math 1zc3 custom courseware 2012

8/11/2019 Math 1ZC3 Custom Courseware 2012

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M c M a s t e r M a t h 1 Z C 3 C u s t o m

C o u r s e w a r e W i n t e r 2 0 1 2

- - R i p p e d f r o m G o o g l e B o o k s - -


2/121

C o n t e n t s : C o n t e n t s : C o n t e n t s : C o n t e n t s :

A d v a n c e d E n g i n e e r i n g M a t h e m a t i c s 3 r d e d . , Z i l l , D . G . & C u l l e n , M . R .

7 . 6 V e c t o r S p a c e s

7 . 7 G r a m - S c h m i d t O r t h o g o n a l i z a t i o n P r o c e s s 8 . 1 M a t r i x A l g e b r a

8 . 2 S y s t e m s o f L i n e a r A l g e b r a i c E q u a t i o n s 8 . 3 R a n k o f M a t r i x

8 . 4 D e t e r m i n a n t s

8 . 5 P r o p e r t i e s o f D e t e r m i n a n t s 8 . 6 I n v e r s e o f a m a t r i x

8 . 7 C r a m e r ' s R u l e C h a p t e r 8 R e v i e w E x e r c i s e s

N u m e r i c a l M a t h e m a t i c s , G r a s s e l l i , M . & P e l i n o v s k y , D ." E l e m e n t s o f t h e L a b o r a t o r y " - I n t r o d u c t i o n t o M a t l a b

S o l u t i o n s t o A d v a n c e d E n g i n e e r i n g M a t h e m a t i c s ( i n t h i s o r d e r ) :

7 . 6 C h a p t e r 7 R e v i e w 7 . 7 ( b r i e f - a l l f u l l s o l u t i o n s n o t a v a i l a b l e )

A l l o f C h a p t e r 8


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Exercises 7.6

Exercises 7.6

1. Not a vector space. Axiom (vi) is not satisfied. 2. Not a vector space. Axiom (i) is not satisfied.

3. Not a vector space. Axiom (x) is not satisfied. 4. A vector space

5. A vector space 6. A vector space

7. Not a vector space. Axiom (ii) is not satisfied. 8. A vector space

9. A vector space 10. Not a vector space. Axiom (i) is not satisfied.

11. A subspace 12. Not a subspace. Axiom (i) is not satisfied.

13. Not a subspace. Axiom (ii) is not satisfied. 14. A subspace

15. A subspace 16. A subspace

17. A subspace 18. A subspace

19. Not a subspace. Neither axioms (i) nor(ii) are satisfied.

20. A subspace

21. Let (x1, y1, z1) and (x2, y2, z2) be inS. Then

(x1, y1, z1) + (x2, y2, z2) = (at1, bt1, ct1) + (at2, bt2, ct2) = (a(t1+t2), b(t1+t2), c(t1+t2))

is in S. Also, for (x,y,z) inSthen k(x,y,z) = (kx,ky,kz) = (a(kt), b(kt), c(kt)) is also in S.

22. Let (x1, y1, z1) and (x2, y2, z2) be in S. Then ax1 + by1 + cz1 = 0 and ax2 + by2 + cz2 = 0. Adding gives

a(x1+x2) + b(y1+ y2) + c(z1+ z2) = 0 and so (x1, y1, z1) + (x2, y2, z2) = (x1+x2, y1+ y2, z1+ z2) is inS. Also,

for (x,y,z) thenax + by+ cz= 0 implies k(ax + by+ cz) = k 0 = 0 and a(kx) + b(ky) + c(kz) = 0. this meansk(x,y,z) = (kx,ky,kz) is inS.

23. (a) c1u1+ c2u2+ c3u3 = 0 if and only ifc1+ c2+ c3 = 0, c2+ c3 = 0,c3 = 0. The only solution of this system

isc1 = 0, c2 = 0, c3 = 0.

(b) Solving the system c1 +c2 +c3 = 3, c2 +c3 =4, c3 = 8 gives c1 = 7, c2 =12, c3 = 8. Thusa= 7u112u2+ 8u3.

24. (a) The assumption c1p1+c2p2 = 0 is equivalent to (c1+c2)x+ (c1c2) = 0. Thus c1+c2 = 0, c1c2 = 0.The only solution of this system is c1 = 0, c2 = 0.

(b) Solving the system c1+c2 = 5, c1c2 = 2 gives c1 = 7

2, c2 = 3

2. Thus p(x) = 7

2p1(x) + 3

2p2(x)25. Linearly dependent since6, 12=3

24,8

26. Linearly dependent since 21, 1+ 30, 1+ (1)2, 5=0, 027. Linearly independent

28. Linearly dependent since for all x (1)1 + (2)(x+ 1) + (1)(x+ 1)2 + (1)x2 = 0.29. f is discontinuous at x =1 and atx =3.

30. (x, sinx) =

2

0

x sinxdx= (x cosx+ sin x)20

=2

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Exercises 7.6

31. x2 =

2

0

x2 dx=1

3x32

0

=8

33 and sox= 2

23

3 . Now

sinx2 =

2

0

sin2 x dx=1

2

2

0

(1cos 2x) dx= 12

x 1

2sin 2x

2

0

=

and so sinx=.32. A basis could be 1, x, ex cos3x, ex sin3x.

33. We need to show that Span{x1, x2, . . . , xn} is closed under vector addition and scalar multiplication. Supposeuand v are in Span{x1, x2, . . . , xn}. Then u = a1x1+a2x2+ +anxn andv = b1x1+b2x2+ +bnxn, sothat

u + v= (a1+b1)x1+ (a2+b2)x2+ + (an+bn)xn,which is in Span{x1, x2, . . . , xn}. Also, for any real number k ,

ku= k(a1x1+a2x2+ +anxn) = ka1x1+ka2x2+ +kanxn,which is in Span

{x1, x2, . . . , xn

}. Thus, Span

{x1, x2, . . . , xn

}is a subspace ofV.

Chapter 7 Review Exercises

1. True

2. False; the points must be non-collinear.

3. False; since a normal to the plane is2, 3,4 which is not a multiple of the direction vector5,2, 1 of theline.

4. True 5. True 6. True 7. True 8. True 9. True

10. True; since a b and c dare both normal to the plane and hence parallel (unless a b= 0 or c d= 0.)

11. 9i + 2j + 2k 12. orthogonal

13. 5(k j) =5(i) = 5i 14. i (i j) = i k= 0

15.

(12)2 + 42 + 62 = 1416. (120)i (20)j + (80)k=21i + 2j + 8k17. 6i +j 7k18. The coordinates of (1,2,10) satisfy the given equation.

19. Writing the line in parametric form, we havex = 1 + t,y =2 + 3t,z =1 + 2t. Substituting into the equationof the plane yields (1+ t)+2(2 + 3t) (1 + 2t) = 13 ort = 3. Thus, the point of intersection isx = 1+ 3 = 4,y=2 + 3(3) = 7, z =1 + 2(3) = 5, or (4, 7, 5).

