪 = 𝑝𝑨 + 𝑞𝑩
2
2
1 𝑎
𝑏
3
) = 𝑝(
)+𝑞(
)
0
2 2
−4 −5
𝑝 + 𝑏𝑞 𝑎𝑝 + 3𝑞
1
)=(
)
2𝑝
− 4𝑞 2𝑝 − 5𝑞
0
So
1 = 𝑝 + 𝑏𝑞
1 = 𝑎𝑝 + 3𝑞
{
2 = 2𝑝 − 4𝑞
0 = 2𝑝 − 5𝑞
(1)
(2)
(3)
(4)
Solving (3) and (4) simultaneously: 𝑝 = 5, 𝑞 = 2
From (1): 1 = 5 + 2𝑏 ⇒ 𝑏 = −2
From (2): 1 = 5𝑎 + 6 ⇒ 𝑎 = −1
37
𝑐𝐌 + 𝑑𝐍 = 𝐈
1 2𝑎
1
4 𝑏+1
𝑐(
)+𝑑(
)=(
−𝑎 3
0
3𝑏
1
𝑐 + 4𝑑
2𝑎𝑐 + 𝑏𝑑 + 𝑑
1
(
)=(
−𝑎𝑐 + 3𝑏𝑑
3𝑐 + 𝑑
0
0
)
1
0
)
1
So
𝑐 + 4𝑑 = 1
2𝑎𝑐 + 𝑏𝑑 + 𝑑 = 0
{
−𝑎𝑐 + 3𝑏𝑑 = 0
3𝑐 + 𝑑 = 1
(1)
(2)
(3)
(4)
Solving (1) and (4) simultaneously: 𝑐 =
3
,𝑑
11
=
2
11
Substituting into (2) and (3):
6
2
2
𝑎+ 𝑏=−
11
11
{11
3
6
− 𝑎+ 𝑏=0
11
11
2
1
Solving simultaneously: 𝑎 = − 7 , 𝑏 = − 7
38 Since the two matrices commute,
1 3 𝑘 6
1 3
𝑘 6
(
)(
)=(
)(
)
2 4 4 −1
4 −1 2 4
𝑘 + 12 3
𝑘 + 12 3𝑘 + 24
(
)=(
)
2𝑘 + 16 8
2
8
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
4
Worked solutions
3 = 3𝑘 + 24 ⇒ 𝑘 = −7
Check in 2𝑘 + 16 = 2: −14 + 16 = 2
So consistent. Therefore 𝑘 = −7.
Note: It is possible that there is no value for 𝑘 for which these matrices commute so it is
important to check that the value of 𝑘 found in the first equation is consistent in any other
equations.
39 Since the two matrices commute,
2 5
𝑐 𝑑
𝑐 𝑑 2 5
(
)(
)=(
)(
)
5 −2 𝑑 −𝑐
𝑑 −𝑐 5 −2
2𝑐 + 5𝑑 2𝑑 − 5𝑐
2𝑐 + 5𝑑 5𝑐 − 2𝑑
(
)=(
)
5𝑐 − 2𝑑 5𝑑 − 2𝑐
2𝑑 − 5𝑐 5𝑑 − 2𝑐
2𝑑 − 5𝑐 = 5𝑐 − 2𝑑
4𝑑 = 10𝑐
𝑑 = 2.5𝑐
40 a
𝑎1
𝐀𝐁 = (𝑎
𝑎2 𝑏1
𝑎4 ) (𝑏3
𝑎 𝑏 + 𝑎2 𝑏3
=( 1 1
𝑎3 𝑏1 + 𝑎4 𝑏3
3
𝑏2
)
𝑏4
𝑎1 𝑏2 + 𝑎2 𝑏4
)
𝑎3 𝑏2 + 𝑎4 𝑏4
𝑏1 𝑏2 𝑐1 𝑐2
)(
)
𝑏3 𝑏4 𝑐3 𝑐4
𝑏 𝑐 + 𝑏2 𝑐3 𝑏1 𝑐2 + 𝑏2 𝑐4
=( 1 1
)
𝑏3 𝑐1 + 𝑏4 𝑐3 𝑏3 𝑐2 + 𝑏4 𝑐4
𝐁𝐂 = (
b
𝑎 𝑏 + 𝑎2 𝑏3 𝑎1 𝑏2 + 𝑎2 𝑏4 𝑐1 𝑐2
(𝐀𝐁)𝐂 = ( 1 1
)(
)
𝑎3 𝑏1 + 𝑎4 𝑏3 𝑎3 𝑏2 + 𝑎4 𝑏4 𝑐3 𝑐4
𝑎 𝑏 𝑐 + 𝑎2 𝑏3 𝑐1 + 𝑎1 𝑏2 𝑐3 + 𝑎2 𝑏4 𝑐3 𝑎1 𝑏1 𝑐2 + 𝑎2 𝑏3 𝑐2 + 𝑎1 𝑏2 𝑐4 + 𝑎2 𝑏4 𝑐4
=( 1 1 1
)
𝑎3 𝑏1 𝑐1 + 𝑎4 𝑏3 𝑐1 + 𝑎3 𝑏2 𝑐3 + 𝑎4 𝑏4 𝑐3 𝑎3 𝑏1 𝑐2 + 𝑎4 𝑏3 𝑐2 + 𝑎3 𝑏2 𝑐4 + 𝑎4 𝑏4 𝑐4
𝑎1
𝐀(𝐁𝐂) = (𝑎
𝑎2 𝑏1 𝑐1 + 𝑏2 𝑐3 𝑏1 𝑐2 + 𝑏2 𝑐4
)(
)
𝑏3 𝑐1 + 𝑏4 𝑐3 𝑏3 𝑐2 + 𝑏4 𝑐4
3 𝑎4
𝑎 𝑏 𝑐 + 𝑎1 𝑏2 𝑐3 + 𝑎2 𝑏3 𝑐1 + 𝑎2 𝑏4 𝑐3 𝑎1 𝑏1 𝑐2 + 𝑎1 𝑏2 𝑐4 + 𝑎2 𝑏3 𝑐2 + 𝑎2 𝑏4 𝑐4
=( 1 1 1
)
𝑎3 𝑏1 𝑐1 + 𝑎3 𝑏2 𝑐3 + 𝑎4 𝑏3 𝑐1 + 𝑎4 𝑏4 𝑐3 𝑎3 𝑏1 𝑐2 + 𝑎3 𝑏2 𝑐4 + 𝑎4 𝑏3 𝑐2 + 𝑎4 𝑏4 𝑐4
So (𝐀𝐁)𝐂 = 𝐀(𝐁𝐂)
Exercise 3B
26 a
det 𝐀 = 0
2𝑘 2 − 4 = 0
𝑘2 = 2
𝑘 = ±√2
1
𝑘 −1
(
)
det 𝐀 −4 2𝑘
1
𝑘 −1
= 2
(
)
2𝑘 − 4 −4 2𝑘
𝐀−1 =
Mathematics for the IB Diploma: Applications & Interpretation
