CHAPTER 3ruby.fgcu.edu/Courses/rschnack/Spring_2005_Classes/ch03.pdf · L NM O QP aa aa bb bb ab a b aba b bb bb aa aa ba b a bab a 11 12 21 22 11 12 21 22 11 11 12 12 21 21 22 22

CHAPTER 3 LinearAlgebra

3.1 Matrices: Sums and Products

!!!! Do They Compute?

1. 22 0 64 2 42 0 2

A =−

−

L

NMMM

O

QPPP

2. A B+ =−

L

NMMM

O

QPPP

21 6 32 3 21 0 3

3. 2C D− , Matrices are not compatible

4. AB =− −

− −

L

NMMM

O

QPPP

1 3 32 7 21 3 1

5. BA =−

L

NMMM

O

QPPP

5 3 92 1 21 0 1

6. CD =−−L

NMMM

O

QPPP

3 1 08 1 29 2 6

7. DC =−LNMOQP

1 16 7

8. DC( ) =−LNMOQP

T 1 61 7

9. C DT , Matrices are not compatible

10. D CT , Matrices are not compatible

11. A22 0 02 1 100 0 2

=−−

−

L

NMMM

O

QPPP

12. AD, Matrices are not compatible

13. A I− =−

−

L

NMMM

O

QPPP

3

2 0 32 0 21 0 0

14. 4 31 12 00 1 00 0 1

3B I− =L

NMMM

O

QPPP

15. C I− 3, Matrices are not compatible

16. AC =L

NMMM

O

QPPP

2 96 70 3

165

166 CHAPTER 3 Linear Algebra

!!!! Rows and Columns in Products

17. (a) 5 columns (b) 4 rows (c) 6 4×

!!!! Products with Transposes

18. (a) A BT =−LNMOQP = −1 4

11

3 (b) ABT = LNMOQP − =

−−LNMOQP

14

1 11 14 4

(c) B AT = − LNMOQP = −1 1

14

3 (d) BAT =−LNMOQP =

− −LNMOQP

11

1 41 41 4

!!!! Reckoning

19. The following proofs are carried out for 2 2× matrices. The proofs for general n n× matricesfollow along the same lines.

A B

A B

− = LNMOQP −LNM

OQP =

− −− −

LNM

OQP

+ −( ) = LNMOQP + −( )LNM

OQP =LNM

OQP +

− −− −LNM

OQP

=+ − + −+ − + −

LNM

OQP =

−

a aa a

b bb b

a b a ba b a b

a aa a

b bb b

a aa a

b bb b

a b a ba b a b

a

11 12

21 22

11 12

21 22

11 11 12 12

21 21 22 22

11 12

21 22

11 12

21 22

11 12

21 22

11 12

21 22

11 11 12 12

21 21 22 22

11

1 1

a f a fa f a f

b a ba b a b

11 12 12

21 21 22 22

−− −

LNM

OQP

20. Compare

A B

B A

+ = LNMOQP +LNM

OQP =

+ ++ +

LNM

OQP

+ = LNMOQP +LNM

OQP =

+ ++ +

LNM

OQP

a aa a

b bb b

a b a ba b a b

b bb b

a aa a

b a b ab a b a

11 12

21 22

11 12

21 22

11 11 12 12

21 21 22 22

11 12

21 22

11 12

21 22

11 11 12 12

21 21 22 22

By commutativity of the real numbers, the matrices A B+ and B A+ are the same.

21. c d c da aa a

c d a c d ac d a c d a

ca da ca daca da ca da

ca caca ca

da dada da

ca aa a

da aa a

c d

+( ) = +( )LNMOQP =

+( ) +( )+( ) +( )LNM

OQP =

+ ++ +

LNM

OQP

= LNMOQP +LNM

OQP =LNM

OQP +LNM

OQP = +

A

A A

11 12

21 22

11 12

21 22

11 11 12 12

21 21 22 22

11 12

21 22

11 12

21 22

11 12

21 22

11 12

21 22

22. c ca b a ba b a b

c a b c a bc a b c a b

ca cb ca cbca cb ca cb

ca caca ca

cb cbcb cb

ca aa a

cb bb b

A B+( ) =+ ++ +

LNM

OQP =

+ ++ +

LNM

OQP =

+ ++ +

LNM

OQP

= LNMOQP +LNM

OQP =LNM

OQP +LNM

OQP

11 11 12 12

21 21 22 22

11 11 12 12

21 21 22 22

11 11 12 12

21 21 22 22

11 12

21 22

11 12

21 22

11 12

21 22

11 12

21 22

a f a fa f a f

!!!! Properties of the Transpose

Rather than grinding out the proofs of Problems 23–26, we make the following observations:

23. A AT Tb g = . Interchanging rows and columns of a matrix two times reproduce the original matrix.

SECTION 3.1 Matrices: Sums and Products 167

24. A B A B+( ) = +T T T. Add two matrices and then interchange the rows and columns of the

resulting matrix. You get the same as first interchanging the rows and columns of the matricesand then adding.

25. k kA A( ) =T T. Demonstrate that it makes no difference whether you multiply each element of

matrix A before or after rearranging them to form the transpose.

26. AB B A( ) =T T T . This identity is not so obvious. Due to lack of space we verify the proof for

2 2× matrices. The verification for 3 3× and higher-order matrices follows along exactly thesame lines.

A

B

AB

AB

B A

= LNMOQP

= LNMOQP

= LNMOQPLNM

OQP =

+ ++ +

LNM

OQP

( ) =+ ++ +

LNM

OQP

= LNMOQP

a aa ab bb ba aa a

b bb b

a b a b a b a ba b a b a b a b

a b a b a b a ba b a b a b a bb bb b

a aa

11 12

21 22

11 12

21 22

11 12

21 22

11 12

21 22

11 11 12 21 11 12 12 22

21 11 22 21 21 12 22 22

11 11 12 21 21 11 22 21

11 12 12 22 21 12 22 22

11 21

12 22

11 21

T

T T

12 22

11 11 12 21 21 11 22 21

11 12 12 22 21 12 22 22aa b a b a b a ba b a b a b a b

LNM

OQP =

+ ++ +

LNM

OQP

Hence, AB B A( ) =T T T for 2 2× matrices.

!!!! Transposes and Symmetry

27. If the matrix A = aij is symmetric, then a aij ji= . Hence AT = aji is symmetric since a aji ij= .

!!!! Symmetry and Products

28. We pick at random the two symmetric matrices

A = LNMOQP

0 22 1

, B = LNMOQP

3 11 1

,

which gives

AB = LNMOQPLNMOQP =LNMOQP

0 22 1

3 11 1

2 27 3

.

This is not symmetric. In fact, if A, B are symmetric matrices, we have

AB B A BA( ) = =T T T ,

which says the only time the product of symmetric matrices A and B is symmetric is when thematrices commute (i.e. AB BA= ).