20. |a|=

42 + 32 + (5)2 = 52 ; u= 15

2(4i + 3j 5k) = 4

5

2i 3

5

2j +

12

k

21. x22 = 3, x2 = 5; y21 = 5, y2 = 6; z27 =4, z2 = 3; P2 = (5, 6, 3)22. (5, 1/2, 5/2)

23. (7.2)(10)cos 135 =362

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8 Matrices

Exercises 8.1

1. 2 4 2. 3 2 3. 3 3 4. 1 3 5. 3 4

6. 8 1 7. Not equal 8. Not equal 9. Not equal 10. Not equal11. Solvingx = y 2, y = 3x 2 we obtain x = 2, y = 4.12. Solvingx2 = 9, y = 4x we obtain x = 3, y = 12 andx = 3, t = 12.13. c23 = 2(0) 3(3) = 9; c12 = 2(3) 3(2) = 1214. c23 = 2(1) 3(0) = 2; c12 = 2(1) 3(0) = 2

15. (a) A + B= 4 2 5 + 66 + 8 9 10= 2 112 1

(b) B A=2 4 6 5

8 + 6 10 9

=

6 114 19

(c) 2A + 3B=

8 10

12 18

+

6 1824 30

=

2 28

12 12

16. (a) A B=

2 3 0 + 1

4 0 1 27 + 4 3 + 2

=

5 1

4 111 5

(b) B A= 3 + 2 1 00 4 2 14 7 2 3

= 5 14 111 5

(c) 2(A + B) = 2

1 14 3

3 1

=

2 28 6

6 2

17. (a) AB=

2 9 12 65 + 12 30 + 8

=

11 617 22

(b) BA= 2 30 3 + 24

6 10 9 + 8 =

32 27

4 1 (c) A2 =

4 + 15 6 1210 20 15 + 16

=

19 1830 31

(d) B2 =

1 + 18 6 + 123 + 6 18 + 4

=

19 6

3 22

18. (a) AB=

4 + 4 6 12 3 + 820 + 10 30 30 15 + 2032 + 12 48 36 24 + 24

=

0 6 510 0 520 12 0

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Exercises 8.1

(b) BA=

4 + 30 24 16 + 60 361 15 + 16 4 30 + 24

=

2 8

2 2

19. (a) BC=

9 24

3 8

(b) A(BC) =

1 22 4

9 24

3 8

=

3 8

6 16

(c) C(BA) =

0 2

3 4

0 0

0 0

=

0 0

0 0

(d) A(B + C) =

1 22 4

6 5

5 5

=

4 58 10

20. (a) AB= [ 5 6 7 ]

3

4

1

= (16)

(b) BA=

3

4

1

[ 5 6 7 ] =

15 18 2120 24 285 6 7

(c) (BA)C=

15 18 2120 24 285 6 7

1 2 4

0 1 13 2 1

=

78 54 99

104 72 132

26 18 33

(d) SinceAB is 1 1 and C is 3 3 the product (AB)C is not defined.

21. (a) AT

A= [ 4 8 10 ]4

810

= (180)

(b) BTB=

2

4

5

[ 2 4 5 ] =

4 8 10

8 16 20

10 20 25

(c) A + BT =

4

8

10

+

2

4

5

=

6

12

5

22. (a) A + BT

= 1 2

2 4

+2 53 7 = 1 75 11

(b) 2AT BT =

2 4

4 8

2 5

3 7

=

4 11 1

(c) AT(AB) =

1 2

2 4

3 13 3

=

3 76 14

23. (a) (AB)T =

7 10

38 75

T=

7 38

10 75

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97/121

Exercises 8.1

45. (a) MY

x

y

z

=

cos sin 0

sin cos 00 0 1

x

y

z

=

x cos + y sin

x sin + y cos z

=

xY

yY

zY

(b) MR=

cos 0

sin

0 1 0

sin 0 cos

; MP 1 0 0

0 cos sin

0 sin cos

(c) MP

1

1

1

=

1 0 0

0 cos 30 sin30

0 sin30 cos 30

1

1

1

=

1 0 0

03

2

1

2

0 12

3

2

1

1

1

=

11

2(

3 + 1)1

2(

3 1)

MRMP

1

1

1

=

cos45 0 sin450 1 0

sin45 0 cos 45

11

2(

3 + 1)1

2(

3 1)

=

2

2 0

2

2

0 1 02

2 0

2

2

11

2(

3 + 1)1

2(

3 1)

=

1

4(3

2

6 )

12 (3 + 1)

1

4(

2 +

6 )

MYMRMP

1

1

1

=

cos60 sin60 0

sin60 cos60 00 0 1

1

4(3

26 )1

2(

3 + 1)1

4(

2 +

6 )

=

1

2

3

2 0

3

2

1

2 0

0 0 1

1

4(3

2

6 )1

2(

3 + 1)1

4(

2 +

6 )

=

1

8(3

2

6 + 6 + 2

3 )1

8(36 + 32 + 23 + 2)

1

4(

2 +

6 )

46. (a) LU=1 0

12 1

2 2

0 3

= 2

2

1 2

= A

(b) LU=

1 02

3 1

6 2

0 13

=

6 2

4 1

= A

(c) LU=

1 0 0

0 1 0

2 10 1

1 2 10 1 2

0 0 21

=

1 2 10 1 2

2 6 1

=A

(d) LU=

1 0 0

3 1 0

1 1 1

1 1 1

0 2 10 0 1

=

1 1 1

3 1 2

1 1 1

=A

47. (a) AB=

A11 A12

A21 A22

B1

B2

=

A11B1+ A12B2

A21B1+ A22B2

=

17 43

3 75

14 51

since

A11B1+ A12B2 =

13 25

9 49

+

4 18

12 26

=

17 43

3 75

and

A21B1+ A22B2= [24 34] + [ 10 17] = [14 51 ] .

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Exercises 8.2

(b) It is easier to enter smaller strings of numbers and the chance of error is decreased. Also, if the large matrix

has submatrices consisting of all zeros or diagonal matrices, these are easily entered without listing all of

the entries.

Exercises 8.2

1.

1 1 114 3 5

4R1+R2

1 1 110 7 49

1

7R2

1 1 110 1 7

R3+R1

1 0 4

0 1 7

The solution is x1 = 4, x2 = 7.

2.

3 2 41 1 2

R12

1 1 23 2 4

3R1+R2

1 1 20 1 10

R2+R1

1 0 8

0 1 10


3. 9 3 52

1

1 1

9R1

1 13 5

9

2 1 12R1+R2

1 13 5

9

0

1

3

1

9

3R21 1

3 5

9

0 1

1

3

13R2+R1

1 0 2

3

0 1 13

The solution is x1 = 23 , x2 = 13 .

4.

10 15 1

3 2 1

1

10R1

1 3

2

1

10

3 2 1

3R1+R2

1 3

2

1

10

0 52 13

10

2

5R2

1 3

2

1

10

0 1 1325

32R2+R1

1 0 17

25

0 1 1325

The solution is x1 = 17

25 , x2 = 13

25 .

5.

1 1 1 32 3 5 7

1 2 3 11

2R1+R2

R1+R3

1 1 1 30 5 7 13

0 1 4 8

15R2

1 1 1 30 1 7

5

13

5

0 1 4 8

R2+R1R2+R3

1 0 25 2

5

0 1 75

13

5

0 0 275 27

5

527R3

1 0 25 2

5

0 1 75

13

5

0 0 1 1

25R3+R1 7

5R3+R2

1 0 0 0

0 1 0 4

0 0 1 1

The solution is x1 = 0, x2 = 4, x3 = 1.

6.

1 2 1 0

2 1 2 91 1 1 3

2R1+R2R1+R3 1 2 1 0

0 3 4 90 3 2 3

13R21 2 1 0

0 1 4

3 30 3 2 3

2R2+R13R2+R3

1 0 53

6

0 1 43 3

0 0 2 6

12R3

1 0 53

6

0 1 43 3

0 0 1 3

53R3+R1

4

3R3+R2

1 0 0 1

0 1 0 1

0 0 1 3


7.

1 1 1 0

1 1 3 0

R1+R2

1 1 1 0

0 0 2 0

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99/121

Exercises 8.2

Sincex3 = 0, setting x2 = t we obtain x1 = t, x2 = t, x3 = 0.

8.

1 2 4 95 1 2 1

5R1+R2

1 2 4 90 11 22 44

1

11R2

1 2 4 90 1 2 4

2R2+R1

1 0 0 1

0 1 2 4

Ifx3 = t, the solution is x1 = 1, x2 = 4 + 2t, x3 = t

9.

1 1 1 81 1 1 3

1 1 1 4

row

operations

1 1 1 80 0 2 50 0 0 12

Since the bottom row implies 0 = 12, the system is inconsistent.

10.

3 1 4

4 3 32 1 11

row

operations

1 13

4

3

0 1 50 0 0


11.

2 2 0 0

2 1 1 03 0 1 0

row

operations

1 1 0 0

0 1 13

0

0 0 1 0

The solution is x1 = x2 = x3 = 0.

12.