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5
Worked solutions
27 a
det 𝐀 = 2— 3𝑐 2
= 2 + 3𝑐 2
> 0 for all real 𝑐
Since det 𝐀 ≠ 0, 𝐀 has an inverse for all real 𝑐
b
1 2 −3𝑐
(
)
1
det 𝐀 𝑐
1
2 −3𝑐
= 2
(
)
1
3𝑐 + 2 𝑐
𝐀−1 =
28 a
1
−2 −4
(
)
8
−16 + 12 3
0.5
1
=(
)
−0.75 −2
𝐀−1 =
b
𝐀𝐁 = 𝐂
𝐁 = 𝐀−1 𝐂
28 20 40
0.5
1
=(
)(
)
−0.75 −2 −13 −7 −18
1 3 2
=(
)
5 −1 6
29 a From the GDC
𝐁−1 =
1 −7 19 16
( 19 32 −10)
117
25 −1 −7
b
𝐀𝐁 = 𝐂
𝐀 = 𝐂𝐁−1
−7 19 16
−9 19 −31 1
=(
)
( 19 32 −10)
25 −10 38 117
25 −1 −7
−3 4 −1
=(
)
5 1 2
30 Let the unknown 3 × 2 matrix be 𝐌.
Then
1 0 −2
1
(3 −1 2 ) 𝐌 = (23
4 1
1
35
1
𝐌 = (23
35
7
= (4
3
−16
20 )
21
−16 1 0 −2 −1
20 ) (3 −1 2 )
21
4 1
1
2
4)
9
Mathematics for the IB Diploma: Applications & Interpretation
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6
Worked solutions
31
4 0
𝐏 −1 𝐐𝐏 = (
)
0 −2
4 0
𝐐𝐏 = 𝐏 (
)
0 −2
4 0
𝐐 = 𝐏(
) 𝐏 −1
0 −2
1 −3 4 0 1 5 3
=(
)(
) (
)
1 5
0 −2 8 −1 1
1.75 2.25
=(
)
3.75 0.25
32
𝐀𝐁𝐂 = 𝐈
𝐁𝐂 = 𝐀−1
𝐂 = 𝐁−1 𝐀−1
6
2 −2 −1 5
1 −4 −1
= (−3 −1 2 ) ( 3
0
2)
−5 −1 2
−1 −1 6
−4.25 5.5 −4.75
= ( 3.75 −4.5 4.25 )
−9.5 12.5 −10.5
33
𝑘𝑎 𝑘𝑏
det(𝑘𝐌) = det (
)
𝑘𝑐 𝑘𝑑
= 𝑘 2 𝑎𝑑 − 𝑘 2 𝑏𝑐
= 𝑘 2 (𝑎𝑑 − 𝑏𝑐)
= 𝑘 2 det 𝐌
1
34 a 𝐌 3 = (0
0
1 2 3
1 7
2 −1) = (0 8
0 4
0 0
1
7𝐌 − 14𝐌 + 8𝐈 = 7 (0
0
7 21
= (0 28
0 0
1 7
= (0 8
0 0
2
38
−28)
64
1 2 2
1 1 2
1 0 0
2 −1) − 14 (0 2 −1) + 8 (0 1 0)
0 4
0 0 4
0 0 1
63
14 14 28
8 0 0
−42) − ( 0 28 −14) + (0 8 0)
112
0
0
56
0 0 8
38
−28)
64
So 𝐌 3 = 7𝐌 2 − 14𝐌 + 8𝐈
b
𝐌 3 = 7𝐌 2 − 14𝐌 + 8𝐈
𝐌 2 = 7𝐌 − 14𝐈 + 8𝐌 −1
8𝐌 −1 = 𝐌 2 − 7𝐌 + 14𝐈
1
𝐌 −1 = (𝐌 2 − 7𝐌 + 14𝐈)
8
(multiplying through by 𝐌 −1 )
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7
𝑒
𝑔
𝑓
)
ℎ
Worked solutions
35 Let 𝐌 −1 = (
Then, since 𝐌𝐌 −1 = 𝐈,
1
𝑎 𝑏 𝑒 𝑓
(
)(
)=(
0
𝑐 𝑑 𝑔 ℎ
𝑎𝑒 + 𝑏𝑔 𝑎𝑓 + 𝑏ℎ
1
(
)=(
𝑐𝑒 + 𝑑𝑔 𝑐𝑓 + 𝑑ℎ
0
0
)
1
0
)
1
So
𝑎𝑒 + 𝑏𝑔 = 1
𝑎𝑓 + 𝑏ℎ = 0
{
𝑐𝑒 + 𝑑𝑔 = 0
𝑐𝑓 + 𝑑ℎ = 1
(1)
(2)
(3)
(4)
From (3): 𝑔 = −
𝑐𝑒
𝑑
(5)
Substituting (5) into (1):
𝑐𝑒
𝑎𝑒 + 𝑏 (− ) = 1
𝑑
𝑒(𝑎𝑑 − 𝑏𝑐) = 𝑑
𝑑
𝑒=
𝑎𝑑 − 𝑏𝑐
Substituting into (5):
−𝑐
𝑔=
𝑎𝑑 − 𝑏𝑐
𝑎𝑓
𝑏
From (2): ℎ = −
(6)
Substituting (6) into (4):
𝑎𝑓
)=1
𝑏
𝑓(𝑏𝑐 − 𝑎𝑑) = 𝑏
𝑐𝑓 + 𝑑 (−
−𝑏
𝑎𝑑 − 𝑏𝑐
𝑓=
Substituting into (6):
𝑎
ℎ=
𝑎𝑑 − 𝑏𝑐
So
𝐌 −1
𝑑
− 𝑏𝑐
= (𝑎𝑑−𝑐
𝑎𝑑 − 𝑏𝑐
−𝑏
1
𝑑
𝑎𝑑 − 𝑏𝑐) =
(
𝑎
−𝑐
𝑎𝑑 − 𝑏𝑐
𝑎𝑑 − 𝑏𝑐
−𝑏
)
𝑎
36 a
(𝐀𝐁)(𝐀𝐁)−1