!!!! Constructing Symmetry

29. We verify the statement that A A+ T is symmetric for any 2 2× matrix. The general prooffollows along the same lines.

A A+ =LNM

OQP +LNM

OQP =

++

LNM

OQP

T a aa a

a aa a

a a aa a a

11 12

21 22

11 21

12 22

11 12 21

21 12 22

22

,

which is clearly symmetric.

!!!! More Symmetry

30. Let

A =L

NMMM

O

QPPP

a aa aa a

11 12

21 22

31 32

.

Hence, we have

A AT = LNMOQPL

NMMM

O

QPPP= LNM

OQP

a a aa a a

a aa aa a

A AA A

11 21 31

12 22 32

11 12

21 22

31 32

11 12

21 22,

A a a aA a a a a a aA a a a a a a

A a a a

11 112

212

312

12 11 12 21 22 31 32

21 11 12 21 22 31 32

22 122

222

322

= + += + += + +

= + + .

Note A A12 21= , which means AAT is symmetric. We could verify the same result for 3 3×

matrices.

!!!! Trace of a Matrix

31. Tr Tr TrA B A B+ = +b g b g b gTr Tr( ) + Tr(A B+ = + + + + = + + + + + =b g b g b g b g b ga b a b a a b bnn nn nn nn11 11 11 11! ! ! A B) .

32. Tr Trc ca ca c a a cnn nnA A( ) = + + = + + = ( )11 11! !a f33. Tr TrTA Ab g = ( ). Taking the transpose of a (square) matrix does not alter the diagonal element, so

Tr Tr TA A( ) = b g.34. Tr

Tr

AB

BA

( ) = + + ⋅ + + + + + + ⋅ + += + + + + + += + + + + + += + + ⋅ + + + + + + ⋅ + + = ( )

a a b b a a b ba b a b a b a bb a b a b a b ab b a a b b a a

n n n nn n nn

n n n n nn nn

n n n n nn nn

n n n nn n nn

11 1 11 1 1 1

11 11 1 1 1 1

11 11 1 1 1 1

11 1 11 1 1 1

! ! ! ! !! ! !! ! !

! ! ! ! !

a f a fa f a f

!!!! Matrices Can Be Complex

35. A B+ =++ −LNM

OQP2

3 02 4 4

ii i

36. AB =− + − ++ −LNM

OQP

3 18 4 5 3

i ii i

37. BA =− −

−LNM

OQP

1 34 1

ii i


38. A2 6 4 66 4 5 8

=+

− − −LNM

OQP

i ii i

39. ii

i iA =

− + −+

LNM

OQP

1 22 3 2

40. A B− =− − +

−LNM

OQP2

1 2 26 4 5

ii i

i

41. BT =− +LNM

OQP

1 21

ii i

42. Tr B( ) = +2 i

!!!! Real and Imaginary Components

43. A =+

−LNM

OQP =LNMOQP + −LNMOQP

1 22 2 3

1 02 2

1 20 3

i ii

i , B =−+

LNM

OQP =LNMOQP +

−LNMOQP

12 1

1 00 1

0 12 1

ii i

i

!!!! Square Roots of Zero

44. If we assume

A = LNMOQP

a bc d

is the square root of

0 00 0LNMOQP ,

then we must have

A22

2

0 00 0

= LNMOQPLNMOQP =

+ ++ +LNM

OQP =LNMOQP

a bc d

a bc d

a bc ab bdac cd bc d

,

which implies the four equations

a bcab bdac cdbc d

2

2

0000

+ =+ =+ =

+ = .

From the first and last equations, we have a d2 2= . We now consider two cases: first we assumea d= . From the middle two preceding equations we arrive at b = 0, c = 0, and hence a = 0,d = 0 . The other condition, a d= − , gives no condition on b and c, so we seek a matrix of theform (we pick a = 1, d = −1 for simplicity)

11

11

1 00 1

bc

bc

bcbc−

LNMOQP −LNMOQP =

++

LNM

OQP .

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

= −1 , and hence

1 1

1−

−

LNMMOQPPcc

,


which gives

1 1

11 1

1

0 00 0

−

−

LNMMOQPP

−

−

LNMMOQPP =LNMOQPc

cc

c.

!!!! Zero Divisors

45. No, AB = 0 does not imply that A = 0 or B = 0 . For example, the product

1 00 0

0 00 1

LNMOQPLNMOQP

is the zero matrix, but neither factor is itself the zero matrix.

!!!! Does Cancellation Work?

46. No. A counter example is: 0 00 1

1 20 4

0 00 1

0 00 4

FHGIKJFHGIKJ =FHGIKJFHGIKJ since

1 20 4

0 00 4

FHGIKJ ≠FHGIKJ .

!!!! Taking Matrices Apart

47. (a) A A A A= = −L

NMMM

O

QPPP

1 2 3

1 5 21 0 32 4 7

, x =L

NMMM

O

QPPP

243

where A1, A2 , and A3 are the three columns of the matrix A and x1 2= , x2 4= , x3 3=

are the elements of x. We can write

A

A A A

x

x x x

= −L

NMMM

O

QPPP

L

NMMM

O

QPPP=

× + × + ×− × + × + ×× + × + ×

L

NMMM

O

QPPP= −L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP

= + +

1 5 21 0 32 4 7

243

1 2 5 4 2 31 2 0 4 3 32 2 4 4 7 3

2112

4504

3237

1 1 2 2 3 3 .

(b) We verify the fact for a 3 3× matrix. The general n n× case follows along the samelines.

Axa a aa a aa a a

xxx

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

a xa xa x

a xa xa x

a xa xa x

xaaa

=L

NMMM

O

QPPP

L

NMMM

O

QPPP=

+ ++ ++ +

L

NMMM

O

QPPP=L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP

=L

11 12 13

21 22 23

31 32 33

1

2

3

11 1 12 2 13 3

21 1 22 2 23 3

31 1 32 2 33 3

11 1

21 1

31 1

12 2

22 2

32 2

13 3

23 3

33 3

1

11

21

31NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP= + +x

aaa

xaaa

x x x2

12

22

32

3

13

23

33

1 1 2 2 3 3A A A


!!!! Diagonal Matrices

48. A =

L

N

MMMM

O

Q

PPPP

aa

ann

11

22

0 00 000 0

!!

! ! !!

,

B =

L

N

MMMM

O

Q

PPPP

bb

bnn

11

22

0 00 000 0

!!

! ! !!

.

We have

AB =

L

N

MMMM

O

Q

PPPP

a ba b

a bnn nn

11 11

22 22

0 00 000 0

!!

! ! !!

is a diagonal matrix.

49. We have

AB BA= =

L

N

MMMM

O

Q

PPPP

a ba b

a bnn nn

11 11

22 22

0 00 000 0

!!

! ! !!

.

However, it is not true that a diagonal matrix commutes with an arbitrary matrix.

!!!! Upper Triangular Matrices

50. (a) Examples are

1 20 3LNMOQP ,

1 3 00 0 50 0 2

L

NMMM

O

QPPP

,

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

L

N

MMMM

O

Q

PPPP.

