equilibrium solution yt { a is stable. b appm 2360 ... the equilibrium solution yt {0 is unstable,...

152 CHAPTER 2 Linearity and Nonlinearity

2.5 Nonlinear Models: Logistic Equation

� Equilibria

Note: Problems 1–6 are all autonomous s equations, so lines of constant slope (isoclines) are horizontal

lines.

1. 2y ay byc � , � �0, 0a b! !

We find the equilibrium points by solving

2 0y ay byc � ,

getting 0y , ab

� . By inspecting

� �y y a byc � ,

we see that solutions have positive slope � �0yc ! when 0y ! or ayb

� � and negative slope

� �0yc � for 0a yb

� � � . Hence, the equilibrium solution � � 0y t { is unstable, and the

equilibrium solution � � ay tb

{ � is stable.

y

y = 0

y = –a/bstable equilibrium

unstable equilibriumt

2. 2y ay byc � , � �0, 0a b! !


2 0y ay byc � ,

getting 0y , ab

. By inspecting

� �y y a byc � ,

APPM 2360 Homework 4 SolutionsFall 2015

SECTION 2.5 Nonlinear Models: Logistic Equation 153

we see that solutions have negative slope � �0yc � when 0y � or ayb

! and positive slope

� �0yc ! for 0 ayb

� � . Hence, the equilibrium solution � � 0y t { is unstable, and the equilibrium

solution � � ay tb

{ is stable.

y

y = 0

y = a/bstable equilibrium

unstable equilibriumt

3. 2y ay byc � � , � �0, 0a b! !


2 0y ay byc � � ,

getting 0y , ab

. By inspecting

� �y y a byc � � ,

we see that solutions have positive slope when 0y � or ayb

! and negative slope for 0 ayb

� � .

Hence, the equilibrium solution � � 0y t { is stable, and the equilibrium solution � � ay tb

{ is

unstable.

y

y = 0

y = a/b unstable equilibrium

stable equilibriumt

Section 2.5


gives

� � � � � �^ `

� �0 0 0

0

2

3

1 2 1 1 1

1 1

rt rt rtL L Ly y y

rtLy

L r e e ey

e

� � �

�

ª º� � � � �¬ ¼cc ª º� �¬ ¼

.

Setting 0ycc , we get

0 0

2 1 1 1 0rt rtL Le ey y

� �ª º§ · § ·� � � � « »¨ ¸ ¨ ¸

« »© ¹ © ¹¬ ¼

or

0

1 1rtL ey

�§ ·� ¨ ¸

© ¹.

Solving for t, we get *

0

1 ln 1Ltr y

§ · �¨ ¸

© ¹. Substituting this value into the analytical solution

for the logistic equation, we get � �*2Ly t .

At *t the rate yc is

212 2 2 4

L L r L rLrL

§ ·§ · § ·� ¨ ¸¨ ¸ ¨ ¸¨ ¸© ¹ © ¹© ¹.

� Fitting the Logistic Law

14. The logistic equation is

1 yy ryL

§ ·c �¨ ¸© ¹

.

If initially the population doubles every hour, we have

ln 2 1dt r

which gives the growth rate 1 1.4ln 2

r | . We are also given 95 10L u . The logistic curve after

4 hrs is calculated from the analytic solution formula,

� � � � � � � �9

90

9 99

5.61.4 45 1010

5 10 5 10 4.9 101 41 11 1 rtL

y

Ly teee �� u

u u | u

�� .


� Semistable Equilibrium

19. � �21y yc �

We draw upward arrows on the y-axis for

1y z

to indicate the solution is increasing. When

1y

we have a constant solution.

–1

2y

6t

y = 1semistable equilibrium

Because the slope lines have positive slope both above and below the constant solution � � 1y t { , we say that the solution � � 1y t { , or the point 1, is semistable (stable from below,

unstable from above). In other words, if a solution is perturbed from the value of 1 to a value below 1, the solution will move back towards 1, but if the constant solution � � 1y t { is perturbed

to a value larger than 1, it will not move back towards 1. Semistable equilibria are customarily

noted with half-filled circles.

� Gompertz Equation � �1 lndy y b ydt

�

20. (a) Letting lnz y and using the chain rule we get

1dz dz dy dydt dy dt y dt

§ · § ·§ · § · ¨ ¸ ¨ ¸¨ ¸ ¨ ¸© ¹ © ¹© ¹ © ¹

.

Hence, the Gompertz equation becomes

dz a bzdt

� .

(b) Solving this DE for z we find

� � bt az t Ceb

� � .

Substituting back lnz y gives

� � bta b Cey t e e�

.

Using the initial condition � � 00y y , we finally get 0ln aC yb

� .

(c) From the solution in part (b), � �lim a b

ty t e

of when 0b ! , � �y t of when 0b � .


(b) For a pure logistic curve, the inflection

point always occurs when 2Ly .

However, if we consider models different from the logistic model that still show similar solutions between 0 and the nonzero equilibrium, it is possible for the inflection point to be closer to 0. When this happens oil recovery reaches the maximum production rate much earlier.