1 1 2 02 4 5 0

6 0 3 0

row

operations

1 1 2 00 1 3

2 0

0 0 0 0

The solution is x1 = 1

2t, x2 = 32 t, x3 = t.

13.

1 2 2 21 1 1 01 3 1 0

rowoperations

1 2 2 20 1 1 20 0 1 4

The solution is x1 = 2, x2 2, x3 = 4.

14.

1 2 1 23 1 2 52 1 1 1

row

operations

1 2 1 20 1 1

5 1

5

0 0 0 2


15.

1 1 1 3

1 1 1 13 1 1 5

rowoperations

1 1 1 3

0 1 1 2

0 0 0 0

Ifx3 = t the solution is x1 = 1, x2 = 2 t, x3 = t.

16.

1 1 2 13 2 1 7

2 3 1 8

row

operations

1 1 2 10 1 1 2

0 0 0 0

Ifx3 = t the solution is x1 = 1 + t, x2 = 2 t, x3 = t.

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100/121

Exercises 8.2

17.

1 0 1 1 10 2 1 1 3

1 1 0 1 11 1 1 1 2

rowoperations

1 0 1 1 10 1 1

2

1

2

3

2

0 0 1 5 10 0 0 1 0

The solution is x1 = 0, x2 = 1, x3 = 1, x4 = 0.

18.

2 1 1 0 3

3 1 1 1 4

1 2 2 3 3

4 5 2 1 16

rowoperations

1 12

1

2 0 3

2

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

The solution is x1 = 1, x2 = 2, x3 = 1, x4 = 0.

19.

1 3 5 1 10 1 1 1 41 2 5 4 21 4 6 2 6

rowoperations

1 3 5 1 10 1 1 1 40 0 1 4 10 0 0 0 1


20.

1 2 0 1 0

4 9 1 12 0

3 9 6 21 0

1 3 1 9 0

rowoperations

1 2 0 1 0

0 1 1 8 0

0 0 1 2 00 0 0 0 0

Ifx4 = t the solution is x1 = 19t, x2 = 10t, x3 = 2t, x4 = t.

21.

1 1 1 4.280

0.2 0.1 0.5 1.978

4.1 0.3 0.12 1.686

row

operations

1 1 1 4.28

0 1 2.333 9.447

0 0 1 4.1

The solution is x1 = 0.3, x2 = 0.12,x3 = 4.1.

22.

2.5 1.4 4.5 2.6170

1.35 0.95 1.2 0.7545

2.7 3.05 1.44 1.4292

row

operations

1 0.56 1.8 1.0468

0 1 6.3402 3.39530 0 1 0.28

The solution is x1 = 1.45, x2 = 1.62, x3 = 0.28.

23. From x1Na + x2H2O x3NaOH +x4H2 we obtain the system x1 = x3, 2x2 = x3 + 2x4, x2 = x3. We seethat x1 = x2 = x3, so the second equation becomes 2x1 = x1+ 2x4 or x1 = 2x4. A solution of the system is

x1 = x2 = x3 = 2t, x4 = t. Letting t = 1 we obtain the balanced equation 2Na + 2H2O 2NaOH + H2.24. Fromx1KClO3 x2KCl+ x3O2 we obtain the systemx1 = x2,x1 = x2, 3x1 = 2x3. Lettingx3 = t we see that

a solution of the system is x1 = x2 = 2

3t, x3 = t. Takingt = 3 we obtain the balanced equation

2KClO3 2KCl + 3O2.

25. From x1Fe3O4+ x2Cx3Fe +x4CO we obtain the system 3x1 = x3, 4x1 =x4, x2 = x4. Lettingx1 = t wesee that x3 = 3tand x2 = x4 = 4t. Taking t = 1 we obtain the balanced equation

Fe3O4+ 4C 3Fe + 4CO.

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Exercises 8.2

26. From x1C5H8+x2O2 x3CO2+x4H2O we obtain the system 5x1 = x3, 8x1 = 2x4, 2x2 = 2x3+x4. Lettingx1 = t we see that x3 = 5t, x4 = 4t, andx2 = 7t. Taking t = 1 we obtain the balanced equation

C5H8+ 7O2 5CO2+ 4H2O.

27. From x1Cu + x2HNO3 x3Cu(NO3)2+ x4H2O + x5NO we obtain the systemx1 = 3, x2 = 2x4, x2 = 2x3+ x5, 3x2 = 6x3+ x4+ x5.

Lettingx4 = t we see that x2 = 2tand

2t= 2x3+ x5

6t= 6x3+ t + x5or

2x3+ x5 = 2t

6x3+ x5 = 5t.

Thenx3 = 3

4tand x5 =

1

2t. Finally, x1 = x3 =

3

4t. Takingt = 4 we obtain the balanced equation

3Cu + 8HNO3 3Cu(NO3)2+ 4H2O + 2NO.

28. From x1Ca3(PO4)2+ x2H3PO4 x3Ca(H2PO4)2 we obtain the system3x1 = x3, 2x1+ x2 = 2x3, 8x1+ 4x2 = 8x3, 3x2= 4x3.

Lettingx1 = t we see from the first equation that x3 = 3t and from the fourth equation that x2 = 4t. These

choices also satisfy the second and third equations. Taking t = 1 we obtain the balanced equation

Ca3(PO4)2+ 4H3PO4 3Ca(H2PO4)2.

29. The system of equations is

i1+ i2 i3 = 010 3i1+ 5i3 = 027

6i2

5i3 = 0

or

i1+ i2 i3 = 03i1 5i3 = 106i2+ 5i3 = 27

Gaussian elimination gives1 1 1 0

3 0 5 100 6 5 27

row

operations

1 1 1 00 1 8/3 10/30 0 1 1/3

.

The solution is i1 =35

9 , i2 =

38

9 , i3 =

1

3.

30. The system of equations isi1 i2 i3 = 0

52 i1 5i2 = 0

10i3+ 5i2 = 0

or

i1 i2 i3 = 0i1+ 5i2 = 52

5i2 10i3 = 0Gaussian elimination gives

1 1 1 01 5 0 52

0 5 10 0

row

operations

1 1 1 00 1 1/6 26/3

0 0 1 4

.

The solution is i1 = 12, i2 = 8, i3 = 4.

31. Interchange row 1 and row in I3. 32. Multiply row 3 by c in I3.

320


102/121

Exercises 8.2

33. Addc times row 2 to row 3 in I3. 34. Add row 4 to row 1 in I4.

35. EA=

a21 a22 a23

a11 a12 a13

a31 a32 a33

36. EA=

a11 a12 a13

a21 a22 a23

ca31 ca32 ca33

37. EA=

a11 a12 a13

a21 a22 a23

ca21+ a31 ca22+ a32 ca23+ a33

38. E1E2A= E1

a11 a12 a13

a21 a22 a23

ca21+ a31 ca22+ a32 ca23+ a33

=

a21 a22 a23

a11 a12 a13

ca21+ a31 ca22+ a32 ca23+ a33

39. The system is equivalent to 1 01

2 1

2 20 3

X=

2

6

.

Letting

Y=

y1

y2

=

2 20 3

X

we have 1 01

2 1

y1

y2

=

2

6

.

This implies y1= 2 and 1

2y1+ y2 = 1 + y2 = 6 ory2 = 5. Then

2 20 3

x1

x2

=

2

5

,

which implies 3x2

= 5 or x2

=

5

3 and 2x1 2x

2

= 2x1

10

3 = 2 or x1

=

8

3. The solution is X =83 ,

5

3

.40. The system is equivalent to

1 02

3 1

6 2

0 13

X=

1

1

.

Letting

Y=

y1

y2

=

6 2

0 13

X

we have 1 02

3 1

y1

y2

=

1

1

.

This implies y1= 1 and

2

3y1+ y2 =

2

3 + y2 = 1 or y2 = 5

3 . Then6 2

0 13

x1

x2

=

1

53

,

which implies13

x2 = 53 or x2 = 5 and 6x1+ 2x2 = 6x1+ 10 = 1 or x1 = 32. The solution is X =3

2, 5

.