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

AB =L

NMMM

O

QPPP

L

NMMM

O

QPPP

a a aa a

a

b b bb b

b

11 12 13

22 23

33

11 12 13

22 23

33

00 0

00 0

below the diagonal are zero.


(c) In the general case, if we multiply two upper-triangular matrices, it yields

AB =

L

N

MMMMMM

O

Q

PPPPPP

×

L

N

MMMMMM

O

Q

PPPPPP

=

L

N

MMMMMM

O

Q

PPPPPP

a a a aa a a

a a

a

b b b bb b b

b b

b

c c c cc c c

c c

c

n

n

n

nn

n

n

n

nn

n

n

n

nn

11 12 13 1

22 23 2

33 3

11 12 13 1

22 23 2

33 3

11 12 13 1

22 23 2

33 3

00 0

0 0 0 0

00 0

0 0 0 0

00 0

0 0 0 0

!!!

! ! ! ! !

!!!

! ! ! ! !

!!!

! ! ! ! !.

We won’t bother to write the general expression for the elements cij ; the important point

is that the entries in the product matrix that lie below the main diagonal are clearly zero.

!!!! Hard Puzzle

51. If

M = LNMOQP

a bc d

is a square root of

A = LNMOQP

0 10 0

,

then

M2 0 10 0

= LNMOQP ,

which leads to the condition a d2 2= . Each of the possible cases leads to a contradiction. How-ever for matrix B because

1 01

1 01

1 00 1α α−

LNMOQP −LNMOQP =LNMOQP

for any α, we conclude that

B =−

LNMOQP

1 01α

is a square root of the identity matrix for any number α.

!!!! Dot Products

52. 2, 1 1 2 0 • − =, , orthogonal

53. − • = −3 0 2, 1 6, , not orthogonal. Because the dot product is negative, this means the angle

between the vectors is greater than 90°.

54. 2, 1 2 3 1 0 5 , , ,• − = . Because the dot product is positive, this means the angle between the

vectors is less than 90°.


55. 1 0 1 1 1 1 0, , , , − • = , orthogonal

56. 5 7, 1 2, 4, 3 3 0, , , 5 • − − − = , orthogonal

57. 7, 1 4, 3 2, 3 30 5 5 , , ,• − = , not orthogonal

!!!! Lengths

58. Introducing the two vectors u = a b, , v = c d, , we have the distance d between the heads of

the vectors

d a c b d= −( ) + −( )2 2 .

But we also have

u v u v u v− = −( ) ⋅ −( ) = −( ) + −( )2 2 2a c b d ,

so d = −u v . This proof can be extended easily to "u and "v in Rn .

!!!! Geometric Vector Operations

59. A C+ lies on the horizontal axis, from 0 to –2.

A C+ = + − − = −1 2 3 2 2,0, ,

1 3–1–1

–2

–3

1

3

2

–2

–3 2

A = 1 2,

C = − −3 2,

A C+ = −2, 0

60. 12

12

1 2 3 1 2 5 2A B+ = + − = −, , . ,

1 3–1–1

–2

–3

1

3

2

–2

–3 2

A = 1 2,B A+ = −

12

25 2. ,

B = −3 1,


61. A B− 2 lies on the horizontal axis, from 0 to 7.

4 82–1

–2

–3

1

3

2

–2

–4 6

A = 1 2,

B = −3 1, A B− =2 7, 0

!!!! Triangles

62. If 3 2, and 2, 3 are two sides of a triangle,their difference 1 1, − or −1 1, is the third

side. If we compute the dot products of thesesides, we see

3 2 2, 3 12, • = , 3 2 1 1 1, , • − = ,

2 1 1 1, , 3 • − = − . None of these angles are

right angles, so the triangle is not a right triangle(see figure).

1 3–1–1

–2

–3

1

3

2

–2

–3 2

2, 3

3 2,

63. 2, 1 2 1 0 1 0 − • − =, , , so these vectors form a

right triangle, since dot produce is zero (see fig-ure).

1 3–1–1

–2

–3

1

3

2

–2

–3 2

12

–2

Right triangle in three-space

!!!! Properties of Scalar Products

We let a = a an1… , b = b bn1… , and c = c cn1… for simplicity.

64. True. a b b a• = • = = = •a a b b a b a b b a b an n n n n n1 1 1 1 1 1 ! ! ! ! .


65. False. Neither a b c• •( ) nor a b c•( )• . Invalid operation, since we ask for the scalar product of a

vector and a scalar.

66. True.

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

a a kb kb kn n n n n n

n n

a b

a bb g

b g• = • = + + = + +

= • = •1 1 1 1 1 1

1 1

! ! ! !

! !

67. True.

a b c

a b a c

• + = • + + = + + + +

= + + + + + = • + •

b g b g b gb g b ga a b c b c a b c a b c

a b a b a c a cn n n n n n

n n n n

1 1 1 1 1 1

1 1 1 1

! ! !

! !

!!!! Markov Chains

68.1 00 1LNMOQP . Markov chain.

Yes; with square matrix with entries between 0 and 1 inclusive,column sums are 1. We draw the Markov tree for two stagesstarting in state 1. 2

11

1

2

2

1

1

0

1

0

0

1

69.05 0505 05. .. .LNMOQP . Markov chain.


11

1

2

2

1

0.5

0.5

0.5

0.5

0.5

0.5

70.0 051 05

.

.LNMOQP . Markov chain.


11

1

2

2

1

0

1

0

1

0.5

0.5

71.0 11 0LNMOQP . Markov chain.

Yes; with square matrix with entries between 0 and 1 inclusive,column sums are 1. We draw the Markov tree for two stagesstarting in state 1. Note that the states on this Markov chainalternate from state 1 to state 2.

2

1

1

1

2

0

1 1

0


72.0 0 11 0 00 1 0

L

NMMM

O

QPPP.

Yes; with square matrix with entries between 0 and 1 inclusive,column sums are 1. We draw the Markov tree for two stagesstarting in state 1. Note that the states on this Markov chainalternate from 1 2 3 1→ → → →… . Of course it is possible to

start the process in any state.

1

13

10

021

23

10

120

33

11

020

0

0

1

73.0 05 05

05 05 005 0 05

. .. .. .

L

NMMM

O

QPPP

.

Yes; with square matrix with entries between 0 and 1 inclusive,column sums are 1. We draw the Markov tree for two stagesstarting in state 1.

1

1

23

10.5

020.5

33

10.5

0.520

0

0.5

0.5

74.0 2 01 00 6 05 00 2 0 4 01

. .

. .

. . .

L

NMMM

O

QPPP.

No; although entries are between 0 and 1 inclusive, the last column does not sum to 1.

75.0 1 01 0 00 0 1

L

NMMM

O

QPPP. Markov chain.

As the first tree diagram indicates, if you start at State 1 (or 2), you will alternate between 1 and2 and never get to State 3. Furthermore, as shown in the second tree diagram, if you are in State3, you stay there, so the only way to get to State 3 is to start there.