Of course the logistic model is a crude model of oil production. For example it doesn’t take into consideration the fact that when oil prices are high, many oil wells are placed into production.

If the inflection point is lower than halfway on an approximately logistic curve, the peak on the yc curve occurs sooner and lower creating an asymmetric curve for yc .

(c) These differences may or may not be significant to people studying oil production; it depends on what they are looking for. The long-term behavior, however, is the same; the peak just occurs sooner. After the peak occurs, if the model holds, it is downhill insofar as oil production is concerned. Typical skew of peak position is presented on the figures above.

� Useful Transformation

30. � �1y ky yc �

Letting 1yzy

�

yields

� �21

1dz dz dy dydt dy dt dty

§ ·§ · § · ¨ ¸¨ ¸ ¨ ¸© ¹ © ¹�© ¹

.

Substituting for dydt

from the original DE yields a new equation

� � � �21 1dzy ky ydt

� � ,

which gives the result

1

dz ky kzdt y

�

.

inflection point for dashed y curve.

inflection point for solid y curve.

170 CHAPTER 2 Linearity and Nonlinearity Solving this first-order equation for � �z z t , yields � � ktz t ce and substituting this in the

transformation 1y

zy

�

, we get 1

ktyce

y

�.

Finally, solving this for y gives � � 11

1 1 ,1 1kt kt

c

y te c e� �

� �

where 10

1 1cy

� .

� Chemical Reactions

31. � �� 100 50x k x x � �

The solutions for the given initial conditionsare shown on the graph. Note that all behaviorsare at equilibrium or flown off scale before

0.1t !

The solution curve for � �0 150x is almost

vertical.

0

200y

0.50t

stable equilibrium

unstable equilibrium

Direction field and equilbrium

(a) A solution starting at � �0 0x

increases and approaches 50.

(b) A solution starting at � �0 75x

decreases and approaches 50.

(c) A solution starting at � �0 150x

increases without bound.

0

200y

0.50t

Solutions for three given initial conditions.

Noting the location of equilibrium and the direction field as shown in a second graph leads to the following conclusions: Any (0) 100x ! causes � �x t to increase without bound and fly off scale very quickly. On the other hand, for any � � � �0 0,100x � the solution will approach an equilibrium

value of 50, which implies the tiniest amount is sufficient to start the reaction.

If you are looking for a different scenario, you might consider some other modeling options that appear in Problem 32.

Section 2.6


2.6 Systems of DEs: A First Look

� Predicting System Behavior

1. (a) 3

x yy x yc c �

This (linear) system has one equilibrium point at the origin, � � � �, 0, 0x y , as do all

linear systems. The X- and h-nullclines are respectively, as shown in part (b).

(b)

h

–2

2y

2–2x

-nullcline

X -nullcline

(c) A few solutions along with the vertical

and horizontal nullclines are drawn.

–2

2y

2–2x

(d) The equilibrium point (0, 0) is unstable.

All solutions tend quickly to 3x

y then move gradually towards �f or �f

asymptotically along that line. Whether the motion is left or right depends on the initial

conditions.

2. (a) 2

1x x y

y x y

c � �

c �

Setting 0xc and 0yc gives

2

-nullcline 1 0

-nullcline 0.

x y

h x y

X � �

�

SECTION 2.6 Systems of DEs: A First Look 181

From the intersection of the two nullclines we find two equilibrium points shown in the following figures. We can locate them graphically far more easily than algebraically!

(b)

2–2x

–2

2y

h -nullclineX -nullcline

(c)

2–2x

–2

2y

(d) The lower equilibrium point at

21 1(1 5) , ( 1 5)4 2ª º� � �« »¬ ¼

is unstable and the upper equilibrium at

21 1(1 5) , ( 5 1)4 2ª º� �« »¬ ¼

is stable.

Most trajectories spiral counterclockwise toward the first quadrant equilibrium point. However, if the initial condition is somewhat left or below the 4th quadrant equilibrium, they shoot down towards �f . We suspect a dividing line between these behaviors, and we will find it in Chapter 6.

3. (a) 2 2

1

1

x x y

y x y

c � �

c � �

Setting 0xc and 0yc gives

2 2-nullcline 1

-nullcline 1.h x y

x yX� �

From the intersection of the two nullclines we find two equilibrium points (0, 1), (1, 0).


(c) The direction field and a few solutions

are drawn. Note how the solutions cross

the vertical and horizontal nullclines

–2

2y

2–2x

(d) We see from the preceding figure that solutions come from infinity along a line (that is,

not a nulllcline), and then if they are not exactly on the line head off either upwards and

to the left or downwards and go to the left on another line. Whether they go up or down

depends on whether they initially start above or below the line. It appears that points that

start exactly on the line will go to (0, 0). We will see later in Chapter 6 when we study

linear systems using eigenvalues and eigenvectors that the solutions come from infinity

on one eigenvector and go to infinity on another eigenvector.