𝐀−1 𝐀𝐁(𝐀𝐁)−1
𝐁(𝐀𝐁)−1
−1
𝐁 𝐁(𝐀𝐁)−1
(𝐀𝐁)−1
=𝐈
= 𝐀−1 𝐈
= 𝐀−1
= 𝐁 −1 𝐀−1
= 𝐁 −1 𝐀−1
(multiplying on the left by 𝐀−1 )
(since 𝐀−1 𝐀 = 𝐈)
(multipying on the left by 𝐁−1 )
(since 𝐁−1 𝐁 = 𝐈)
Mathematics for the IB Diploma: Applications & Interpretation
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8
Worked solutions
b
(𝐀−1 𝐁)−1 𝐀−1 = 𝐁 −1 (𝐀−1 )−1 𝐀−1 (by part a)
= 𝐁 −1 𝐀𝐀−1
(since (𝐀−1 )−1 = 𝐀)
−1
=𝐁
Exercise 3C
7
5 −4 𝑥
7 a (
)( ) = ( )
−1
3 −2 𝑦
𝑥
7
−4
) , 𝐱 = (𝑦) and 𝐛 = ( )
−1
−2
5
where 𝐀 = (
3
b
1
−2 4
(
)
−10 − (−12) −3 5
1 −2 4
= (
)
2 −3 5
𝐀−1 =
c
𝐱 = 𝐀−1 𝐛
1 −2 4
𝑥
7
(𝑦) = (
)( )
2 −3 5 −1
−9
=(
)
−13
i.e. 𝑥 = −9, 𝑦 = −13
8 a
𝑥
1 −1 1
1
(5 3
2 ) (𝑦) = (1)
0 4 −3 𝑧
4
𝑥
1 −1 1
1
where 𝐀 = (5 3
2 ) , 𝐱 = (𝑦) and 𝐛 = (1)
𝑧
0 4 −3
4
b From the GDC,
𝐀−1 = −
1 −17 1 −5
( 15 −3 3 )
12
20 −4 8
c
𝐱 = 𝐀−1 𝐛
1 −17 1 −5 1
( 15 −3 3 ) (1)
12
4
20 −4 8
3
= (−2)
−4
=−
i.e. 𝑥 = 3, 𝑦 = −2, 𝑧 = −4
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
9
Worked solutions
9 Let the quantity of raspberries be 𝑟 kg and the quantity of strawberries be 𝑠 kg.
Then
8.5 5.5 𝑟
26.5
(
)( ) = (
)
𝑠
8
6
27
So
𝑟
8.5 5.5 −1 26.5
( )=(
) (
)
𝑠
8
6
27
1 6 −5.5 26.5
= (
)(
)
27
7 −8 8.5
1.5
=( )
2.5
i.e. the restaurant owner buys 1.5 kg or raspberries and 2.5 kg of strawberries.
10 a Let the amount of money invested in Allshare be $𝐴, the amount invested in Baserate be
$𝐵 and the amount invested in Chartwise be $𝐶.
Total invested $50 000, so 𝐴 + 𝐵 + 𝐶 = 50 000
$10 000 more invested in 𝐴 than 𝐶, so 𝐴 − 𝐶 = 10 000
1 year returns of 5.2%, 3.1% and 6.5% giving a total of $52 447.50, so
1.052𝐴 + 1.031𝐵 + 1.065𝐶 = 52447.5
Hence
1
1
1
𝐴
50 000
( 1
0
−1 ) (𝐵) = ( 10 000 )
1.052 1.031 1.065 𝐶
52 447.5
b
−1
𝐴
1
1
1
50 000
(𝐵) = ( 1
0
−1 ) ( 10 000 )
𝐶
1.052 1.031 1.065
52 447.5
22 500
= (15 000)
12 500
i.e. amounts invested were, 𝐴 = $22 500, 𝐵 = $15 000, 𝐶 = $12 500
11 a
𝑘
(
𝑘−2
−1
3 𝑥
)( ) = ( )
1
4 𝑦
𝑥
−1
𝑘
3
where 𝐀 = (
) , 𝐱 = (𝑦) and 𝐛 = ( )
1
𝑘−2 4
b Equations do not have a solution when 𝐀−1 doesn’t exist, i.e. when det 𝐀 = 0:
4𝑘 − 3(𝑘 − 2) = 0
𝑘+6=0
𝑘 = −6
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
10
Worked solutions
c
𝐱 = 𝐀−1 𝐛
1
𝑥
4
−3 −1
(𝑦) =
(
)( )
1
𝑘+6 2−𝑘 𝑘
1
−7
=
(
)
𝑘 + 6 2𝑘 − 2
7
i.e. 𝑥 = − 𝑘+6 , 𝑦 =