1

1

23

11

020

3

0

0

1

3

1

2

3

10

1203

0

1

0


!!!! Best-of-Three-Game Series

76. (a) (from this state)2 0 2 1 1 0 1 1 0 0 0 1 1 2 0 2

1 0.61 0.6

0.60.4 0.6

0.40.4 1

0.4 1

2 02 11 01 10 00 11 20 2

- - - - - - - ---------

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

L

N

MMMMMMMMMM

O

Q

PPPPPPPPPP

(b) The Bulls will win the series in one of three ways: if they win the first two (WW); losethe first and win the next two (LWW); or win the first, lose the second and win the third(WLW). Because the probabilities of these disjoint events is

P WW

P LWW

P WLW

( ) = ( )

( ) = ( )( )

( ) = ( ) ( )

06

04 04

06 0 4

2

2

2

.

. .

. .

we haveP Bulls win the series P WW P LWW WLW( ) = ( ) + ( ) + ( )

= ( ) + ( )( ) + ( ) ( ) =06 0 4 06 06 04 06482 2 2. . . . . .

Hence, the probability the Knicks will win the three-game series is 0.352.

!!!! Flipping Coins

77. We designate the state of the system as the number of dollars that Jerry has in his pocket. This isan absorbing random walk problem. Once Jerry reaches state $0 or $5 the game is over.

For each player there are six states: $0, $1, $2, $3, $4, $5. From Jerry’s point of view,

x0

000100

=

L

N

MMMMMMMM

O

Q

PPPPPPPP

, P =

L

N

MMMMMMMM

O

Q

PPPPPPPP

1 0 5 0 0 0 00 0 0 5 0 0 00 0 5 0 0 5 0 00 0 0 5 0 0 5 00 0 0 0 5 0 00 0 0 0 0 5 1

..

. .. .

..

P x50 tells us that there is a 22% chance Jerry will be broke after 5 tosses and a 38% chance he

will win everything from Sheryl.


!!!! Genetics Problem

78. If we square the transition matrix

Red Pink White NextRedPinkWhite

P =L

NMMM

O

QPPP

0 5 0 25 00 5 0 5 0 50 0 25 0 5

. .

. . .. .

we get

P2

0 375 0 25 01250 5 0 5 0 50125 0 25 0 375

=L

NMMM

O

QPPP

. . .

. . .

. . ..

This tells us two things: the Markov chain is regular and that if the initial mixture of roses is theproportion 0 5 0 5 0. : . : , then the next generation should expect to be

0 375 0 25 012505 05 050125 0 250 0 375

0 50 50

031250 5001875

. . .

. . .

. . .

.

....

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP.

To find the limiting ratio of genotypes, instead of solving the usual system of equations, we cancompute powers of P until there is no change. We find

P P5 6025 025 0 25050 050 050025 025 0 25

= =L

NMMM

O

QPPP

. . .

. . .

. . ..

This tells us the powers of the transition matrix have stabilized to a matrix. When this happensthe steady state probability vector s is the common column of this matrix. In this case it iss = 025 050 025. , . , . . In other words, after a few generations, the ratios of the three genotypeswill be 0 25 0 50 0 25. : . : . .

!!!! Estimating a Transition Matrix

79. Among the 19 ones in the sequence there are 6 that are followed by ones and 13 followed by

twos so the probability of moving from one to one is 619

032≈ . , and the probability of moving to

State 2, given we are in State 1, is 1319

068≈ . . On the other hand, among the 20 twos (we don’t

count the very last 2), 12 are followed by a one and eight are followed by a two. Hence, we

would say that the probability of moving to State 1, given we are in State 2, is 1220

060= . , and the


probability of moving to State 2, given we are in State 2, is 820

040= . . Hence, we have the

transition matrix1 2

12

P = LNMOQP

0 32 0 600 68 0 40. .. .

.

!!!! Directed Graphs

80. (a) A =

L

N

MMMMMM

O

Q

PPPPPP

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

(b) A2

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

=

L

N

MMMMMM

O

Q

PPPPPPThe ijth entry in A2 gives the number of paths of length 2 from node i to node j.

!!!! Tournament Play

81. The tournament graph had adjacency matrix

T =

L

N

MMMMMM

O

Q

PPPPPP

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

.

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.

Second-order dominance can be determined from

T2

0 1 0 1 21 0 1 0 10 0 0 1 10 2 1 0 20 0 0 0 0

=

L

N

MMMMMM

O

Q

PPPPPP


For example, T2 tells us that Player 1 can dominate Player 5 in two second-order ways (bybeating either Player 2 or Player 4, both of whom beat Player 5). The sum

T T+ =

L

N

MMMMMM

O

Q

PPPPPP

2

0 2 1 1 31 0 1 1 20 1 0 1 21 2 2 0 30 0 0 0 0

,

gives the number of ways one player has beaten another both directly and indirectly. Rerankingplayers by sums of row elements of T T+ 2 can sometimes break a tie: In this case it does so andranks the players in order 4, 1, 2, 3, 5.

!!!! Suggested Journal Entry

82. Student Project

SECTION 3.2 Systems of Linear Equations 181

3.2 Systems of Linear Equations

!!!! Matrix-Vector Form

1.1 22 13 2

101

−L

NMMM

O

QPPPLNMOQP =L

NMMM

O

QPPP

xy

2.1 2 1 31 3 3 0

21

1

2

3

4

−LNM

OQPL

N

MMMM

O

Q

PPPP= LNMOQP

iiii

Augmented matrix = −L

NMMM

O

QPPP

1 2 12 1 03 2 1

Augmented matrix =−LNM

OQP

1 2 1 3 21 3 3 0 1

3.1 2 11 3 30 4 5

113

−−

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

rst

4. 1 2 3 01

2

3

−L

NMMM

O

QPPP=

xxx

Augmented matrix = −−

L

NMMM

O

QPPP

1 2 1 11 3 3 10 4 5 3

Augmented matrix = −1 2 3 0|

!!!! Solutions in R2

5. (A) 6. (B) 7. (C) 8. (B) 9. (A)

!!!! A Special Solution Set in R3

10. The three equationsx y zx y zx y z

+ + =+ + =+ + =

12 2 2 23 3 3 3

are equivalent to the single plane x y z+ + = 1, which can be written in parametric form by lettingy s= , z t= . We then have the parametric form 1− −( )s t s t s t, , : , any real numbersk p .

!!!! Reduced Row Echelon Form

11. RREF 12. Not RREF (not all zeros above leading ones)

13. Not RREF (leading nonzero element in row 2 is not 1; not all zeros above the leading ones)

14. Not RREF (row 3 does not have a leading one, nor does it move to the right; plus pivot columnshave nonzero entries other than the leading ones)

15. RREF 16. Not RREF (not all zeros above leading ones)


17. Not RREF (not all zeros above leading ones)

18. RREF 19. RREF

!!!! Gauss-Jordan Elimination

20. RREF. Starting with1 3 8 00 1 2 10 1 2 4

L

NMMM

O

QPPP

R R R3 3 21∗ = + −( )

1 3 8 00 1 2 10 0 0 3

L

NMMM

O

QPPP

R R3 313

∗ =

1 3 8 00 1 2 10 0 0 1

L

NMMM

O

QPPP.