7. (a) 1x x y

y x y

c � �

c �

Setting 0x yc c and finding the intersection of the nullclines:

-nullcline-nullcline 1h y x

y xX

�

we find one equilibrium point 1 1, 2 2

§ ·¨ ¸© ¹

. The arrows indicate that it is a stable equilib-

rium.

(b)

2–2x

–2

2y

(c)

2–2x

–2

2y

(d) The equilibrium is stable; all other solutions spiral into it.

SECTION 2.6 Systems of DEs: A First Look 189

In this system the equilibria on the axes are all unstable, so the populations always head toward a

coexistence equilibrium at 200300ª º« »¬ ¼

. See Figure, where x and y are measured in hundreds.

� Finding the Model

Example of appropriate models are as follows, with real positive coefficients.

13. 2x ax bx dxy fx

y cy dxyc � � �c � �

14.

2

x ax bxyy cy dxy eyz

z fz gx hyz

c �c � �

c � �

15. 2

2

x ax bx cxy dxz

y ey fy gxyz hz kxz

c � � �

c � �c � �

� Host-Parasite Models

16. (a) A suggested model is

1HH aH cP

P bP dHP

c ��

c � �

where a, b, c, and d are positive parameters. Here a species of beetle (parasite) depends on a certain species of tree (host) for survival. Note that if the beetle were so effective as to wipe out the entire population of trees, then it would die out itself, which is reflected in our model (note the differential equation in P). On the other hand, in the absence of the beetle, the host tree may or may not die out depending on the size of the parameters a and c. We would probably pick a c! , so the host population would increase in the absence of

the parasite. Note too that model says that when the parasite (P) population gets large, it


will not destroy the host entirely, as 1HP�

becomes small. The modeler might want to

estimate the values of parameters a, b, c, d so the solution fits observed data. The modeler would also like to know the qualitative behavior of � �,P H in the PH plane.

Professor Larry Turyn of Wright State University argues for a different model,

CHHHP

c � ,

to better account for the case of very small P.

(b) Many bacteria are parasitic on external and internal body surfaces; some invading inner tissue causing diseases such as typhoid fever, tuberculosis, and pneumonia. It is important to construct models of the dynamics of these complex organisms.

� Competition

17. (a) (4 2 )(4 2 )

x x x yy y x yc � �c � �

Setting 0xc and 0yc we find

-nullclines 2 4, 0-nullclines 2 4, 0.

x y xh x y yX �

�

Equilibrium points: (0, 0), (0, 2), (2, 0), 4 4, 3 3

§ ·¨ ¸© ¹

. The directions of the solution

curves are shown in the figure.

50x

y

(b) It can be seen from the figure, that the equilibrium points (0, 0), (0, 2) and (2, 0) are

unstable. Only the point 4 4, 3 3

§ ·¨ ¸© ¹

is stable because all solution curves nearby point

toward it.

(c) Some solution curves are shown in the figure.

(d) Because all the solution curves eventually reach the stable equilibrium at 4 4, 3 3

§ ·¨ ¸© ¹

, the

two species described by this model can coexist.

201

CHAPTER 3 LinearAlgebra

3.1 Matrices: Sums and Products

� Do They Compute?