𝑘
12 a (
2
2𝑘−2
𝑘+6
𝑥
2
5
) (𝑦) = ( )
−1
𝑘−3
𝑥
2
5
) , 𝐱 = (𝑦) and 𝐛 = ( )
−1
𝑘−3
𝑘
where 𝐀 = (
2
b Equations do not have a solution when 𝐀−1 doesn’t exist, i.e. when det 𝐀 = 0:
𝑘(𝑘 − 3) − 10 = 0
𝑘 2 − 3𝑘 − 10 = 0
𝑘 = 5 or − 2
c
𝐱 = 𝐀−1 𝐛
1
𝑥
2
𝑘 − 3 −5
(𝑦) = 2
(
)( )
−1
𝑘
𝑘 − 3𝑘 − 10 −2
1
2𝑘 − 1
= 2
(
)
𝑘 − 3𝑘 − 10 −𝑘 − 4
2𝑘−1
−𝑘−4
i.e. 𝑥 = 𝑘 2 −3𝑘−10 , 𝑦 = 𝑘 2 −3𝑘−10
Exercise 3D
10 a i
det(𝐌 − 𝜆𝐈) = 0
(7 − 𝜆)(4 − 𝜆) − 4 = 0
𝜆2 − 11𝜆 + 24 = 0
ii
(𝜆 − 3)(𝜆 − 8) = 0
𝜆 = 3 or 8
b 𝜆1 = 3:
𝑥
7 2 𝑥
(
) (𝑦) = 3 (𝑦)
2 4
4𝑥 + 2𝑦 = 0 ⇒ 𝑦 = −2𝑥
1
𝑣1 = ( )
−2
𝜆2 = 8:
𝑥
7 2 𝑥
(
) ( ) = 8 (𝑦)
2 4 𝑦
−𝑥 + 2𝑦 = 0 ⇒ 𝑥 = 2𝑦
Mathematics for the IB Diploma: Applications & Interpretation
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11
Worked solutions
2
𝑣2 = ( )
1
−2 2
11 𝐌 = (
)
2 1
det(𝐌 − 𝜆𝐈) = 0
(−2 − 𝜆)(1 − 𝜆) − 4 = 0
𝜆2 + 𝜆 − 6 = 0
(𝜆 − 2)(𝜆 + 3) = 0
𝜆 = 2 or − 3
𝜆1 = 2:
𝑥
−2 2 𝑥
(
) ( ) = 2 (𝑦)
2 1 𝑦
−4𝑥 + 2𝑦 = 0 ⇒ 2𝑥 = 𝑦
1
𝑣1 = ( )
2
𝜆2 = −3:
𝑥
−2 2 𝑥
(
) (𝑦) = −3 (𝑦)
2 1
𝑥 + 2𝑦 = 0 ⇒ 𝑥 = −2𝑦
−2
𝑣2 = ( )
1
4
12 𝐌 = (
6
−5
)
−9
det(𝐌 − 𝜆𝐈) = 0
(4 − 𝜆)(−9 − 𝜆) + 30 = 0
𝜆2 + 5𝜆 − 6 = 0
(𝜆 − 1)(𝜆 + 6) = 0
𝜆 = 1 or − 6
𝜆1 = 1:
𝑥
4 −5 𝑥
(
) (𝑦) = (𝑦)
6 −9
3𝑥 − 5𝑦 = 0 ⇒ 3𝑥 = 5𝑦
5
𝑣1 = ( )
3
𝜆2 = −6:
𝑥
4 −5 𝑥
(
) (𝑦) = −6 (𝑦)
6 −9
2𝑥 − 𝑦 = 0 ⇒ 𝑦 = 2𝑥
1
𝑣2 = ( )
2
3 1 1
1
13 a (
)( ) = 𝜆( )
4 𝑎 2
2
(1)
3+2=𝜆
{
4 + 2𝑎 = 2𝜆 (2)
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
12
Worked solutions
From (1): 𝜆 = 5
Substituting into (2): 4 + 2𝑎 = 10 ⇒ 𝑎 = 3
b i
ii
From 𝑎), 𝜆1 = 5
det 𝐀 = 𝜆1 𝜆2
9 − 4 = 5𝜆2
𝜆2 = 1
𝑥
3 1 𝑥
(
) (𝑦) = 1 (𝑦)
4 3
2𝑥 + 𝑦 = 0 ⇒ 𝑦 = −2𝑥
1
𝑣2 = ( )
−2
14 a
det(𝐀 − 𝜆𝐈) = 0
(−5 − 𝜆)(−7 − 𝜆) − 24 = 0
𝜆2 + 12𝜆 + 11 = 0
(𝜆 + 1)(𝜆 + 11) = 0
𝜆 = −1 or − 11
𝜆1 = −1:
𝑥
𝑥
−5 8
(
) (𝑦) = − (𝑦)
3 −7
−4𝑥 + 8𝑦 = 0 ⇒ 𝑥 = 2𝑦
2
𝑣1 = ( )
1
𝜆2 = −11:
𝑥
𝑥
−5 8
(
) (𝑦) = −11 (𝑦)
3 −7
6𝑥 + 8𝑦 = 0 ⇒ 3𝑥 = −4𝑦
4
𝑣2 = ( )
−3
2 4
−1
0
b 𝐂=(
),𝐃 = (
)
1 −3
0 −11
15 a
det(𝐌 − 𝜆𝐈) = 0
(8 − 𝜆)(5 − 𝜆) − (−2) = 0
𝜆2 − 13𝜆 + 42 = 0
(𝜆 − 6)(𝜆 − 7) = 0
𝜆 = 6 or 7
𝜆1 = 6:
𝑥
8 2 𝑥
(
) (𝑦) = 6 (𝑦)
−1 5
2𝑥 + 2𝑦 = 0 ⇒ 𝑥 = −𝑦
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
13
Worked solutions
−1
𝑣1 = ( )
1
𝜆2 = 7:
𝑥
8 2 𝑥
(
) (𝑦) = 7 (𝑦)
−1 5
𝑥 + 2𝑦 = 0 ⇒ 𝑥 = −2𝑦
−2
𝑣2 = ( )
1
−1 −2
6
So, 𝐏 = (
),𝐃 = (
1