41. The system is equivalent to

1 0 0

0 1 0

2 10 1

1 2 10 1 2

0 0 21

X=

2

11

.

321


103/121

Exercises 8.2

Letting

Y=

y1

y2

y3

=

1 2 10 1 2

0 0 21

X

we have 1 0 0

0 1 0

2 10 1

y1

y2

y3

=

2

11

.

This implies y1= 2, y2 = 1, and 2y1+ 10y2+ y3 = 4 10 + y3 = 1 ory3 = 7. Then

1 2 10 1 2

0 0 21

x1

x2

x3

=

2

17

,

which implies 21x3 = 7 orx3 = 13, x2 + 2x3 = x2 23 = 1 orx2= 13, andx12x2 + x3 = x1 + 23 13 = 2orx1 =

5

3. The solution is X = 53 ,

1

3,

1

3.42. The system is equivalent to

1 0 0

3 1 0

1 1 1

1 1 1

0 2 10 0 1

X=

0

1

4

.

Letting

Y=

y1

y2

y3

=

1 1 1

0 2 10 0 1

X

we have

1 0 0

3 1 0

1 1 1

y1

y2

y3

= 0

1

4

.This implies y1= 0, 3y1+ y2 = y2 = 1, and y1+ y2+ y3 = 0 + 1 + y3 = 4 or y3 = 3. Then

1 1 1

0 2 10 0 1

x1

x2

x3

=

0

1

3

,

which implies x3 = 3,2x2 x3 = 2x2 3 = 1 orx2 = 2, and x1+x2+x3 =x1 2 + 3 = 0 or x1 = 1.The solution is X = (1,2, 3).

43. Using the Solve function in Mathematicawe find

(a) x1 = 0.0717393 1.43084c, x2 = 0.332591 + 0.855709c, x3 = c, wherec is any real number(b) x1 = c/3, x2 = 5c/6, x3= c, where c is any real number

(c) x1 = 3.76993,x2= 1.09071,x3 = 4.50461,x4 = 3.12221(d) x1 =

8

3 7

3b + 2

3c, x2 =

2

3 1

3b 1

3c, x3 = 3, x4 = b, x5 = c, whereb and c are any real numbers.

322


104/121

Exercises 8.3

Exercises 8.3

1.

3 11 3

row

operations

1 3

0 1

; The rank is 2.

2.

2 20 0

row

operations

1 10 0

; The rank is 1.

3.

2 1 3

6 3 9

1 12 3

2

row

operations

1 12

3

2

0 0 0

0 0 0

; The rank is 1.

4.

1 1 2

1 2 41 0 3

row

operations

1 1 2

0 1 5

0 0 1

; The rank is 3.

5.

1 1 1

1 0 4

1 4 1

rowoperations

1 1 1

0 1 30 0 1

; The rank is 3.

6.

3 1 2 06 2 4 5

row

operations

1 1

3

2

3 0

0 1 0 54

; The rank is 2.

7.

1 23 67 14 5

rowoperations

1 20 1

0 0

0 0

; The rank is 2.

8.

1 2 3 41 4 6 8

0 1 0 0

2 5 6 8

rowoperations

1 2 3 40 1 0 0

0 0 1 43

0 0 0 0

; The rank is 3.

9.

0 2 4 2 2

4 1 0 5 1

2 1 23

3 13

6 6 6 12 0

rowoperations

1 12

1

3

3

2

1

6

0 1 43

1 13

0 0 1 0 2

0 0 0 0 0

; The rank is 3.

10.

1 2 1 8 1 1 1 60 0 1 3 1 1 1 50 0 1 3 1 2 10 80 0 0 0 0 1 1 3

1 2 1 8 1 1 2 6

row

operations

1 2 1 8 1 1 1 60 0 1 3 1 1 1 50 0 0 0 0 1 9 3

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0

; The rank is 4.

11.

1 2 3

1 0 1

1 1 5

row

operations

1 2 3

0 1 1

0 0 1

;

Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.

323


105/121

Exercises 8.3

12.

2 6 3

1 1 43 2 1

2 5 4

rowoperations

1 1 40 1 5

8

0 0 1

0 0 0

Since the rank of the matrix is 3 and there are 4 vectors, the vectors are linearly dependent.

13.

1 1 3 11 1 4 21 1 5 7

row

operations

1 1 3 10 0 1 3

0 0 0 1


14.

2 1 1 5

2 2 1 1

3 1 6 11 1 1 1

rowoperations

1 1 1 10 1 1 70 0 1 30 0 0 1


15. Since the number of unknowns is n = 8 and the rank of the coefficient matrix is r = 3, the solution of the

system hasn r= 5 parameters.16. (a) The maximum possible rank ofA is the number of rows inA, which is 4.

(b) The system is inconsistent if rank(A)< rank(A/B) = 2 and consistent if rank(A) = rank(A/B) = 2.

(c) The system has n= 6 unknowns and the rank ofA is r = 3, so the solution of the system has n r = 3parameters.

17. Since 2v1+ 3v2 v3 = 0 we conclude that v1, v2, and v3 are linearly dependent. Thus, the rank ofA is atmost 2.

18. Since the rank ofA is r = 3 and the number of equations is n = 6, the solution of the system has n r = 3parameters. Thus, the solution of the system is not unique.

19. The system consists of 4 equations, so the rank of the coefficient matrix is at most 4, and the maximum number

of linearly independent rows is 4. However, the maximum number of linearly independent columns is the same

as the maximum number of linearly independent rows. Thus, the coefficient matrix has at most 4 linearly

independent columns. Since there are 5 column vectors, they must be linearly dependent.

20. Using theRowReducein Mathematicawe find that the reduced row-echelon form of the augmented matrix is

1 0 0 0 0 8342215

261443

0 1 0 0 0 18182215

282

443

0 0 1 0 0 13

443 6

443

0 0 0 1 0 42142215

130443

0 0 0 0 1 60792215

677

443

.

We conclude that the system is consistent and the solution is x1 = 226443 8342215 c, x2 = 282443 18182215 c,x3 = 6443 13443c, x4 = 130443 42142215c, x5 = 677433 + 60792215 c, x6 = c.

324


106/121

Exercises 8.4

Exercises 8.4

1. M12 =

1 2

2 5

= 9 2. M32 =

2 4

1 2

3. C13 = (1)1+3 1 12 3

= 1 4. C22 = (1)2+2 2 42 5

= 18

5. M33 =

0 2 0

1 2 3

1 1 2

= 2 6. M41 =

2 4 0

2 2 31 0 1

= 24

7. C34 = (1)3+4

0 2 4

1 2 21 1 1

= 10 8. C23 = (1)2+3

0 2 0

5 1 11 1 2

= 22

9. 7 10. 2 11. 17 12. 1/213. (1 )(2 ) 6 = 2 3 4 14. (3 )(5 ) 8 = 2 2 23

15.

0 2 0

3 0 1

0 5 8

= 3 2 05 8

= 48 16.

5 0 0

0 3 00 0 2

= 53 00 2

= 5(3)(2) = 30

17.

3 0 2

2 7 1

2 6 4

= 3

7 16 4 + 2

2 72 6 = 3(22) + 2(2) = 62

18.

1

1

1

2 2 21 1 9

= 2 21 9 2 1 11 9 + 1 12 2 = 20 2(8) + 4 = 40

19.

4 5 3

1 2 3

1 2 3

= 4 2 32 3

5 1 31 3

+ 3 1 21 2

= 0

20.

1

4 6 0

1

3 8 0

1

2 9 0

= 0, expanding along the third column.

21.2 1 43 6 13 4 8

= 2 6 14 8 + 3

1 4

4 8 3 1 46 1 = 2(44) + 3(24) 3(25) = 85

22.

3 5 1

1 2 57 4 10

= 3 2 54 10

51 57 10

+1 27 4

= 3(40) 5(45) + (10) = 335

23.

1 1 1

x y z

2 3 4

= y z3 4

x z2 4

+x y2 3

= (4y 3z) (4x 2z) + (3x 2y) = x + 2y z

325


107/121

Exercises 8.4

24.