This matrix is in row echelon form. To further reduce it to RREF we carry out the followingelementary row operations

R R R1 1 23∗ = + −( ) , R R R2 2 31∗ = + −( )

1 0 2 00 1 2 00 0 0 1

L

NMMM

O

QPPP← RREF.

Hence, we see the leading ones in this RREF form are in columns 1, 2, and 4, so the pivotcolumns of the original matrix are columns 1, 2, and 4 shown in bold and underlined as follows:

1 3 00 1 10 1 4

822

L

NMMM

O

QPPP.

21.0 0 2 2 22 2 6 14 4

−LNM

OQP

R R1 2↔

2 2 6 14 40 0 2 2 2−LNM

OQP


R R1 112

∗ =

1 1 3 7 20 0 2 2 2−LNM

OQP

R R2 212

∗ =

1 1 3 7 20 0 1 1 1−LNM

OQP .

The matrix is in row echelon form. To further reduce it to RREF we carry out the followingelementary row operation.

R R R1 1 23∗ = + −( )

1 1 0 4 50 0 1 1 1−LNM

OQP← RREF

pivot columns of the original matrix are first and third.

0 22 6

0 2 22 14 4

−LNM

OQP.

22.

1 0 02 4 65 8 120 8 12

L

N

MMMM

O

Q

PPPP

R R R2 2 12∗ = + −( ) , R R R3 3 15∗ = + −( )

1 0 00 4 60 8 120 8 12

L

N

MMMM

O

Q

PPPP

R R2 214

∗ =

1 0 00 1 3

20 8 120 8 12

L

N

MMMM

O

Q

PPPP


R R R3 3 28∗ = + −( ) , R R R4 4 28∗ = + −( )

1 0 00 1 3

20 0 00 0 0

L

N

MMMM

O

Q

PPPP← RREF .

This matrix is in both row echelon form and RREF form.

1 02 45 80 8

06

1212

L

N

MMMM

O

Q

PPPP

pivot columns of the original matrix are first and second.

23.1 2 3 13 7 10 42 4 6 2

L

NMMM

O

QPPP

R R R2 2 13∗ = + −( ) , R R R3 3 12∗ = + −( )

1 2 3 10 1 1 10 0 0 0

L

NMMM

O

QPPP← row echelon form.

The matrix is in row echelon form. To further reduce it to RREF, we carry out the followingelementary row operation.

R R R1 1 22∗ = + −( )

1 0 1 10 1 1 10 0 0 0

−L

NMMM

O

QPPP← RREF

pivot columns of the original matrix are first and second.1 23 72 4

3 110 46 2

L

NMMM

O

QPPP

.


!!!! Solving Systems

24.1 1 41 1 0−LNM

OQP

R R R2 2 11∗ = + −( )

1 1 40 2 4− −LNM

OQP

R R2 212

* = −

1 1 40 1 2LNMOQP

R R R1 1 21∗ = + −b g1 0 20 1 2LNM

OQP

x = 2, y = 2 (unique solution)

25.2 1 01 1 3−− −LNM

OQP

R R1 2↔

1 1 32 1 0

− −−LNM

OQP

R R R2 2 12∗ = + −( )

1 1 30 1 6

− −LNM

OQP

R R R1 1 21∗ = + b g

RREF 1 0 30 1 6LNMOQP

unique solution; x = 3, y = 6

26.1 1 1 00 1 1 1LNM

OQP

R R R1 1 21∗ = + −( )

RREF 1 0 0 10 1 1 1

−LNM

OQP

arbitrary (infinitely many solutions); x = −1, y z= −1


27.2 4 2 05 3 0 0

−LNM

OQP

R R1 112

∗ =

1 2 1 05 3 0 0

−LNM

OQP

R R R2 2 15∗ = + −( )

1 2 1 00 7 5 0

−−LNM

OQP

R R2 217

∗ = −

1 2 1 00 1 5

70

−−

LNMM

OQPP

R R R1 1 22∗ = + −( )

RREF 1 0 3

70

0 1 57

0−

L

NMMM

O

QPPP

nonunique solutions; x z= −37

, y z=57

, z is arbitrary

28.1 1 2 12 3 1 25 4 2 4

− −L

NMMM

O

QPPP

R R R2 2 12∗ = + −( ) , R R R3 3 15∗ = + −( )

1 1 2 10 5 5 00 9 12 1

− −

−

L

NMMM

O

QPPP

R R2 215

∗ =

1 1 2 10 1 1 00 9 12 1

− −

−

L

NMMM

O

QPPP


R R R1 1 2∗ = + , R R R3 3 29∗ = + −b g

1 0 1 10 1 1 00 0 3 1

−

−

L

NMMM

O

QPPP

R R3 313

∗ =

1 0 1 10 1 1 00 0 1 1

3

−

−

L

N

MMMM

O

Q

PPPPR R R1 1 3∗ = + , R R R2 2 31∗ = + −b g

RREF

1 0 0 23

0 1 0 13

0 0 1 13

−

L

N

MMMMMM

O

Q

PPPPPP

x = 23

, y = 13

, z = − 13

unique solution

29.1 4 5 02 1 8 9

−−LNM

OQP

R R R2 2 12∗ = + −( )

1 4 5 00 9 18 9

−−LNM

OQP

R R2 219

∗ = −

1 4 5 00 1 2 1

−− −

LNM

OQP

R R R1 1 24∗ = + −( )

RREF 1 0 3 40 1 2 1− −LNM

OQP

nonunique solutions; x x1 34 3= − , x x2 31 2= − + , x3 is arbitrary


30.1 0 1 22 3 5 43 2 1 4−

−

L

NMMM

O

QPPP

R R R2 2 12∗ = + −( ) , R R R3 3 13∗ = + −( )

1 0 1 20 3 3 00 2 4 2−

− −

L

NMMM

O

QPPP

R R2 213

∗ = −

1 0 1 20 1 1 00 2 4 2

−− −

L

NMMM

O

QPPP

R R R3 3 22∗ = + −( )

1 0 1 20 1 1 00 0 2 2

−− −

L

NMMM

O

QPPP

R R3 312

∗ = −

1 0 1 20 1 1 00 0 1 1

−L

NMMM

O

QPPP

R R R1 1 31∗ = + −( ) , R R R2 2 3∗ = +

RREF 1 0 0 10 1 0 10 0 1 1

L

NMMM

O

QPPP

unique solution;x y z= = = 1

31.1 1 1 01 1 0 01 2 1 0

−

−

L

NMMM

O

QPPP

R R R2 2 11∗ = + −( ) , R R R3 3 11∗ = + −( )

1 1 1 00 2 1 00 3 2 0

−−−

L

NMMM

O

QPPP


R R2 212

∗ =

1 1 1 00 1 1

20

0 3 2 0

−−

−

L

NMMM

O

QPPP

R R R1 1 2∗ = + , R R R3 3 23∗ = + −b g

1 0 12

0

0 1 12

0

0 0 12

0

−

−

L

N

MMMMMM

O

Q

PPPPPPR R3 32∗ = −( )