1. 2 0 6

2 4 2 42 0 2

�ª º« » « »« »�¬ ¼

A 2. 1 6 3

2 2 3 21 0 3

ª º« »� « »« »�¬ ¼

A B

3. 2 �C D , Matrices are not compatible

4. 1 3 32 7 21 3 1

� �ª º« » « »« »� �¬ ¼

AB 5. 5 3 92 1 21 0 1

ª º« » « »« »�¬ ¼

BA 6. 3 1 08 1 29 2 6

�ª º« » �« »« »¬ ¼

CD

7. 1 16 7

�ª º « »¬ ¼

DC 8. � �T 1 61 7

ª º « »�¬ ¼

DC

9. TC D , Matrices are not compatible 10. TD C , Matrices are not compatible

11. 2

2 0 02 1 100 0 2

�ª º« » �« »« »�¬ ¼

A 12. AD, Matrices are not compatible

13. 3

2 0 32 0 21 0 0

�ª º« »� « »« »�¬ ¼

A I 14. 3

1 12 04 3 0 1 0

0 0 1

ª º« »� « »« »¬ ¼

B I

15. 3�C I , Matrices are not compatible 16. 2 96 70 3

ª º« » « »« »¬ ¼

AC

Section 3.1

201




1. 2 0 6

2 4 2 42 0 2

�ª º« » « »« »�¬ ¼

A 2. 1 6 3

2 2 3 21 0 3

ª º« »� « »« »�¬ ¼

A B


4. 1 3 32 7 21 3 1

� �ª º« » « »« »� �¬ ¼

AB 5. 5 3 92 1 21 0 1

ª º« » « »« »�¬ ¼

BA 6. 3 1 08 1 29 2 6

�ª º« » �« »« »¬ ¼

CD

7. 1 16 7

�ª º « »¬ ¼

DC 8. � �T 1 61 7

ª º « »�¬ ¼

DC


11. 2

2 0 02 1 100 0 2

�ª º« » �« »« »�¬ ¼


13. 3

2 0 32 0 21 0 0

�ª º« »� « »« »�¬ ¼

A I 14. 3

1 12 04 3 0 1 0

0 0 1

ª º« »� « »« »¬ ¼

B I


ª º« » « »« »¬ ¼

AC

201




1. 2 0 6

2 4 2 42 0 2

�ª º« » « »« »�¬ ¼

A 2. 1 6 3

2 2 3 21 0 3

ª º« »� « »« »�¬ ¼

A B


4. 1 3 32 7 21 3 1

� �ª º« » « »« »� �¬ ¼

AB 5. 5 3 92 1 21 0 1

ª º« » « »« »�¬ ¼

BA 6. 3 1 08 1 29 2 6

�ª º« » �« »« »¬ ¼

CD

7. 1 16 7

�ª º « »¬ ¼

DC 8. � �T 1 61 7

ª º « »�¬ ¼

DC


11. 2

2 0 02 1 100 0 2

�ª º« » �« »« »�¬ ¼


13. 3

2 0 32 0 21 0 0

�ª º« »� « »« »�¬ ¼

A I 14. 3

1 12 04 3 0 1 0

0 0 1

ª º« »� « »« »¬ ¼

B I


ª º« » « »« »¬ ¼

AC

201




1. 2 0 6

2 4 2 42 0 2

�ª º« » « »« »�¬ ¼

A 2. 1 6 3

2 2 3 21 0 3

ª º« »� « »« »�¬ ¼

A B


4. 1 3 32 7 21 3 1

� �ª º« » « »« »� �¬ ¼

AB 5. 5 3 92 1 21 0 1

ª º« » « »« »�¬ ¼

BA 6. 3 1 08 1 29 2 6

�ª º« » �« »« »¬ ¼

CD

7. 1 16 7

�ª º « »¬ ¼

DC 8. � �T 1 61 7

ª º « »�¬ ¼

DC


11. 2

2 0 02 1 100 0 2

�ª º« » �« »« »�¬ ¼


13. 3

2 0 32 0 21 0 0

�ª º« »� « »« »�¬ ¼

A I 14. 3

1 12 04 3 0 1 0

0 0 1

ª º« »� « »« »¬ ¼

B I


ª º« » « »« »¬ ¼

AC

202 CHAPTER 3 Linear Algebra

� More Multiplication Practice

17. > @ 1 0 2 21 0 2

1 0 2 2

a ba c e a e

c db d f b f

e f

ª º� � � � � �ª º ª º« »� « » « »« » � � � � � �¬ ¼ ¬ ¼« »¬ ¼

18. 0

0a b d b ad bc ab ba ad bcc d c a cd dc db da ad bc

� � � � �ª º ª º ª º ª º « » « » « » « »� � � � �¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

19.

1 10 2 0 02 0 1 02 21 1 1 1 1 0 11 1

2 2 2

ª º ª º� �« » « »ª º ª º « » « »« » « »

¬ ¼ ¬ ¼« » « »� �« » « »¬ ¼ ¬ ¼

20. > @ > @0 1 0a b cd e f d e fg h k

ª º« » « »« »¬ ¼

21. > @ > @0 1 1 1 0a b cd e fª º« »¬ ¼

= > @> @1 1 0d e f not possible

22. > @ > @ > @1 11 1 0

1 1

a ba c b d a c b dc d

e f

ª ºª º ª º« » � � � � �« » « »« » ¬ ¼ ¬ ¼« »¬ ¼

� Rows and Columns in Products 23. (a) 5 columns (b) 4 rows (c) 6 4u � Which Rules Work for Matrix Multiplication? 24. Counterexample:

A = 1 11 0ª º« »¬ ¼

B = 2 10 1

�ª º« »¬ ¼

(A + B)(A � B) = 3 0 1 21 1 1 1

�ª º ª º« » « »�¬ ¼ ¬ ¼

= 3 6

0 1�ª º« »¬ ¼

A2 � B2 = 2 1 4 3 2 41 1 0 1 1 0

� �ª º ª º ª º� « » « » « »

¬ ¼ ¬ ¼ ¬ ¼

25. Counterexample: Also due to the fact that AB z BA for most matrices

A = 1 11 0ª º« »¬ ¼

B = 2 10 1

�ª º« »¬ ¼

(A + B)2 = 9 04 1ª º« »¬ ¼

AB = 2 02 1ª º« »�¬ ¼

A2 + 2AB + B2 = 2 1 2 0 4 3

21 1 2 1 0 1

�ª º ª º ª º� �« » « » « »�¬ ¼ ¬ ¼ ¬ ¼

= 10 25 0

�ª º« »¬ ¼

SECTION 3.1 Matrices: Sums and Products 203

26. Proof (I + A)2 = (I + A)(I + A) = I(I + A) + A(I + A) distributive property = I2 + IA + AI + A2 = I + A + A + A2 identity property = I + 2A + A2 27. Proof (A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B) distributive property = A2 + AB + BA + B2 distributive property � Find the Matrix

28. Set 1 2 3 2 4 0 03 4 3 2 4 0 0

a b a b a bc d c d c d

� �ª º ª º ª º ª º « » « » « » « »� �¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

a + 3b = 0 a + 3b = 0 c + 3d = 0 c + 3d = 0 2a + 4b = 0 a + 2b = 0 2c � 4d = 0 c + 2d = 0 b = 0 d = 0 ? a = 0 ? c = 0 Therefore no nonzero matrix A will work.