1
0
0
)
7
b
𝐌 3 = 𝐏𝐃3 𝐏−1
−1 −2 6 0 3 −1 −2 −1
=(
)(
) (
)
1
1
0 7
1
1
−1 −2 63 0
1
2
=(
)(
)(
)
1
1
0 73 −1 −1
−216 −686
1
2
=(
)(
)
216
343
−1 −1
470 254
=(
)
−127 89
16 a
det(𝐌 − 𝜆𝐈) = 0
(7 − 𝜆)(1 − 𝜆) − (−8) = 0
𝜆2 − 8𝜆 + 15 = 0
(𝜆 − 3)(𝜆 − 5) = 0
𝜆 = 3 or 5
𝜆1 = 3:
𝑥
7 4 𝑥
(
) (𝑦) = 3 (𝑦)
−2 1
4𝑥 + 4𝑦 = 0 ⇒ 𝑥 = −𝑦
1
𝑣1 = ( )
−1
𝜆2 = 5:
𝑥
7 4 𝑥
(
) ( ) = 5 (𝑦)
−2 1 𝑦
2𝑥 + 4𝑦 = 0 ⇒ 𝑥 = −2𝑦
−2
𝑣2 = ( )
1
1 −2
3
So, 𝐏 = (
),𝐃 = (
−1 1
0
0
)
5
b
𝐌 𝑛 = 𝐏𝐃𝑛 𝐏 −1
1
2
3 0 𝑛 1
2 −1
=(
)(
) (
)
−1 −1 0 5
−1 −1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
14
Worked solutions
1
2
3𝑛 0 −1 −2
=(
)(
)(
)
−1 −1 0 5𝑛
1
1
3𝑛 2(5𝑛 ) −1 −2
=( 𝑛
)(
)
1
1
−3
−5𝑛
2(5𝑛 ) − 3𝑛 2(5𝑛 ) − 2(3𝑛 )
=( 𝑛
)
3 − 5𝑛
2(3𝑛 ) − 5𝑛
17 a
det(𝐀 − 𝜆𝐈) = 0
(4 − 𝜆)(−4 − 𝜆) − (−12) = 0
𝜆2 − 4 = 0
(𝜆 − 2)(𝜆 + 2) = 0
𝜆 = 2 or − 2
𝜆1 = 2:
𝑥
4 −4 𝑥
(
) (𝑦) = 2 (𝑦)
3 −4
2𝑥 − 4𝑦 = 0 ⇒ 𝑥 = 2𝑦
2
𝑣1 = ( )
1
𝜆2 = −2:
𝑥
4 −4 𝑥
(
) ( ) = −2 (𝑦)
3 −4 𝑦
6𝑥 − 4𝑦 = 0 ⇒ 3𝑥 = 2𝑦
2
𝑣2 = ( )
3
2 2
2
So, 𝐏 = (
),𝐃 = (
1 3
0
0
)
−2
b
𝐀𝑛 = 𝐏𝐃𝑛 𝐏 −1
2 2 2 0 𝑛 2 2 −1
=(
)(
) (
)
1 3 0 −2
1 3
1 3 −2
0
2 2 2𝑛
=(
)(
) (
)
𝑛
1 3 0 (−2) 4 −1 2
1 2(2𝑛 ) 2(−2)𝑛
3 −2
= ( 𝑛
)(
)
3(−2)𝑛 −1 2
4 2
1 6(2𝑛 ) − 2(−2)𝑛 −4(2𝑛 ) + 4(−2)𝑛
= (
)
4 3(2𝑛 ) − 3(−2)𝑛 −2(2𝑛 ) + 6(−2)𝑛
i
When 𝑛 is odd (−2)𝑛 = −2𝑛 , so
1 6(2𝑛 ) + 2(2𝑛 ) −4(2𝑛 ) − 4(2𝑛 )
𝐀𝑛 = (
)
4 3(2𝑛 ) + 3(2𝑛 ) −2(2𝑛 ) − 6(2𝑛 )
2(2𝑛 ) −2(2𝑛 )
=(
)
3(2𝑛−1 ) −2(2𝑛 )
4 −4
= 2𝑛−1 (
)
3 −4
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
15
When 𝑛 is even (−2)𝑛 = 2𝑛
Worked solutions
ii
1 6(2𝑛 ) − 2(2𝑛 ) −4(2𝑛 ) + 4(2𝑛 )
𝐀𝑛 = (
)
4 3(2𝑛 ) − 3(2𝑛 ) −2(2𝑛 ) + 6(2𝑛 )
2𝑛 0
=(
)
0 2𝑛
1 0
= 2𝑛 (
)
0 1
1
1
18 a For eigenvector ( ) with eigenvalue 1 and eigenvector ( ) with eigenvalue – 2,
3
1
1 1 1 0
1 1 −1
𝐌1 = (
)(
)(
)
3 1 0 −2 3 1
1 1 −2
1 −1
=− (
)(
)
2 3 −2 −3 1
−3.5 1.5
=(
)
−4.5 2.5
1
1
For eigenvector ( ) with eigenvalue −2 and eigenvector ( ) with eigenvalue 1,
3
1
1 1 −2 0 1 1 −1
𝐌2 = (
)(
)(
)
3 1
0 1 3 1
1 −2 1
1 −1
=− (
)(
)
2 −6 1 −3 1
2.5 −1.5
=(
)
4.5 −3.5
b
−3.5 1.5 2.5 −1.5
𝐌1 𝐌2 = (
)(
)
−4.5 2.5 4.5 −3.5
−2 0
=(
)
0 −2
= −2𝐈
So
𝐌1 𝐌2 𝐯 = −2𝐈𝐯
(𝐌1 𝐌2 )𝐯 = (−2 × 1)𝐯
Therefore any eigenvector of the matrix 𝐌 = 𝐌1 𝐌2 will have eigenvector −2, which is
the product of the original eigenvectors.