1 1 1

x y z

2 + x 3 + y 4 + z

= y z3 + y 4 + z

x z2 + x 4 + z

+ x y2 + x 3 + y

= (4y+ yz 3z yz) (4x + xz 2z xz) + (3x + xy 2y xy) = x + 2y z

25.

1 1 3 01 5 3 2

1 2 1 04 8 0 0

= 2

1 1 31 2 14 8 0

= 2(4) 1 32 1

2(8) 1 31 1

= 8(5) 16(4) = 104

26.

2 1 2 10 5 0 4

1 6 1 0

5 1 1 1

= 5

2 2 11 1 0

5 1 1

+ 4

2 1 21 6 1

5 1 1

= 5(0) + 4(80) = 320

27. Expanding along the first column in the original matrix and each succeeding minor, we obtain 3(1)(2)(4)(2) = 48.

28. Expanding along the bottom row we obtain

1

2 0 0 21 6 0 5

1 2 1 12 1 2 3

+

2 2 0 0

1 1 6 0

1 0 2 12 0 1 2

= 1(48) + 0 = 48.

29. Solving2 2 15 20 = 2 2 35 = ( 7)( + 5) = 0 we obtain = 7 and5.30. Solving3 + 32 2= ( 2)( 1) = 0 we obtain = 0, 1, and 2.

Exercises 8.5

1. Theorem 8.11 2. Theorem 8.14

3. Theorem 8.14 4. Theorem 8.12 and 8.11

5. Theorem 8.12 (twice) 6. Theorem 8.11 (twice)

7. Theorem 8.10 8. Theorem 8.12 and 8.9

9. Theorem 8.8 10. Theorem 8.11 (twice)

11. det A= 5 12. det B= 2(3)(5) = 30

13. det C= 5 14. det D= 5

15. det A= 6(23

)(4)(5) = 80 16. det B= a13a22a31

17. det C= (5)(7)(3) = 105 18. det D= 4(7)(2) = 56

19. det A= 14 = det AT 20. det A= 96 = det T

326


108/121

Exercises 8.5

21. det AB=

0 2 2

10 7 23

8 4 16

= 80 = 20(4) = det A det B22. From Problem 21, (det A)2 = det A2 = det I= 1, so det A=

1.

23. Using Theorems 8.14, 8.12, and 8.9, det A=

a 1 2

b 1 2

c 1 2

= 2

a 1 1

b 1 1

c 1 1

= 0.24. Using Theorems 8.14 and 8.9,

det A=

1 1 1

x y z

x + y+ z x + y+ z x + y+ z

= (x + y+ z)

1 1 1

x y z

1 1 1

= 0.

25.

1 1 5

4 3 6

0 1 1

=

1 1 5

0 1 140 1 1

=

1 1 5

0 1 140 0 15

= 1(1)(15) = 15

26.

2 4 5

4 2 0

8 7 2

=

2 4 5

0 6 100 9 22

= 2

2 4 5

0 3 5

0 9 22

= 2

2 4 5

0 3 5

0 0 7

= 2(2)(3)(7) = 84

27.

1 2 34 5 29 9 6

=

1 2 30 3 10

0 9 33

=

1 2 30 3 10

0 0 3

= 1(3)(3) = 9

28.

2 2 6

5 0 1

1 2 2

=

1 2 25 0 1

2 2 6

=

1 2 20 10 90 2 2

=

1 2 20 10 90 0 19

5

= 1(10)(19

5) = 38

29.

1 2 2 12 1 2 33 4 8 13 11 12 2

=

1 2 2 10 5 6 10 10 14 20 5 6 1

=

1 2 2 10 5 6 10 0 2 40 0 0 0

= 1(5)(2)(0) = 0

30.

0 1 4 5

2 5 0 1

1 2 2 0

3 1 3 2

=

1 2 2 0

2 5 0 1

0 1 4 5

3 1 3 2

=

1 2 2 0

0 1

4 1

0 1 4 5

0 5 3 2

=

1 2 2 0

0 1

4 1

0 0 8 4

0 0 23 7

=

1 2 2 0

0 1

4 1

0 0 8 4

0 0 23 372

= (1)(1)(8)(37

2) = 148

31.

1 2 3 4

1 3 5 7

2 3 6 7

1 5 8 20

=

1 2 3 4

0 1 2 3

0 1 0 10 3 5 16

=

1 2 3 4

0 1 2 3

0 0 2 2

0 0 1 7

=

1 2 3 4

0 1 2 3

0 0 2 2

0 0 0 8

= 1(1)(2)(8) = 16

327


109/121

Exercises 8.5

32.

2 9 1 8

1 3 7 4

0 1 6 5

3 1 4 2

=

1 3 7 4

2 9 1 8

0 1 6 5

3 1 4 2

=

1 3 7 4

0 3 13 00 1 6 5

0 8 17 10

=

1 3 7 4

0 1 6 5

0 3 13 00 8 17 10

=

1 3 7 40 1 6 5

0 0 31 150 0 31 30

=

1 3 7 40 1 6 5

0 0 31 150 0 0 15

= 1(1)(31)(15) = 465

33. We first use the second row to reduce the third row. Then we use the first row to reduce the second row.1 1 1

a b c

0 b2 ab c2 ac

=

1 1 1

0 b a c a0 b(b a) c(c a)

= (b a)(c a)

1 1 1

0 1 1

0 b c

.Expanding along the first row gives (b a)(c a)(c b).

34. In order, we use the third row to reduce the fourth row, the second row to reduce the third row, and the firstrow to reduce the second row. We then pull out a common factor from each column.

1 1 1 1

a b c d

a2 b2 c2 d2

a3 b3 c3 d3

=

1 1 1 1

0 b a c a d a0 b2 ab c2 ac d2 ac0 b3 ab2 c3 ac2 d3 ad2

= (b a)(c a)(d a)

1 1 1 1

0 1 1 1

0 b c d

0 b2 c2 d2

.

Expanding along the first column and using Problem 33 we obtain (b a)(c a)(d a)(c b)(d b)(d c).35. SinceC11 = 4, C12 = 5, and C13 = 6, we havea21C11+ a22C12+ a23C13 = (1)(4) + 2(5) + 1(6) = 0. Since

C12 = 5, C22 = 7, andC23 = 3, we have a13C12+ a23C22+ a33C32 = 2(5) + 1(7) + 1(3) = 0.

36. SinceC11+7, C12 = 8, andC13 = 10 we have a21C11+ a22C12+ a23C13 = 2(7)+ 3(8) 1(10) = 0.SinceC12 = 8, C22 = 19, and C32 = 7 we have a13C12+ a23C22+ a33C32 = 5(8) 1(19) 3(7) = 0.

37. det(A + B) =

10 00 3 = 30; det A + det B= 10 31 = 21

38. det(2A) = 25 det A= 32(7) = 22439. Factoring1 out of each row we see that det(A) = (1)5 det A = det A. Then det A = det(A) =

det AT = det A and det A= 0.

40. (a) Cofactors: 25! 1.55(1025); Row reduction: 253/3 5.2(103)(b) Cofactors: about 90 billion centuries; Row reduction: about 1

10 second

328


110/121

Exercises 8.6

Exercises 8.6

1. AB=

3 2 1 + 16 6 2 + 3

=

1 0

0 1

2. AB=

2 1 1 + 1 2 + 26 6 3 + 4 6 62 + 1 3 1 1 + 2 2 + 2 3

= 1 0 00 1 0

0 0 1

3. det A= 9. A is nonsingular. A1 =1

9

1 1

4 5

=

1

9

1

9

49

5

9

4. det A= 5. A is nonsingular. A1 =1

5

3 1

4 13

=

3

5

1

5

45

1

15

5. det A= 12. A is nonsingular. A1 = 1

12

2 0

3 6

=

1

6 0

1

4

1

2


2

=

1

3 1

3

13

2

3


8 8 82 4 6

6 4 2

=

1

2

1

2

1

2

18

1

4 3

8

3

8 1

4

1

8

8. det A= 0. A is singular.