29. B must be 3 u 2 Set 1 2 3 1 0

0 10 1 0

a bc de f

ª ºª º ª º« » « » « »« » ¬ ¼¬ ¼ « »¬ ¼

a + 2c + 3e = 1 c = 0 b + 2d + 3f = 0 d = 1

B is any matrix of the form 1 3 2 3

0 1e f

e f

� � �ª º« »« »« »¬ ¼

for any real numbers e and f.

30. Set 1 2 2 04 1 1 4

a bc d

ª º ª º ª º « » « » « »

¬ ¼ ¬ ¼ ¬ ¼

a + 2c = 2 b + 2d = 0 4a + c = 1 4b + d = 4

Thus c = 1, a = 0 b = 87

, d = 47

� and

82 7417

a bc d

ª º« »ª º

« »« »¬ ¼ « »�« »¬ ¼

.

� Commuters

31. 0 1 0

0 0 1a

aa

ª º ª º « » « »

¬ ¼ ¬ ¼ so the matrix commutes with every 2 u 2 matrix.

a = 1 � 3e b = �2 � 3f �


26. Proof (I + A)2 = (I + A)(I + A) = I(I + A) + A(I + A) distributive property = I2 + IA + AI + A2 = I + A + A + A2 identity property = I + 2A + A2 27. Proof (A + B)2 = (A + B)(A + B) = A(A + B) + B(A + B) distributive property = A2 + AB + BA + B2 distributive property � Find the Matrix

28. Set 1 2 3 2 4 0 03 4 3 2 4 0 0

a b a b a bc d c d c d

� �ª º ª º ª º ª º « » « » « » « »� �¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

a + 3b = 0 a + 3b = 0 c + 3d = 0 c + 3d = 0 2a + 4b = 0 a + 2b = 0 2c � 4d = 0 c + 2d = 0 b = 0 d = 0 ? a = 0 ? c = 0 Therefore no nonzero matrix A will work.

29. B must be 3 u 2 Set 1 2 3 1 0

0 10 1 0

a bc de f

ª ºª º ª º« » « » « »« » ¬ ¼¬ ¼ « »¬ ¼

a + 2c + 3e = 1 c = 0 b + 2d + 3f = 0 d = 1

B is any matrix of the form 1 3 2 3

0 1e f

e f

� � �ª º« »« »« »¬ ¼

for any real numbers e and f.

30. Set 1 2 2 04 1 1 4

a bc d

ª º ª º ª º « » « » « »

¬ ¼ ¬ ¼ ¬ ¼

a + 2c = 2 b + 2d = 0 4a + c = 1 4b + d = 4

Thus c = 1, a = 0 b = 87

, d = 47

� and

82 7417

a bc d

ª º« »ª º

« »« »¬ ¼ « »�« »¬ ¼

.

� Commuters

31. 0 1 0

0 0 1a

aa

ª º ª º « » « »

¬ ¼ ¬ ¼ so the matrix commutes with every 2 u 2 matrix.

a = 1 � 3e b = �2 � 3f �


� Trace of a Matrix 47. � � � � � �Tr Tr Tr� �A B A B � � � � � � � � � �11 11 11 11Tr Tr( )+Tr( )nn nn nn nna b a b a a b b� � � � � � � � � � A B A B" " " . 48. � � � � � �11 11Tr Trnn nnc ca ca c a a c � � � � A A" " 49. � � � �TTr Tr A A . Taking the transpose of a (square) matrix does not alter the diagonal element,

so � � � �TTr Tr A A .

50. � � > @ > @ > @ > @

� � � �� > @ > @ > @ > @ � �

11 1 11 1 1 1

11 11 1 1 1 1

11 11 1 1 1 1

11 1 11 1 1 1

Tr

Tr

n n n nn n nn

n n n n nn nn

n n n n nn nn

n n n nn n nn

a a b b a a b b

a b a b a b a b

b a b a b a b a

b b a a b b a a

� � � � � � � � � � � �

� � � � � �

� � � � � �

� � � � � � � � � � � �

AB

BA

" " " " "

" " "

" " "

" " " " "