19 a
det(𝐌 − 𝜆𝐈) = 0
(0.25 − 𝜆)(0.5 − 𝜆) − 0.375 = 0
4𝜆2 − 3𝜆 − 1 = 0
(4𝜆 + 1)(𝜆 − 1) = 0
𝜆 = −0.25 or 1
𝜆1 = −0.25:
𝑥
0.25 0.5 𝑥
(
) (𝑦) = −0.25 (𝑦)
0.75 0.5
0.5𝑥 + 0.5𝑦 = 0 ⇒ 𝑥 = −𝑦
−1
𝑣1 = ( )
1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
16
Worked solutions
𝜆2 = 1:
𝑥
0.25 0.5 𝑥
(
) (𝑦) = 1 (𝑦)
0.75 0.5
−0.75𝑥 + 0.5𝑦 = 0 ⇒ 3𝑥 = 2𝑦
2
𝑣2 = ( )
3
−1 2
−0.25 0
b So, 𝐏 = (
),𝐃 = (
)
1 3
0
1
c
𝐌 𝑛 = 𝐏𝐃𝑛 𝐏 −1
−1 2 −0.25 0 𝑛 −1 2 −1
=(
)(
) (
)
1 3
1 3
0
1
−1 2 (−0.25)𝑛 0 1 −3 2
=(
)(
) (
)
1 3
0
1𝑛 5 1 1
1 −(−0.25)𝑛 2 −3 2
= (
)(
)
1 1
5 (−0.25)𝑛 3
1 3(−0.25)𝑛 + 2 −2(−0.25)𝑛 + 2
= (
)
5 −3(−0.25)𝑛 + 3 2(−0.25)𝑛 + 3
Since (−0.25)𝑛 approaches 0 as 𝑛 becomes large,
1 2 2
0.4 0.4
𝐌 𝑛 approaches (
)=(
)
5 3 3
0.6 0.6
20 a
det(𝐌 − 𝜆𝐈) = 0
(𝑎 − 𝜆)2 − 𝑏 2 = 0
𝜆=𝑎±𝑏
b 𝜆1 = 𝑎 + 𝑏:
𝑎
(
𝑏
𝑥
𝑏 𝑥
) (𝑦) = (𝑎 + 𝑏) (𝑦)
𝑎
−𝑥 + 𝑦 = 0 ⇒ 𝑥 = 𝑦
1
𝑣1 = ( )
1
𝜆2 = 𝑎 − 𝑏:
𝑎
(
𝑏
𝑥
𝑏 𝑥
) (𝑦) = (𝑎 − 𝑏) (𝑦)
𝑎
𝑥 + 𝑦 = 0 ⇒ 𝑦 = −𝑥
1
𝑣2 = ( )
−1
1
c 𝐌=(
1
1
𝑎+𝑏
)(
−1
0
1 1 −1
0
)(
)
𝑎 − 𝑏 1 −1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
17
Worked solutions
d
1 1
𝑎+𝑏
0 𝑛 1 1 −1
𝐌𝑛 = (
)(
) (
)
1 −1
1 −1
0
𝑎−𝑏
1 1 1
(𝑎 + 𝑏)𝑛
0
1 1
=(
)(
) (
)
𝑛
(𝑎 − 𝑏) 2 1 −1
0
1 −1
1 1 1
(𝑎 + 𝑏)𝑛 (𝑎 + 𝑏)𝑛
= (
)(
)
2 1 −1 (𝑎 − 𝑏)𝑛 −(𝑎 − 𝑏)𝑛
1 (𝑎 + 𝑏)𝑛 + (𝑎 − 𝑏)𝑛 (𝑎 + 𝑏)𝑛 − (𝑎 − 𝑏)𝑛
= (
)
2 (𝑎 + 𝑏)𝑛 − (𝑎 − 𝑏)𝑛 (𝑎 + 𝑏)𝑛 + (𝑎 − 𝑏)𝑛
1 1 + 0.2𝑛
e 𝐌𝑛 = 2 (
1 − 0.2𝑛
1 − 0.2𝑛
)
1 + 0.2𝑛
Since 0.2𝑛 approaches 0 as 𝑛 becomes large,
1 1 1
0.5 0.5
𝐌 𝑛 approaches (
)=(
)
2 1 1
0.5 0.5
Mixed Practice 3
1 a
𝐏=𝐑−𝐅
42 30
110 85
= (154 102) − (65 54 )
88 106
130 175
68 55
= (89 48)
42 69
b 𝐐 = 1.05𝐑 − 1.02𝐅
c
42 30
110 85
𝐐 = 1.05 (154 102) − 1.02 (65 54 )
88 106
130 175
72.66 58.65
= (95.40 52.02)
46.74 75.63
2
2 −3 2𝑘 −𝑘
𝐀𝐁 = (
)(
)
0 4
𝑘
1
4𝑘 − 3𝑘 −2𝑘 − 3
=(
)
4𝑘
4
𝑘 −2𝑘 − 3
=(
)
4𝑘
4
3 a If A and B are commutative, then
2𝑘
(
𝑘−1
2 0
2 0
0
2𝑘
)(
)=(
)(
3 −1 𝑘 − 1
−1 3 −1
4𝑘
0
4𝑘
0
(
)=(
)
5𝑘 + 1 1
2𝑘 − 5 1
0
)
−1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
18
Worked solutions
So
2𝑘 − 5 = 5𝑘 + 1
3𝑘 = −6
𝑘 = −2
b For example, 𝐂 = (
1
𝐂𝐃 = (
0
0
𝐃𝐂 = (
1
0 0
)(
0 1
0 1
)(
0 0
1 0
0 0
),𝐃 = (
)
0 0
1 0
0
0 0
)=(
)
0
0 0
0
0 0
)=(
)
0
1 0
So, 𝐂𝐃 ≠ 𝐃𝐂
4 𝐀−1 =
=
1
−2
(
det 𝐀 −5
1
)
3
1
−2 1
(
)
−6 − (−5) −5 3
2
=(
5
−1
)
−3
5
𝐀𝐂 = 𝐁
𝐂 = 𝐀−𝟏 𝐁
1
2 −3 1
=
(
)(
−10 − (−6) 2 −5 𝑘
1 2 − 3𝑘 −6
=− (
)
4 2 − 5𝑘 −14
1 3𝑘 − 2 6
= (
)
4 5𝑘 − 2 14
3
)
4
6 a
1 1 −𝑘
(
)
det 𝐀 0 3
1 1 −𝑘
= (
)
3 0 3
𝐀−1 =
b
𝐂𝐀 = 𝐁
𝐂 = 𝐁𝐀−𝟏
3 𝑘 1 1 −𝑘
=(
) (
)
6 1 3 0 3
1
0
=(
)