14 13 162 4 2

4

7

4

=

7

15 13

30 8

15

1

15 2

15

1

15

2

15

7

30

2

15

10. det A= 78. A is nonsingular. A1 =

1

78

8 20 2

2 5 1912 9 3

4

39

10

39

1

39

139 5

78

19

78

2

13 3

26

1

26


12 0 0

0 6 00 0 18

=

1

3 0 0

0 16

0

0 0 12


16

0 0 2

8 0 0

0 16 0

=

0 0 18

1

2 0 0

0 1 0


27

6 21 9 361 1 6 310 17 6 51

4 4 3 12

=

29

79 13 43

127

1

27

2

9 1

9

10

27

17

27 2

9 17

9

4

27 4

27

1

9

4

9


0 1 3 30 1 3 90 2 0 0

6 1 3 15

=

0 16

1

2 1

2

0 16 1

2

3

2

0 13

0 0

1 16

1

2 5

2

329


111/121

Exercises 8.6

15.

6 2 1 00 4 0 1

1

6R1

1

4R2

1 1

3

1

6 0

0 1 0 14

1

3R2+R1

1 0 1

6

1

12

0 1 0 14

; A1 =

1

6

1

12

0 14

16. 8 0 1 0

0 12

0 1 1

8R1

2R2 1 0 1

8 0

0 1 0 2 ; A1 =

1

8 0

0 2 17.

1 3 1 0

5 3 0 1

5R1+R2

1 3 1 0

0 12 5 1

1

12R2

1 3 1 0

0 1 512 1

12

3R2+R1

1 0 1

4

1

4

0 1 512 1

12

;

A1 =

1

4

1

4

5

12 1

12

18.

2 3 1 02 4 0 1

1

2R1

1 3

2

1

2 0

2 4 0 1

2R1+R2

1 3

2

1

2 0

0 1 1 1

3

2R2+R1

1 0 2 3

2

0 1 1 1

;

A1 = 2 3

2

1 1 19.

1 2 3 1 0 0

4 5 6 0 1 0

7 8 9 0 0 1

row

operations

1 2 3 1 0 0

0 1 2 43 1

3 0

0 0 0 1 2 1

; Ais singular.

20.

1 0 1 1 0 00 2 1 0 1 02 1 3 0 0 1

row

operations

1 0 0 59 1

9

2

9

0 1 0 29 5

9

1

9

0 0 1 49 1

9

2

9

; A1 =

5

9 1

9

2

9

29 5

9

1

9

49 1

9

2

9

21.

4 2 3 1 0 0

2 1 0 0 1 0

1 2 0 0 0 1

R13

1 2 0 0 0 1

2 1 0 0 1 0

4 2 3 1 0 0

row

operations

1 0 0 0 23

1

3

0 1 0 0

1

3

2

3

0 0 1 13 2

3 0

;

A1 =

0 23

1

3

0 13 2

3

1

3 2

3 0

22.

2 4 2 1 0 04 2 2 0 1 08 10 6 0 0 1

row

operations

1 2 1 12

0 0

0 1 13

1

3 1

6 0

0 0 0 2 1 1

; Ais singular.

23.1 3 0 1 0 0

3 2 1 0 1 00 1 2 0 0 1

rowoperations1 3 0 1 0 0

0 1 1 1 1 00 0 1 1 1 1

rowoperations1 0 0 5 6 3

0 1 0 2 2 10 0 1 1 1 1

;

A1 =

5 6 32 2 1

1 1 1

24.

1 2 3 1 0 0

0 1 4 0 1 0

0 0 8 0 0 1

row

operations

1 0 0 1 2 58

0 1 0 0 1 12

0 0 1 0 0 18

; A1 =

1 2 58

0 1 12

0 0 18

330


112/121

Exercises 8.6

25.

1 2 3 1 1 0 0 0

1 0 2 1 0 1 0 02 1 3 0 0 0 1 01 1 2 1 0 0 0 1

rowoperations

1 2 3 1 1 0 0 0

0 1 52

1 12

1

2 0 0

0 0 1 23

1

3 1 2

3 0

0 0 0 1 12

1 12

1

2

rowoperations

1 0 0 0

1

2 2

3 1

6

7

6

0 1 0 0 1 13

1

3 4

3

0 0 1 0 0 13 1

3

1

3

0 0 0 1 12

1 12

1

2

; A1 =

1

2 2

3 1

6

7

6

1 13

1

3 4

3

0 13 1

3

1

3

12

1 12

1

2

26.

1 0 0 0 1 0 0 0

0 0 1 0 0 1 0 0

0 0 0 1 0 0 1 0

0 1 0 0 0 0 0 1

rowinterchange

1 0 0 0 1 0 0 0

0 1 0 0 0 0 0 1

0 0 1 0 0 1 0 0

0 0 0 1 0 0 1 0

; A1 =

1 0 0 0

0 0 0 1

0 1 0 0

0 0 1 0

27. (AB)1 =B1A1 = 1

3

1

3

1 10

3

28. (AB)1 =B1A1 =

1 4 20

2 6 303 6 32

29. A= (A1)1 =

2 33 4

30. AT =

1 2

4 10

; (AT)1 =

5 12 1

2

; A1 =

5 21 1

2

; (A1)T =

5 12 1

2

31. Multiplying 4 3x 4

4 3x 4 =

16 3x 00 16 3x we see that x = 5.

32. A1 =

sin cos cos sin

33. (a) AT =

sin cos cos sin

= A1 (b) AT =

13

13

13

0 12 1

2

26

16

16

=A1

34. Since det A det A1 = det AA1 = det I = 1, we see that det A1 = 1/ det A. IfA is orthogonal, det A =det AT = det A1 = 1/ det A and (det A)2 = 1, so det A= 1.

35. Since A and B are nonsingular, det AB= det A det B = 0, andAB is nonsingular.36. SupposeA is singular. Then det A= 0, det AB= det A det B= 0, and AB is singular.37. Since det A det A1 = det AA1 = det I= 1, det A1 = 1/ det A.38. Suppose A2 = A and A is nonsingular. ThenA2A1 = AA1, and A = I. Thus, ifA2 = A, either A is

singular or A = I.

39. IfA is nonsingular, then A1 exists, andAB = 0 implies A1AB= A10, soB = 0.

40. IfA is nonsingular, A1 exists, andAB = AC implies A1AB= A1AC, so B = C.

331


113/121

Exercises 8.6

41. No, consider A =

1 0

0 0

and B =

0 0

0 1

.

42. A is nonsingular ifa11a22a33 = 0 or a11, a22, anda33 are all nonzero.

A1

=

1/a11 0 0

0 1/a22 00 0 1/a33

For any diagonal matrix, the inverse matrix is obtaining by taking the reciprocals of the diagonal entries and

leaving all other entries 0.

43. A1 =

1

3

1

3

2

3 1

3

; A1

4

14

=

6

2

; x1 = 6, x2 = 2

44. A1 =

2

3

1

6

13

1

6

; A1

2

5

=

12

32

; x1 =

1

2, x2 = 3

2

45. A1

= 116 381

814

; A

1 61

= 341

2

; x1 =

3

4, x2 = 1

2

46. A1 =

2 13

2 1

2

; A1

4

3

=

1115

2

; x1 = 11, x2 = 15

2

47. A1 =

1

5

1

5

1

5

1 1 06

5 1

5 1

5

; A1

4

0

6

=

2

4

6

; x1 = 2, x2 = 4, x3 = 6

48. A1 =

5

12 1

12

1

4

23

1

3 0

112

5

12 1

4

; A1

1

2

3

=

12

03

2

; x1 = 12

, x2 = 0, x3 =3

2

49. A1 =

2 3 21

4 1

4 0

5

4

7

4 1

; A1

1

37

=

21

1

11

; x1 = 21, x2 = 1, x3 = 11

50. A1 =

2 1 1 11 2 1 1

1 1 1 11 1 1 0

; A1

2

1

53

=

1

2

14

; x1 = 1, x2 = 2, x3 = 1, x4 = 4

51.