� Matrices Can Be Complex

51. 3 0

22 4 4

ii i

�ª º� « »� �¬ ¼

A B 52. 3 1

8 4 5 3i ii i

� � � �ª º « »� �¬ ¼

AB

53. 1 34 1ii i� �ª º

« »�¬ ¼BA 54. 2 6 4 6

6 4 5 8i ii i

�ª º « »� � �¬ ¼

A

55. 1 22 3 2i

ii i

� � �ª º « »�¬ ¼

A 56. 1 2 2

26 4 5i i

ii

� � �ª º� « »�¬ ¼

A B

57. T 1 21i

i iª º

« »� �¬ ¼B 58. � �Tr 2 i �B

� Real and Imaginary Components

59. 1 2 1 0 1 2

2 2 3 2 2 0 3i i

ii

�ª º ª º ª º �« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼

A , 1 1 0 0 12 1 0 1 2 1

ii

i i� �ª º ª º ª º

�« » « » « »�¬ ¼ ¬ ¼ ¬ ¼B

� Square Roots of Zero 60. If we assume

a bc dª º

« »¬ ¼

A

is the square root of

0 00 0ª º« »¬ ¼

,

then we must have

2

22

0 00 0

a b a b a bc ab bdc d c d ac cd bc d

ª º� �ª º ª º ª º « »« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼¬ ¼

A ,




50. � � > @ > @ > @ > @

� � � �� > @ > @ > @ > @ � �

11 1 11 1 1 1

11 11 1 1 1 1

11 11 1 1 1 1

11 1 11 1 1 1

Tr

Tr

n n n nn n nn

n n n n nn nn

n n n n nn nn

n n n nn n nn

a a b b a a b b

a b a b a b a b

b a b a b a b a

b b a a b b a a

� � � � � � � � � � � �

� � � � � �

� � � � � �

� � � � � � � � � � � �

AB

BA

" " " " "

" " "

" " "

" " " " "


51. 3 0

22 4 4

ii i

�ª º� « »� �¬ ¼

A B 52. 3 1

8 4 5 3i ii i

� � � �ª º « »� �¬ ¼

AB

53. 1 34 1ii i� �ª º

« »�¬ ¼BA 54. 2 6 4 6

6 4 5 8i ii i

�ª º « »� � �¬ ¼

A

55. 1 22 3 2i

ii i

� � �ª º « »�¬ ¼

A 56. 1 2 2

26 4 5i i

ii

� � �ª º� « »�¬ ¼

A B

57. T 1 21i

i iª º

« »� �¬ ¼B 58. � �Tr 2 i �B


59. 1 2 1 0 1 2

2 2 3 2 2 0 3i i

ii

�ª º ª º ª º �« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼

A , 1 1 0 0 12 1 0 1 2 1

ii


�« » « » « »�¬ ¼ ¬ ¼ ¬ ¼B


a bc dª º

« »¬ ¼

A


0 00 0ª º« »¬ ¼

,

then we must have

2

22

0 00 0


ª º� �ª º ª º ª º « »« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼¬ ¼

A ,




50. � � > @ > @ > @ > @

� � � �� > @ > @ > @ > @ � �

11 1 11 1 1 1

11 11 1 1 1 1

11 11 1 1 1 1

11 1 11 1 1 1

Tr

Tr

n n n nn n nn

n n n n nn nn

n n n n nn nn

n n n nn n nn

a a b b a a b b

a b a b a b a b

b a b a b a b a

b b a a b b a a

� � � � � � � � � � � �

� � � � � �

� � � � � �

� � � � � � � � � � � �

AB

BA

" " " " "

" " "

" " "

" " " " "


51. 3 0

22 4 4

ii i

�ª º� « »� �¬ ¼

A B 52. 3 1

8 4 5 3i ii i

� � � �ª º « »� �¬ ¼

AB

53. 1 34 1ii i� �ª º

« »�¬ ¼BA 54. 2 6 4 6

6 4 5 8i ii i

�ª º « »� � �¬ ¼

A

55. 1 22 3 2i

ii i

� � �ª º « »�¬ ¼

A 56. 1 2 2

26 4 5i i

ii

� � �ª º� « »�¬ ¼

A B

57. T 1 21i

i iª º

« »� �¬ ¼B 58. � �Tr 2 i �B


59. 1 2 1 0 1 2

2 2 3 2 2 0 3i i

ii

�ª º ª º ª º �« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼

A , 1 1 0 0 12 1 0 1 2 1

ii


�« » « » « »�¬ ¼ ¬ ¼ ¬ ¼B


a bc dª º

« »¬ ¼

A


0 00 0ª º« »¬ ¼

,

then we must have

2

22

0 00 0


ª º� �ª º ª º ª º « »« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼¬ ¼

A , 208 CHAPTER 3 Linear Algebra

which implies the four equations

2

2

000

0.

a bcab bdac cdbc d

� � �

�

From the first and last equations, we have 2 2a d . We now consider two cases: first we assume

a d . From the middle two preceding equations we arrive at 0b , 0c , and hence 0a , 0d . The other condition, a d � , gives no condition on b and c, so we seek a matrix of the

form (we pick 1a , 1d � for simplicity)

1 1 1 0

1 1 0 1b b bc

c c bc�ª º ª º ª º

« » « » « »� � �¬ ¼ ¬ ¼ ¬ ¼.

Hence, in order for the matrix to be the zero matrix, we must have 1bc

� , and hence

11

1c

c

ª º�« »« »

�¬ ¼

,

which gives

1 1 0 01 1

0 01 1c c

c c

ª º ª º� � ª º« » « » « »« » « » ¬ ¼� �¬ ¼ ¬ ¼

.