2 1 − 2𝑘
7 Let the original 2 × 4 matrix be 𝐌.
Then
9 18 −13 3
−3 −6
4
−1
68 127 −130 22
𝐌(
)
)=(
7 12 −15 2
119 218 −240 38
1
1 −13 1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
19
Worked solutions
9 18 −13 3
68 127 −130 22 −3 −6
4
−1
𝐌=(
)(
)
119 218 −240 38
7 12 −15 2
1
1 −13 1
5 2 4 1
=(
)
7 3 9 2
−1
i.e. the two 4-digit ID numbers are 5241 and 7392
−3 1
8 a 𝐌2 = (
)
−7 2
1
−3 1
−3 1
𝐌 −1 =
(
)=(
)
−7 2
−6— 7 −7 2
So 𝐌 2 = 𝐌 −1
b
𝐌 2 = 𝐌 −1
𝐌 2 𝐌 = 𝐌 −1 𝐌
𝐌3 = 𝐈
1
i.e. 𝐌 3 = (
0
0
)
1
c
𝑥
7
𝐌 (𝑦) = ( )
23
𝑥
7
(𝑦) = 𝐌 −1 ( )
23
−3 1 7
=(
)( )
−7 2 23
2
=( )
−3
i.e. 𝑥 = 2, 𝑦 = −3
9
24.9
1.00 1.80 2.50 𝐴
(1.10 2.00 2.10) (𝐵) = (22.9)
24.3
1.20 1.70 2.40 𝐶
𝐴
1.00 1.80 2.50 −1 24.9
(𝐵) = (1.10 2.00 2.10) (22.9)
𝐶
24.3
1.20 1.70 2.40
2
= (3)
7
i.e. he orders 2 kg of 𝐴, 3 kg of 𝐵 and 7 kg of 𝐶.
10 a i
det(𝐌 − 𝜆𝐈) = 0
(3 − 𝜆)(2 − 𝜆) − 20 = 0
𝜆2 − 5𝜆 − 14 = 0
ii
(𝜆 − 7)(𝜆 + 2) = 0
𝜆 = 7 or − 2
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
20
Worked solutions
b 𝜆1 = 7:
𝑥
3 −5 𝑥
(
) (𝑦) = 7 (𝑦)
−4 2
−4𝑥 − 5𝑦 = 0 ⇒ 4𝑥 = −5𝑦
−5
𝑣1 = ( )
4
𝜆2 = −2:
𝑥
3 −5 𝑥
(
) (𝑦) = −2 (𝑦)
−4 2
5𝑥 − 5𝑦 = 0 ⇒ 𝑥 = 𝑦
1
𝑣2 = ( )
1
11
det(𝐌 − 𝜆𝐈) = 0
(3 − 𝜆)(4 − 𝜆) = 0
𝜆 = 3 or 4
𝜆1 = 3:
𝑥
3 −2 𝑥
(
) (𝑦) = 3 (𝑦)
0 4
−2𝑦 = 0 ⇒ 𝑦 = 0
1
𝑣1 = ( )
0
𝜆2 = 4:
𝑥
3 −2 𝑥
(
) (𝑦) = 4 (𝑦)
0 4
−𝑥 − 2𝑦 = 0 ⇒ 𝑥 = −2𝑦
−2
𝑣2 = ( )
1
12 𝐀𝐁 + 𝑘𝐂 = 𝐈
𝑎
(
2
−1 3
)(
1
0
3𝑎 + 𝑘
(
6 + 2𝑘
1 4
1
𝑏
)+𝑘(
)=(
2 𝑐
0
−4
𝑎𝑏 + 4 + 4𝑘
1
)=(
2𝑏 − 4 + 𝑐𝑘
0
0
)
1
0
)
1
So
3𝑎 + 𝑘 = 1
6 + 2𝑘 = 0
{
𝑎𝑏 + 4 + 4𝑘 = 0
2𝑏 − 4 + 𝑐𝑘 = 1
(1)
(2)
(3)
(4)
From (2): 𝑘 = −3
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
21
4
3
Worked solutions
Substituting into (1): 3𝑎 − 3 = 1 ⇒ 𝑎 =
4
3
Substituting into (3): 𝑏 + 4 − 12 = 0 ⇒ 𝑏 = 6
7
Substituting into (4): 12 − 4 − 3𝑐 = 1 ⇒ 𝑐 = 3
13 a i
det 𝐌 = 0
24 − 5𝑘 = 0
𝑘 = 4.8
ii
1
8 −𝑘
(
)
det 𝐌 −5 3
1
8 −𝑘
=
(
)
24 − 5𝑘 −5 3
𝐌 −1 =
b
3 7 𝑥
−1
(
)( ) = ( )
5 8 𝑦
13
𝑥
3 7 −1 −1
(𝑦) = (
) ( )
5 8
13
1
8 −7 −1
=
(
)( )
13
24 − 35 −5 3
1 −99
=− (
)
11 44
9
=( )
−4
i.e. 𝑥 = 9, 𝑦 = −4
14 a
𝐀𝐁𝐂 = 𝐈
𝐀−1 𝐀𝐁𝐂 = 𝐀−1 𝐈
𝐁𝐂 = 𝐀−1
𝐁−1 𝐁𝐂 = 𝐁−1 𝐀−1
𝐂 = 𝐁−1 𝐀−1
𝐂𝐀 = 𝐁−1 𝐀−1 𝐀
𝐂𝐀 = 𝐁−1
(multiplying on the left by 𝐀−1 )
(multiplying on the left by 𝐁−1 )
(multiplying on the right by 𝐀)
b
𝐁−1 = 𝐂𝐀
0 −1
4
5
=(
)(