7 23 2

x1

x2

=

b1

b2

; A1 =

1

10

1

10

320

7

20

; X= A1

5

4

=

9

10

13

20

; X= A1

10

50

=

6

16

;

X= A1

020

=

27

52.

1 2 5

2 3 8

1 1 2

x1

x2

x3

=

b1

b2

b3

; A1 =

2 1 112 7 25 3 1

; X= A1

1

4

6

=

1252

23

;

X= A1

3

3

3

=

0

9

3

; =A1

0

54

=

1

27

11

332


114/121

Exercises 8.7

53. det A= 18 = 0, so the system has only the trivial solution.54. det A= 0, so the system has a nontrivial solution.

55. det A= 0, so the system has a nontrivial solution.

56. det A= 12= 0, so the system has only the trivial solution.

57. (a)

1 1 1

R1 R2 00 R2 R3

i1

i2

i3

=

0

E2 E1E3 E2

(b) det A= R1R2+ R1R3+ R2R3 > 0, so A is nonsingular.

(c) A1 = 1

R1R2+ R1R3+ R2R3

R2R3 R2 R3 R2R1R3 R3 R1R1R2 R2 R1+ R2

;

A1

0

E2

E1

E3 E2

= 1

R1R2+ R1R3+ R2R3

R2E1 R2E3+ R3E1 R3E2R1E2

R1E3

R3E1+ R3E2

R1E2+ R1E3 R2E1+ R2E3

58. (a) We write the equations in the form

4u1+ u2+ u4= 200u1 4u2+ u3= 300u2 4u3+ u4= 300u1+ u3 4u4= 200.

In matrix form this becomes

4 1 0 1

1 4 1 00 1

4 1

1 0 1 4

u1

u2

u3

u4

=

200300

300

200

.

(b) A1 =

7

24 1

12 1

24 1

12

112 7

24 1

12 1

24

124 1

12 7

24 1

12

112 1

24 1

12 7

24

; A1

200300300200

=

225

2

275

2

275

2

225

2

; u1 = u4 =

225

2 , u2 = u3 =

275

2

Exercises 8.7

1. det A= 10, det A1 = 6, det A2 = 12; x1 = 6

10 = 3

5, x2 = 12

10 = 6

5

2. det A= 3, det A1 = 6, det A2 = 6; x1 = 63 = 2, x2 = 63 = 23. det A= 0.3, det A1 = 0.03, det A2= 0.09; x1 = 0.030.3 = 0.1 , x2 = 0.090.3 = 0.3

4. det A= 0.015, det A1 = 0.00315, detA2 = 0.00855; x1 = 0.003150.015 = 0.21, x2 = 0.008550.015 = 0.575. det A= 1, det A1 = 4, det A2 = 7; x= 4, y = 76. det A= 70, det A1 = 14, det A2 = 35; r= 1470 = 15, s= 3570 = 127. det A= 11, det A1 = 44, det A2 = 44, det A3 = 55; x1 = 4411 = 4, x2 = 4411 = 4, x3 = 5511 = 5

333


115/121


116/121

Year Phytoplankton Water Zooplankton

0 0.00 100.00 0.00

1 2.00 97.00 1.00

2 3.70 94.26 2.04

3 5.14 91.76 3.10

4 6.36 89.47 4.17

5 7.39 87.37 5.24

6 8.25 85.46 6.30

7 8.97 83.70 7.33

8 9.56 82.10 8.34

9 10.06 80.62 9.32

10 10.46 79.28 10.26

11 10.79 78.04 11.17

12 11.06 76.90 12.04


5. (a) The initial state and the transfer matrix are

X0 =

0

100

0

and T=

0.88 0.02 0

0.06 0.97 0.05

0.06 0.01 0.95

.

(b)

6. From TX = 1X we see that the equilibrium state vector X is the eigenvector of the transfer matrix T corre-

sponding to the eigenvalue 1. It has the properties that its components add up to the sum of the components

of the initial state vector.


1.

2 3 43 4 5

4 5 6

5 6 7

2. 4 3

3. AB=

3 4

6 8

; BA= [11 ]

4. A1 = 12

4 23 1

=

2 13

2 1

2

5. False; consider A = 1 0

0 1

and B = 0 1

1 0

6. True

7. det1

2A

=1

2

3(5) = 5

8; det(AT) = (1)3(5) = 5

8. det AB1 = det A/ det B= 6/2 = 3

9. 0

10. det C= (1)3/ det B= 1/103(2) = 1/200011. False; an eigenvalue can be 0.

369


117/121


12. True

13. True

14. True, since complex roots of real polynomials occur in conjugate pairs.

15. False; if the characteristic equation of ann

nmatrix has repeated roots, there may not benlinearly independent

eigenvectors.

16. True

17. True

18. True

19. False; A is singular and thus not orthogonal.

20. True

21. A= 12

(A + AT) + 12

(A AT) where 12

(A + AT) is symmetric and 12

(AAT) is skew-symmetric.

22. Since det A

2

= (det A)

2

0 and det0 1

1 0

= 1, there is no A such that A2

= 0 1

1 0

.

23. (a)

1 1

1 1

is nilpotent.

(b) Since det An = (det A)n = 0 we see that det A= 0 and A is singular.

24. (a) xy =

i 0

0 i

= yx; xz =

0 11 0

= zx; yz =

0 i

i 0

= zy

(b) We first note that for anticommuting matrices AB =BA, so C = 2AB. Then Cxy =

2i 0

0 2i

,

Cyz =

0 2i

2i 0

, andCzx =

0 2

2 0

.

25.

5 1 1 92 4 0 27

1 1 5 9

R13

1 1 5 9

2 4 0 27

5 1 1 9

row

operations

1 1 5 9

0 1 5 92

0 0 1 12

row

operations

1 0 0 12

0 1 0 7

0 0 1 12

.

The solution is X = [12

7 12

]T

.

26.

1 1 1 6

1 2 3 22 0 3 3

row

operations

1 1 1 6

0 1 23

4

3

0 0 1 1

row

operations

1 0 0 3

0 1 0 2

0 0 1 1

.


27. Multiplying the second row byabc we obtain the third row. Thus the determinant is 0.

28. Expanding along the first row we see that the result is an expression of the form ay +bx2 +cx+d= 0, which

is a parabola since, in this case a= 0 and b= 0. Letting x = 1 and y = 2 we note that the first and secondrows are the same. Similarly, whenx = 2 and y = 3, the first and third rows are the same; and when x = 3

and y = 5, the first and fourth rows are the same. In each case the determinant is 0 and the points lie on the

parabola.

29. 4(2)(3)(1)(2)(5) = 24030. (3)(6)(9)(1) = 162

370


118/121


31. Since

1 1 15 1 11 2 1

= 18 = 0, the system has only the trivial solution.

32. Since

1 1 15 1 11 2 1

= 0, the system has infinitely many solutions.33. From x1I2+ x2HNO3 x3HIO3+ x4NO2+ x5H2O we obtain the system 2x1 = x3, x2 = x3+ 2x5, x2 = x4,

3x2 = 3x3 +2x4 +x5. Lettingx4 = x2in the fourth equation we obtain x2= 3x3 +x5. Takingx1 = t we see that

x3 = 2t, x2 = 2t+ 2x5, and x2 = 6t+x5. From the latter two equations we get x5 = 4t. Taking t = 1 we have

x1 = 1,x2 = 10,x3 = 2,x4 = 10, andx5 = 4. The balanced equation is I2+10HNO3 2HIO3+10NO2+4H2O.34. Fromx1Ca + x2H3PO4 x3Ca3P2O8+ x4H2 we obtain the system x1 = 3x3, 3x2 = 2x4,x2 = 2x3, 4x2 = 8x3.