� Zero Divisors 61. No, AB 0 does not imply that A 0 or B 0 . For example, the product

1 0 0 00 0 0 1ª º ª º« » « »¬ ¼ ¬ ¼

is the zero matrix, but neither factor is itself the zero matrix. � Does Cancellation Work?

62. No. A counterexample is: 0 0 1 2 0 0 0 00 1 0 4 0 1 0 4§ · § · § · § ·

¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸© ¹ © ¹ © ¹ © ¹

since 1 2 0 00 4 0 4§ · § ·

z¨ ¸ ¨ ¸© ¹ © ¹

.


which implies the four equations

2

2

000

0.

a bcab bdac cdbc d

� � �

�

From the first and last equations, we have 2 2a d . We now consider two cases: first we assume

a d . From the middle two preceding equations we arrive at 0b , 0c , and hence 0a , 0d . The other condition, a d � , gives no condition on b and c, so we seek a matrix of the

form (we pick 1a , 1d � for simplicity)

1 1 1 0

1 1 0 1b b bc

c c bc�ª º ª º ª º

« » « » « »� � �¬ ¼ ¬ ¼ ¬ ¼.

Hence, in order for the matrix to be the zero matrix, we must have 1bc

� , and hence

11

1c

c

ª º�« »« »

�¬ ¼

,

which gives

1 1 0 01 1

0 01 1c c

c c

ª º ª º� � ª º« » « » « »« » « » ¬ ¼� �¬ ¼ ¬ ¼

.

� Zero Divisors 61. No, AB 0 does not imply that A 0 or B 0 . For example, the product

1 0 0 00 0 0 1ª º ª º« » « »¬ ¼ ¬ ¼

is the zero matrix, but neither factor is itself the zero matrix. � Does Cancellation Work?

62. No. A counterexample is: 0 0 1 2 0 0 0 00 1 0 4 0 1 0 4§ · § · § · § ·

¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸© ¹ © ¹ © ¹ © ¹

since 1 2 0 00 4 0 4§ · § ·

z¨ ¸ ¨ ¸© ¹ © ¹

.


� Taking Matrices Apart

63. (a) 1 2 3

1 5 21 0 32 4 7

ª º« »ª º �¬ ¼ « »« »¬ ¼

A A A A , 243

ª º« » « »« »¬ ¼

xG

where 1A , 2A , and 3A are the three columns of the matrix A and 1 2x , 2 4x , 3 3x are the elements of x

G . We can write

1 1 2 2 3 3

1 5 2 2 1 2 5 4 2 3 1 5 21 0 3 4 1 2 0 4 3 3 2 1 4 0 3 32 4 7 3 2 2 4 4 7 3 2 4 7

.x x x

u � u � uª º ª º ª º ª º ª º ª º« » « » « » « » « » « » � � u � u � u � � �« » « » « » « » « » « »« » « » « » « » « » « »u � u � u¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

� �

Ax

A A A

G

(b) We verify the fact for a 3 3u matrix. The general n nu case follows along the same lines.

11 12 13 1 11 1 12 2 13 3 11 1 12 2 13 3

21 22 23 2 21 1 22 2 23 3 21 1 22 2 23 3

31 32 33 3 31 1 32 2 33 3 31 1 32 2 33 3

1

a a a x a x a x a x a x a x a xa a a x a x a x a x a x a x a xa a a x a x a x a x a x a x a x

x

� �ª º ª º ª º ª º ª º ª º« » « » « » « » « » « » � � � �« » « » « » « » « » « »« » « » « » « » « » « »� �¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

AxG

11 12 13

21 2 22 3 23 1 1 2 2 3 3

31 32 33

a a aa x a x a x x xa a a

ª º ª º ª º« » « » « »� � � �« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

A A A

� Diagonal Matrices

64.

11

22

0 00 000 0 nn

aa

a

ª º« »« » « »« »¬ ¼

A

""

" " ""

,

11

22

0 00 000 0 nn

bb

b

ª º« »« » « »« »¬ ¼

B

""

" " ""

.

By multiplication we get

11 11

22 22

0 00 000 0 nn nn

a ba b

a b

ª º« »« » « »« »¬ ¼

AB

""

" " ""

,

which is a diagonal matrix.


65. By multiplication of the general matrices, and commutativity of resulting individual elements, we have

11 11 11 11

22 22 22 22

0 0 0 00 0 0 00 0

0 0 0 0nn nn nn nn

a b b aa b b a

a b a b

ª º ª º« » « »« » « »« » « » « » « »« » « »« » « »¬ ¼ ¬ ¼

AB BA

" "" "

" " " " " "# #

" "

.

However, it is not true that a diagonal matrix commutes with an arbitrary matrix. � Upper Triangular Matrices 66. (a) Examples are

1 20 3ª º« »¬ ¼

, 1 3 00 0 50 0 2

ª º« »« »« »¬ ¼

,

2 7 9 00 3 8 10 0 4 20 0 0 6

ª º« »« »« »« »¬ ¼

.