)
−3 4
−2 −3
2
3
=(
)
−20 −27
𝐁 = (𝐁−1 )−1
1 −27 −3
= (
)
2
6 20
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
22
Worked solutions
15
𝐌 + 𝐌 −1 = 𝐈
1
2
0 −2
1 0
)+
(
)=(
)
0
0 1
−2𝑦 −𝑦 𝑥
𝑥
(
𝑦
1
𝑥
(
𝑦+
2+𝑦
1
2
𝑥
−2𝑦
1 0
)=(
)
0 1
So
𝑥=1
1
2+𝑦 =0
1
𝑦+2=0
𝑥
{−2𝑦 = 1
(1)
(2)
(3)
(4)
1
From (3): 𝑦 = − 2
1
Check 𝑥 = 1, 𝑦 = − 2 in (2) and (4):
2+
−
1
−12
=0
1
2(−12)
=−
1
=1
−1
1
Both consistent, so 𝑥 = 1, 𝑦 = − 2
16
2 −1 2 0 2 −1 −1
𝐌=(
)(
)(
)
1 3
0 3 1 3
2 −1 2 0 1 3 1
=(
)(
) (
)
1 3
0 3 7 −1 2
1 4 −3
3 1
= (
)(
)
−1 2
7 2 9
1 15 −2
= (
)
7 −3 20
17 a
det(𝐌 − 𝜆𝐈) = 0
(3 − 𝜆)(9 − 𝜆) − (−8) = 0
𝜆2 − 12𝜆 + 35 = 0
(𝜆 − 5)(𝜆 − 7) = 0
𝜆 = 5 or 7
𝜆1 = 5:
𝑥
3 4 𝑥
(
) (𝑦) = 5 (𝑦)
−2 9
−2𝑥 + 4𝑦 = 0 ⇒ 𝑥 = 2𝑦
2
𝑣1 = ( )
1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
23
Worked solutions
𝜆2 = 7:
𝑥
3 4 𝑥
(
) (𝑦) = 7 (𝑦)
−2 9
−4𝑥 + 4𝑦 = 0 ⇒ 𝑥 = 𝑦
1
𝑣2 = ( )
1
2 1
5
So, 𝐏 = (
),𝐃 = (
1 1
0
0
)
7
b
𝐌 𝑛 = 𝐏𝐃𝑛 𝐏 −1
2 1 5 0 𝑛 2 1 −1
=(
)(
) (
)
1 1 0 7
1 1
2 1 5𝑛 0
1 −1
=(
)(
)(
)
1 1 0 7𝑛 −1 2
1 −1
2(5𝑛 ) 7𝑛
=( 𝑛
)(
)
5
7𝑛 −1 2
2(5𝑛 ) − 7𝑛 2(7𝑛 ) − 2(5𝑛 )
=( 𝑛
)
5 − 7𝑛
2(7𝑛 ) − 5𝑛
18 a i
det(𝐌 − 𝜆𝐈) = 0
(𝑎 − 𝜆)(𝑏 − 𝜆) − (1 − 𝑎)(1 − 𝑏) = 0
𝜆2 − (𝑎 + 𝑏)𝜆 − (1 − 𝑎 − 𝑏) = 0
(𝜆 − 1)(𝜆 + (1 − 𝑎 − 𝑏)) = 0
𝜆 = 1 or 𝑎 + 𝑏 − 1
ii
𝜆1 = 1:
𝑎
(
1−𝑎
𝑥
1−𝑏 𝑥
) (𝑦) = 1 (𝑦)
𝑏
(𝑎 − 1)𝑥 + (1 − 𝑏)𝑦 = 0 ⇒ (𝑎 − 1)𝑥 = (𝑏 − 1)𝑦
𝑏−1
𝑣1 = (
)
𝑎−1
𝜆2 = 𝑎 + 𝑏 − 1:
𝑎
(
1−𝑎
𝑥
1−𝑏 𝑥
) (𝑦) = (𝑎 + 𝑏 − 1) (𝑦)
𝑏
(1 − 𝑏)𝑥 + (1 − 𝑏)𝑦 = 0 ⇒ 𝑥 = −𝑦
1
𝑣2 = ( )
−1
𝑏−1
b So, 𝐏 = (
𝑎−1
1
1
),𝐃 = (
0
−1
0
)
𝑎+𝑏−1
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
24
Worked solutions
c
𝐌 𝑛 = 𝐏𝐃𝑛 𝐏 −1
𝑛
1
0
𝑏−1 1
𝑏 − 1 1 −1
=(
)(
) (
)
𝑎 − 1 −1 0 𝑎 + 𝑏 − 1
𝑎 − 1 −1
𝑛
1
1
0
−1
−1
𝑏−1 1
=(
)(
(
)
𝑛)
(𝑎
+ 𝑏 − 1) 1 − 𝑏 − 𝑎 + 1 1 − 𝑎 𝑏 − 1
𝑎 − 1 −1 0
1
𝑏 − 1 (𝑎 + 𝑏 − 1)𝑛
−1
−1
=
(
)
𝑛) (
𝑎
−
1
−(𝑎
+
𝑏
−
1)
1
−
𝑎
𝑏
−1
2−𝑎−𝑏
1
1 − 𝑏 + (1 − 𝑎)(𝑎 + 𝑏 − 1)𝑛 1 − 𝑏 − (1 − 𝑏)(𝑎 + 𝑏 − 1)𝑛
=
(
)
2 − 𝑎 − 𝑏 1 − 𝑎 − (1 − 𝑎)(𝑎 + 𝑏 − 1)𝑛 1 − 𝑎 + (1 − 𝑏)(𝑎 + 𝑏 − 1)𝑛
Note: 0 < 𝑎 < 1 and 0 < 𝑏 < 1 ⇒ −1 < 𝑎 + 𝑏 − 1 < 1
So, as 𝑛 becomes large (𝑎 + 𝑏 − 1)𝑛 approaches 0 and therefore 𝐌 𝑛 approaches
1
1−𝑏
(
2−𝑎−𝑏 1 − 𝑎
1−𝑏
).
1−𝑎
Mathematics for the IB Diploma: Applications & Interpretation
© Paul Fannon, Vesna Kadelburg, Ben Woolley, Steven Ward 2021
25