1 0 12

0

0 1 12

0

0 0 1 0−

L

N

MMMMMM

O

Q

PPPPPP

R R R1 1 312

∗ = + −FHGIKJ , R R R2 2 3

12

∗ = +

RREF 1 0 0 00 1 0 00 0 1 0

L

NMMM

O

QPPP

unique solution; x y z= = = 0

32.1 1 2 02 1 1 04 1 5 0

−L

NMMM

O

QPPP

R R R2 2 12∗ = + −( ) , R R R3 3 14∗ = + −( )

1 1 2 00 3 3 00 3 3 0− −− −

L

NMMM

O

QPPP

R R2 213

∗ = −

1 1 2 00 1 1 00 3 3 0− −

L

NMMM

O

QPPP


R R R1 1 21∗ = + −( ) , R R R3 3 23∗ = + ( )

RREF 1 0 1 00 1 1 00 0 0 0

L

NMMM

O

QPPP

x z y z= − = −, , z is arbitrary

33.1 1 2 12 1 1 24 1 5 4

−L

NMMM

O

QPPP

R R R2 2 12∗ = + −( ) , R R R3 3 14∗ = + −( )

1 1 2 10 3 3 00 3 3 0

− −− −

L

NMMM

O

QPPP

R R2 213

∗ = −

1 1 2 10 1 1 00 3 3 0− −

L

NMMM

O

QPPP

R R R1 1 21∗ = + −( ) , R R R3 3 23∗ = + ( )

RREF 1 0 1 10 1 1 00 0 0 0

L

NMMM

O

QPPP

nonunique solutions; x z= −1 , y z= − , z is arbitrary

!!!! The RREF Example

34. Starting with the augmented matrix, we carry out the following steps

1 0 2 0 1 4 80 2 0 2 4 6 60 0 1 0 0 2 23 0 0 1 5 3 120 2 0 0 0 0 6

− − −

− −

L

N

MMMMMM

O

Q

PPPPPP


R R R4 4 13∗ = + −( )

1 0 2 0 1 4 80 2 0 2 4 6 60 0 1 0 0 2 20 0 6 1 2 9 120 2 0 0 0 0 6

− − −

− − −− −

L

N

MMMMMM

O

Q

PPPPPP

R R2 212

∗ ∗=

1 0 2 0 1 4 80 1 0 1 2 3 30 0 1 0 0 2 20 0 6 1 2 9 120 2 0 0 0 0 6

− − −

− − −− −

L

N

MMMMMM

O

Q

PPPPPP(we leave these last steps for the reader)

1 0 0 0 1 0 40 1 0 0 0 0 30 0 1 0 0 2 20 0 0 1 2 3 00 0 0 0 0 0 0

L

N

MMMMMM

O

Q

PPPPPP

.

!!!! More Equations Than Variables

35. Converting the augmented matrix to RREF yields

3 5 0 13 7 3 80 5 0 50 2 3 71 4 1 1

1 0 0 20 1 0 10 0 1 30 0 0 00 0 0 0

−

L

N

MMMMMM

O

Q

PPPPPP

→−

L

N

MMMMMM

O

Q

PPPPPPconsistent system; unique solution x y= = −2, 1 , z = 3.

!!!! Consistency

36. A homogeneous system Ax = 0 always has at least one solution, namely the zero vector x = 0 .

!!!! Homogeneous Systems

37. The equations are

w x zy z

− + =+ =

2 5 02 0


If we let x r z s= = and , we can solve y s= −2 , w r s= −2 5 . The solution is a plane in R4 given

by

wxyz

r sr

ss

r s

L

N

MMMM

O

Q

PPPP=

−

−

L

N

MMMM

O

Q

PPPP=

L

N

MMMM

O

Q

PPPP+

−

−

L

N

MMMM

O

Q

PPPP

2 5

2

2100

5021

,

r, s any real numbers.

38. The equations are

x zy

+ ==

2 00

If we let z s= , we have x s= −2 and hence the solution is a line in R3 given by

xyz

s

ss

L

NMMM

O

QPPP=

−L

NMMM

O

QPPP=

−L

NMMM

O

QPPP

20

201

.

39. The equation is

x x x x1 2 3 44 3 0 0− + + = .

If we let x r2 = , x s3 = , x t4 = , we can solve

x x x r s1 2 34 3 4 3= − = − .

Hence

xxxx

r srst

r s t

1

2

3

4

4 3 4100

3010

0001

L

N

MMMM

O

Q

PPPP=

−L

N

MMMM

O

Q

PPPP=

L

N

MMMM

O

Q

PPPP+

−L

N

MMMM

O

Q

PPPP+

L

N

MMMM

O

Q

PPPP

where r, s, t are any real numbers.

!!!! Seeking Consistency

40. k ≠ 4

41. Any k will produce a consistent system

42. k ≠ ±1

43. The system is inconsistent for all k because the last two equations are parallel and distinct.


!!!! Homogeneous versus Nonhomogeneous

44. For the nonhomogeneous equation of Problem 33, we can write the solution as

x =L

NMMM

O

QPPP=

−−L

NMMM

O

QPPP+L

NMMM

O

QPPP

xyz

c111

100

.

For the homogeneous equation of Problem 32, we can write the solution as

xh =−−L

NMMM

O

QPPP

c111

where c is an arbitrary constant. In other words, the general solution of the nonhomogeneousalgebraic system, Problem 33 is the sum of the solutions of the associated homogeneous equationplus a particular solution just as it was in Problem 32 for nonhomogeneous linear differentialequations.

!!!! Equivalence of Systems

45. Inverse of R Ri j↔ : The operation that puts the system back the way it was is R Rj i↔ . In other

words, the operation R R3 1↔ will undo the operation R R1 3↔ .

Inverse of R cRi i= : The operation that puts the system back the way it was is Rc

Ri i=1 .

In other words, the operation R R1 113

= will undo the operation R R1 13= .

Inverse of R R cRi i j= + : The operation that puts the system back is R R cRi i j= − . Inother words R R cRi i j= − will undo the operation R R cRi i j= + .This is clear because if we add

cRj to row i and then subtract cRj from row i, then row i will be unchanged. For example,

RR

1

2

1 2 32 1 1LNM

OQP , R R R1 1 22∗ = + ,

5 4 52 1 1LNMOQP , R R R1 1 22∗ = + −( ) ,

1 2 32 1 1LNM

OQP .

!!!! Electrical Circuits

46. (a) There are four junctions in this multicircuit, and Kirchhoff’s current law states that thesum of the currents flowing in and out of any junction is zero. The given equationssimply state this fact for the four junctions J1, J2 , J3, and J4 , respectively. Keep in

mind that if a current is negative in sign, then the actual current flows in the directionopposite the indicated arrow.