Letting x3 = t we see that x1 = 3t, x2 = 2t, and x4 = 3t. Taking t = 1 we obtain the balanced equation

3Ca + 2H3PO4 Ca3P2O8+ 3H2.35. det A=

84, det A1 = 42, det A2 =

21, det A3 =

56; x1 =

42

84=

1

2, x2 =

2184

=1

4, x3 =

5684

=2

3

36. det = 4, det A1 = 16, det A2= 4, det A3 = 0; x1 = 164

= 4, x2 =4

4 = 1, x3 = 0

4= 0

37. det A= cos2 + sin2 , det A1 = Xcos Ysin , det A2 = Y cos +Xsin ;x1 = Xcos Ysin , y = Y cos +Xsin

38. (a) i1 i2 i3 i4 = 0, i2R1 = E, i2R1 i3R2 = 0, i3R2 i4R3 = 0

(b) det A=

1 1 1 10 R1 0 0

0 R1 R2 00 0 R2 R3

=R1R2R3;

det A1 =

0 1 1 1E R1 0 0

0 R1 R2 00 0 R2 R3

= E[R2R3 R1(R3+R2)] = E(R2R3+R1R3+R1R2);

i1 =det A1

det A =

E(R2R3+R1R3+R1R2)

R1R2R3=E

1

R1+

1

R2+

1

R3

39. AX= B is

2 3 11 2 0

2 0 1

x1

x2

x3

=

6

39

. Since A1 = 1

3

2 3 21 0 14 6 7

, we have

X= A1

B=

7

523

.

40. (a) A1B=

3

2 1

4 9

4

1 12

3

2

1

2 1

4 1

4

1

1

1

=

1

1

0

(b) A1B=

3

2 1

4 9

4

1 12

3

2

1

2 1

4 1

4

2

1

3

=

10

7

2

371


119/121


41. From the characteristic equation 2 4 5 = 0 we see that the eigenvalues are 1 =1 and 2 = 5. For1 = 1 we have 2k1 + 2k2 = 0, 4k14k2= 0 andK1 =

11

. For2 = 5 we have 4k1 + 2k2 = 0, 4k12k2 = 0

andK2 =

1

2

.

42. From the characteristic equation 2 = 0 we see that the eigenvalues are 1 = 2 = 0. For1 = 2 = 0 we have

4k1 = 0 and K1=

0

1

is a single eigenvector.

43. From the characteristic equation3 + 62 + 15+ 8 = (+ 1)2( 8) = 0 we see that the eigenvalues are1 = 2 = 1 and3 = 8. For 1 = 2 = 1 we have

4 2 4 0

2 1 2 0

4 2 4 0

row

operations

1 12

1 0

0 0 0 0

0 0 0 0

.

Thus K1 = [ 1 2 0 ]T andK2 = [ 1 0 1 ]T. For3 = 8 we have

5 2 4 0

2 8 2 04 2 5 0

rowoperations

1

2

5 4

5 0

0 1 12

0

0 0 0 0

.

Thus K3 = [ 2 1 2 ]T

.

44. From the characteristic equation3 +18299+162 = (9)(6)(3) = 0 we see that the eigenvaluesare1 = 9, 2 = 6, and 3 = 3. For 1 = 9 we have

2 2 0 02 3 2 0

0 2 4 0

row

operations

1 1 0 0

0 1 2 00 0 0 0

.

Thus K1 = [

2 2 1 ]

T. For2 = 6 we have

1 2 0 0

2 0 2 00 2 1 0

row

operations

1 2 0 00 1 1

2 0

0 0 0 0

.

Thus K2 = [ 2 1 2 ]T

. For3 = 3 we have

4 2 0 02 3 2 0

0 2 2 0

row

operations

1 12

0 0

0 1 1 0

0 0 0 0

.

Thus K3 = [ 1 2 2 ]T.45. From the characteristic equation3 2 + 21+ 45 = (+ 3)2( 5) = 0 we see that the eigenvalues are

1 = 2 = 3 and3 = 5. For 1 = 2 = 3 we have

1 2 3 02 4 6 0

1 2 3 0

row

operations

1 2 3 00 0 0 0

0 0 0 0

.

Thus K1 = [2 1 0 ]T andK2 = [ 3 0 1 ]T. For3= 5 we have7 2 3 0

2 4 6 01 2 5 0

row

operations

1 27

3

7 0

0 1 2 0

0 0 0 0

.

372


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Thus K3 = [1 2 1 ]T.46. From the characteristic equation3 + 2 + 2= ( +1)( 2) = 0 we see that the eigenvalues are 1 = 0,

2 = 1, and 3 = 2. For 1 = 0 we have k3 = 0, 2k1+ 2k2+ k3 = 0 and K1 = [ 1 1 0 ]T. For2 = 1 wehave 1 0 0 00 1 1 0

2 2 2 0

rowoperations

1 0 0 00 1 1 00 0 0 0

.

Thus K2 = [ 0 1 1 ]T. For3 = 2 we have2 0 0 0

0 2 1 02 2 1 0

row

operations

1 0 0 0

0 1 12

0

0 0 0 0

.

Thus K3 = [ 0 1 2 ]T

.

47. Let X1 = [ a b c ]T

be the first column of the matrix. Then XT1 [ 12 0 1

2]T

= 12

(c a) = 0 andXT1 [ 13 1

3

13]T = 1

3 (a + b + c) = 0. Also XT1X1 = a2 + b2 + c2 = 1. We see thatc = a and b = 2afrom

the first two equations. Then a2 + 4a2 +a2 = 6a2 = 1 anda = 16

. ThusX1 = [ 16 2

6

16

]T

.

48. (a) Eigenvalues are 1 = 2 = 0 and 3 = 5 with corresponding eigenvectors K1 = [ 0 1 0 ]T

, K2 =

[ 2 0 1 ]T

, and K3 = [1 0 2 ]T. SinceK1 = 1,K2 =

5, andK3 =

5 , we have

P=

0 25 1

5

1 0 0

0 15

25

and P1 =PT =

0 1 025

0 15

15

0 25

.

(b) P1AP=

0 0 0

0 0 0

0 0 5

49. We identify A =

1 3

2

3

2 1

. Eigenvalues are1 = 12 and 2 = 52 so D =

12

0

0 52

and the equation becomes

[ X Y ] D

X

Y

= 1

2X2 + 5

2Y2 = 1. The graph is a hyperbola.

50. We measure years in units of 10, with 0 corresponding to 1890. Then Y = [ 63 76 92 106 123 ]T

and

A=

0 1 2 3 4

1 1 1 1 1

T, soATA=

30 10

10 5

. Thus

X= (ATA)1ATY= 1

50

5 10

10 30 ATY=

15

62 ,

and the least squares line is y = 15t+ 62. Att = 5 (corresponding to 1940) we have y = 137. The error in the

predicted population is 5 million or 3.7%.

51. The encoded message is

B= AM =

10 1

9 1

19 1 20 5 12 12 9 20 5 0 12 1 21

14 3 8 5 4 0 15 14 0 6 18 9 0

=

204 13 208 55 124 120 105 214 50 6 138 19 210

185 12 188 50 112 108 96 194 45 6 126 18 189

.

373


121/121


52. The encoded message is

B= AM =

10 1

9 1

19 5 3 0 1 7 14 20 0 1 18

18 22 19 0 20 21 5 19 0 1 13

= 208 72 49 0 30 91 145 219 0 11 193

189 67 46 0 29 84 131 199 0 10 175

.

53. The decoded message is

M= A1B=

3 2 1

1 0 0

2 1 1

19 0 15 14 0 20

35 10 27 53 1 54

5 15 3 48 2 39

=

8 5 12 16 0 9

19 0 15 14 0 20

8 5 0 23 1 25

.

From correspondence (1) we obtain: HELP IS ON THE WAY.

54. The decoded message is

M= A1B=

3 2 1

1 0 0

2 1 1

5 2 21

27 17 40

21 13 2

=

18 15 19

5 2 21

4 0 0

.

From correspondence (1) we obtain: ROSEBUD .

55. (a) The parity is even so the decoded message is [ 1 1 0 0 1 ]

(b) The parity is odd; there is a parity error.

56. From

c1

c2

c3

1 1 0 1

1 0 1 1

0 1 1 1

1

0

0

1

=

0

0

1

we obtain the codeword [ 0 0 1 1 0 0 1] .

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