(b) By direct computation, it is easy to see that all the entries in the matrix product

11 12 13 11 12 13

22 23 22 23

33 33

0 00 0 0 0

a a a b b ba a b b

a b

ª º ª º« » « » « » « »« » « »¬ ¼ ¬ ¼

AB

below the diagonal are zero. (c) In the general case, if we multiply two upper-triangular matrices, it yields

11 12 13 1 11 12 13 1 11 12 13 1

22 23 2 22 23 2 22 23 2

33 3 33 3 33 3

0 0 00 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

n n n

n n n

n n n

nn nn nn

a a a a b b b b c c c ca a a b b b c c c

a a b b c c

a b c

ª º ª º ª º« » « » « »« » « » « »« » « » « » u « » « » « »« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

AB

" " "" " "" " "

" " " " " " " " " " " " " " ".

We won’t bother to write the general expression for the elements ijc ; the important point is that the entries in the product matrix that lie below the main diagonal are clearly zero.

� Hard Puzzle 67. If

a bc dª º

« »¬ ¼

M

is a square root of

0 10 0ª º

« »¬ ¼

A ,

then

2 0 10 0ª º

« »¬ ¼

M ,


which leads to the condition 2 2a d . Each of the possible cases leads to a contradiction. How-ever for matrix B because

1 0 1 0 1 0

1 1 0 1D Dª º ª º ª º

« » « » « »� �¬ ¼ ¬ ¼ ¬ ¼

for any D, we conclude that

1 0

1Dª º

« »�¬ ¼B

is a square root of the identity matrix for any number D. � Orthogonality

68. 1 1

20 3kª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= 1 � 1 + 2 � k + 3 � 0 = 0

2k = �1

k = 12

�

69. 1

2 04

k

k

ª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= k � 1 + 2 � 0 + k � 4 = 0

5k = 0 k = 0

70. 2

10 2

3

k

k

ª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= k � 1 + 0 � 2 + k2 � 3 = 0

3k2 + k = 0 k(5k + 1) = 0

k = 0, 13

�

71. 2

1 12 1

1k

�ª º ª º« » « »�« » « »« » « »�¬ ¼ ¬ ¼

= 1 � (�1) + 2 � 1 + k2(�1) = 0

1 � k2 = 0 k = r1

� Orthogonality Subsets

72. Set 101

abc

ª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= 0

a � 1 + b � 0 + c � 1 = 0 a + c = 0 c = �a

Orthogonal set = : ,ab a ba

½ª º° °« » �® ¾« »° °« »�¬ ¼¯ ¿

\

73. Set 101

abc

ª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= 0 to get c = �a

Set 210

abc

ª º ª º« » « »�« » « »« » « »¬ ¼ ¬ ¼

= 0

2a + b � 1 + c � 0 = 0 2a + b = 0 b = �2a

Orthogonal set = 2 :aa aa

½ª º° °« »� �® ¾« »° °« »�¬ ¼¯ ¿

\


87. > @ > @2, 1, 2 1, 0, 1 0� x � so in 3-space these vectors form a right angle, since dot product is zero. � Properties of Scalar Products We let > @1 na a a

G ! , > @1 nb b bG

! , and > @1 nc c cG ! for simplicity.

88. True. > @ > @ > @ > @1 1 1 1 1 1 n n n n n na a b b a b a b b a b ax x xa b b aG GG G" " " " .

89. False. Neither � �x xa b c

GG G nor � �x xa b c

GG G. Invalid operation, since problem asks for the scalar

product of a vector and a scalar, which is not defined. 90. True.

� � > @ > @

> @ > @ � �1 1 1 1 1 1

1 1

n n n n n n

n n

k ka ka b b ka b ka b a kb a kb

a a kb kb k

x x � � � �

x x

a b

a b

GG " " " "GG" "

91. True.

� � > @ > @ � � � �

� � � �1 1 1 1 1 1

1 1 1 1

n n n n n n

n n n n

a a b c b c a b c a b c

a b a b a c a c

x � x � � � � � �

� � � � � x � x

a b c

a b a c

GG G " " "GG G G" "

� Directed Graphs

92. (a)

0 1 1 0 10 0 1 0 00 0 0 0 10 0 0 0 00 0 1 1 0

ª º« »« »« » « »« »« »¬ ¼

A

(b) 2

0 0 2 1 10 0 0 0 10 0 1 1 00 0 0 0 00 0 0 0 1

ª º« »« »« » « »« »« »¬ ¼

A

The ijth entry in 2A gives the number of paths of length 2 from node i to node j. � Tournament Play 93. The tournament graph had adjacency matrix

0 1 1 0 10 0 0 1 10 1 0 0 11 0 1 0 10 0 0 0 0

ª º« »« »« » « »« »« »¬ ¼

T .

Ranking players by the number of games won means summing the elements of each row of T, which in this case gives two ties: 1 and 4, 2 and 3, 5. Players 1 and 4 have each won 3 games. Players 2 and 3 have each won 2 games. Player 5 has won none.

equilibrium solution yt { a is stable. b appm 2360 ... the equilibrium solution yt {0 is unstable,...

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