(b) The augmented system is

1 1 1 0 0 0 00 1 0 1 1 0 00 0 1 1 0 1 01 0 0 0 1 1 0

− −−

− −−

L

N

MMMM

O

Q

PPPP.

Carrying out the three elementary row operations, we can transform this system to RREF

1 0 0 0 1 1 00 1 0 1 1 0 00 0 1 1 0 1 00 0 0 0 0 0 0

− −−

− −

L

N

MMMM

O

Q

PPPP.

Solving for the lead variables I1, I2 , I3 in terms of the free variables I4 , I5, I6, we have

I I I1 5 6= + , I I I2 4 5= − + , I I I3 4 6= + . In matrix form, this becomes

IIIIII

I I I

1

2

3

4

5

6

4 5 6

011100

110010

101001

L

N

MMMMMMM

O

Q

PPPPPPP

=

−L

N

MMMMMMM

O

Q

PPPPPPP

+

L

N

MMMMMMM

O

Q

PPPPPPP

+

L

N

MMMMMMM

O

Q

PPPPPPP

where I1, I2 , and I3 are arbitrary. In other words, we need three of the six currents to

uniquely specify the remaining ones.

!!!! More Circuit Analysis

47. I I II I I1 2 3

1 2 3

00

− − =− + + =

48. I I I II I I I1 2 3 4

1 2 3 4

00

− − − =− + + + =

49. I I I II I I

I I I

1 2 3 4

1 2 5

3 4 5

000

− − − =− + + =

+ − =

50. I I II I I

I I II I I

1 2 3

2 4 5

3 4 6

1 5 6

0000

− − =− − =+ − =

− + + =

!!!! Solutions in Tandem

51. There is nothing surprising here. By placing the two right-hand sides in the last two columns ofthe augmented matrix, the student is simply organizing the material effectively. Neither of thelast two columns affects the other column, so the last two columns will contain the respectivesolutions.


!!!! Tandem with a Twist

52. (a) We place the right-hand sides of the two systems in the last two columns of theaugmented matrix

1 1 0 3 50 2 1 2 4LNM

OQP .

Reducing this matrix to RREF, yields

1 0 12

2 3

0 1 12

1 2

−L

NMMM

O

QPPP

.

Hence, the first system has solutions x z= +2 12

, y z= −1 12

, z arbitrary, and the second

system has solutions x z= +3 12

, y z= −2 12

, z arbitrary.

(b) If you look carefully, you will see that the matrix equation

1 1 00 2 1

3 52 4

11 12

21 22

31 32

LNM

OQPL

NMMM

O

QPPP= LNMOQP

x xx xx x

is equivalent to the two systems of equations

1 1 00 2 1

32

1 1 00 2 1

54

11

21

31

12

22

32

LNM

OQPL

NMMM

O

QPPP= LNMOQP

LNM

OQPL

NMMM

O

QPPP= LNMOQP

xxxxxx

.

We saw in part (a) that the solution of the system on the left was

x x11 312 12

= + , x x21 311 12

= − , x31 arbitrary,

and the solution of the system on the right was

x x12 323 12

= + , x x22 322 12

= − , x32 arbitrary.


Putting these solutions in the columns of our unknown matrix X and calling x31 = α ,

x32 = β , we have

X =L

NMMM

O

QPPP=

+ +

− −

L

N

MMMMMM

O

Q

PPPPPP

x xx xx x

11 12

21 22

31 32

2 12

3 12

1 12

2 12

α β

α β

α β

.

!!!! Two Thousand Year Old Problem

53. Letting A1 and A2 be the areas of the two fields in square yards, we are given the two equations

A A

A A1 2

1 2

1800 squar23

12

1100 bushe

+ =

+ =

e yards

ls

The areas of the two fields are 1200 and 600 square yards.

!!!! Computerizing

54. 2 2× Case. To solve the 2 2× system

a x a x ba x a x b

11 1 12 2 1

21 1 22 2 2

+ =+ =

we start by forming the augmented matrix

A b =LNM

OQP

a aa a

bb

11 12

21 22

1

2

.

Step 1: If a11 1≠ , factor it out of row 1. If a11 0= , interchange the rows and then factor the new

element in the 11 position out of the first row. (This gives a 1 in the first position of the firstrow.)

Step 2: Subtract from the second row the first row times the element in the 21 position ofthe new matrix. (This gives a zero in the first position of the second row).

Step 3: Factor the element in the 22 position from the second row of the new matrix. Ifthis element is zero and the element in the 23 position is nonzero, there are no solutions. If boththis element is zero and the element in the 23 position is zero, then there are an infinite numberof solutions. To find them write out the equation corresponding to the first row of the finalmatrix. (This gives a 1 in the first nonzero position of the second row).


Step 4: Subtract from the first row the second row times the element in the 12 position ofthe new matrix. This operation will yield a matrix of the form matrix

1 00 1

1

2

rr

LNM

OQP

where x r1 1= , x r2 2= . (This gives a zero in the second position of the first row.)

55. The basic idea is to formalize a strategy like that used in Example 3. The augmented matrix forAx b= is

a a a ba a a ba a a b

11 12 13 1

21 22 23 2

31 32 33 3

L

NMMM

O

QPPP

.

A pseudocode might begin:

1. To get a one in first place in row 1, multiply every element of row 1 by 1

11a.

2. To get a zero in first place in row 2, replace row 2 by

row row 12 21− ( )a .$

!!!! Suggested Journal Entry I

56. Student Project

!!!! Suggested Journal Entry II

57. Student Project


3.3 The Inverse of a Matrix

!!!! Checking Inverses

1.5 32 1

1 32 5

5 1 3 2 5 3 3 52 1 1 2 2 3 1 5

1 00 1

LNMOQP−

−LNMOQP =

( ) −( ) + ( )( ) ( )( ) + ( ) −( )( ) −( ) + ( )( ) ( )( ) + ( ) −( )LNM

OQP =LNMOQP

2.2 42 0

0 12

14

14

2 0 4 14

2 12

4 14

2 0 0 14

2 12

0 14

1 00 1

−LNMOQP −

L

NMMM

O

QPPP=( )( ) + −( ) −FH IK ( ) + −( )

( )( ) + ( ) −FHIK ( ) + ( )

L

N

MMM

O

Q

PPP= LNMOQP

3. Direct multiplication as in Problems 1–2.

4. Direct multiplication as in Problems 1–2.

!!!! Matrix Inverses

5. We reduce A I to RREF.

2 0 1 01 1 0 1LNM

OQP

R R1 112

∗ =

1 0 12

01 1 0 1

LNMM

OQPP

R R R2 2 11∗ = + −( )

1 0 12

0

0 1 12

1−

L

NMMM

O

QPPP

.

Hence, A− =−

L

NMMM

O

QPPP

1

12

0

12

1.

6. We reduce A I to RREF.

1 3 1 02 5 0 1LNM

OQP

SECTION 3.3 The Inverse of a Matrix 199

R R R2 2 12∗ = + −( )

1 3 1 00 1 2 1− −LNM

OQP

R R2 21∗ = −( )

1 3 1 00 1 2 1−LNM

OQP

R R R1 1 23∗ = + −( )

1 0 5 30 1 2 1

−−

LNM

OQP.

Hence, A− =−

−LNMOQP

1 5 32 1

.

7. Starting with

A I = −− −

L

NMMM

O

QPPP

0 1 1 1 0 05 1 1 0 1 03 3 3 0 0 1

R R1 2↔

5 1 1 0 1 00 1 1 1 0 03 3 3 0 0 1

−

− −

L

NMMM

O

QPPP

R R1 115

∗ =

1 15

15

0 15

00 1 1 1 0 03 3 3 0 0 1

−

− −

L

N

MMMM

O

Q

PPPP

R R R3 3 13∗ = + −( )

1 15

15

0 15

0

0 1 1 1 0 00 18

5125

0 35

1

−

− − −

L

N

MMMMM

O

Q

PPPPP


R R R1 1 215

∗ = + −FH IK , R R R3 3 2185

∗ = +

1 0 25

15

15

0

0 1 1 1 0 00 0 6

5185

35

1

− −

−

L

N

MMMMM

O

Q

PPPPP

R R3 356

∗ =

1 0 25

15

15

00 1 1 1 0 00 0 1 3 1

256

− −

−

L

N

MMMM

O

Q

PPPP

R R R1 1 325

∗ = + , R R R2 2 31∗ = + −( )

1 0 0 1 0 13

0 1 0 2 12

56

0 0 1 3 12

56

− −

−

L

N

MMMMMM

O

Q

PPPPPP

.

Hence, A− = − −

−

L

N

MMMMMM

O

Q

PPPPPP

1

1 0 13

2 12

56

3 12

56

.

8. Interchanging the first and third rows, we get

A I

I A

A

=L

NMMM

O

QPPP

B

=L

NMMM

O

QPPP

=L

NMMM

O

QPPP

−

−

0 0 1 1 0 00 1 0 0 1 01 0 0 0 0 1

1 0 0 0 0 10 1 0 0 1 00 0 1 1 0 00 0 10 1 01 0 0

1

1


9. Dividing the first row by k gives

A I

I A

=L

NMMM

O

QPPP

B

=

L

N

MMMM

O

Q

PPPP−

k

k

0 0 1 0 00 1 0 0 1 00 0 1 0 0 1

1 0 0 1 0 00 1 0 0 1 00 0 1 0 0 1

1

Hence A− =

L

N

MMMM

O

Q

PPPP1

1 0 00 1 00 0 1

k.

10. A I = −L

NMMM

O

QPPP→ − − −L

NMMM

O

QPPP→ −L

NMMM

O

QPPP

→ −− −

L

NMMM

O

QPPP→ −

− −

L

NMMM

O

QPPP→

−−

− −

1 0 1 1 0 01 1 0 0 1 00 2 1 0 0 1

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

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

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

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

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

L

NMMM

O

QPPP

Hence A− =−−

− −

L

NMMM

O

QPPP

11 2 11 1 12 2 1

.

11.

1 0 0 0 1 0 0 00 1 0 0 1 0 00 0 1 0 0 0 1 00 0 0 1 0 0 0 1

1 0 0 0 1 0 0 00 1 0 0 0 1 00 0 1 0 0 0 1 00 0 0 1 0 0 0 1

k kL

N

MMMM

O

Q

PPPP→

−L

N

MMMM

O

Q

PPPP

Hence

A− =−

L

N

MMMM

O

Q

PPPP1

1 0 0 00 1 00 0 1 00 0 0 1

k.


12.

1 0 1 1 1 0 0 00 0 1 0 0 1 0 01 1 1 0 0 0 1 01 0 0 2 0 0 0 1

1 0 1 1 1 0 0 00 0 1 0 0 1 0 00 1 0 1 1 0 1 00 0 1 1 1 0 0 1

1 0 1 1 1 0 0 00 1 0 1 1 0 1 00 0 1 0 0 1 0 00 0 1 1 1 0 0 1

1 0 0 1 1 1 0 00 1 0 1 1 0 1 00 0 1 0 0 1 0 00 0 0 1 1 1 0 1

L

N

MMMM

O

Q

PPPP→

− −− −

L

N

MMMM

O

Q

PPPP→

− −

− −

L

N

MMMM

O

Q

PPPP

→

−− −

−

L

N

MMMM

O

Q

PPPP→

− −−

−

L

N

MMMM

O

Q

PPPP

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

Hence

A− =

− −−

−

L

N

MMMM

O

Q

PPPP1

2 2 0 12 1 1 10 1 0 01 1 0 1

.

13. Starting with the augmented matrix

A I =−

−−

L

N

MMMM

O

Q

PPPP→

−−

− −

L

N

MMMM

O

Q

PPPP

→−

−− −

L

N

MMMM

O

Q

PPPP→

−−

−

1 0 0 0 1 0 0 00 1 0 0 0 1 0 00 1 2 0 0 0 1 01 1 3 3 0 0 0 1

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

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

1 0 0 0 1 0 0 00 1 0 0 0 1 0 00 0 2 0 0 1 1 00 0 3 3 1 −

L

N

MMMM

O

Q

PPPP

→−− −

− −

L

N

MMMM

O

Q

PPPP→

−− −

−

L

N

MMMMMM

O

Q

PPPPPP

→−− −

−

L

N

MMMMMM

O

Q

PPP

1 0 1

1 0 0 0 1 0 0 00 1 0 0 0 1 0 00 0 1 0 0 1

212

00 0 3 3 1 1 0 1

1 0 0 0 1 0 0 00 1 0 0 0 1 0 00 0 1 0 0 1

212

0

0 0 0 3 1 12

32

1

1 0 0 0 1 0 0 00 1 0 0 0 1 0 00 0 1 0 0 1

212

0

0 0 0 1 13

16

12

13

PPP

.


Hence

A− =−− −

−

L

N

MMMMMM

O

Q

PPPPPP

1

1 0 0 00 1 0 00 1

212

013

16

12

13

.

!!!! Inverse of the 2 2× Matrix

14. Verify A A I AA− −= =1 1 . We have

A A I

AA I

−

−

=−

−−LNM

OQPLNMOQP = −

−−

LNM

OQP =

=LNMOQP −

−−LNM

OQP = −

−−

LNM

OQP =

1

1

1 1 00

1 1 00

ad bcd bc a

a bc d ad bc

ad bcad bc

a bc d ad bc

d bc a ad bc

ad bcad bc

Note that we must have A = − ≠ad bc 0.

!!!! Brute Force

15. To find the inverse of

1 31 2LNMOQP,

we seek the matrix

a bc dLNMOQP

that satisfies

a bc dLNMOQPLNMOQP =LNMOQP

1 31 2

1 00 1

.

Multiplying this out we get the equationsa b

a bc d

c d

+ =+ =+ =+ =

13 2 0

03 2 1.

The top two equations involve a and b, and the bottom two involve c and d, so we write the twosystems


1 13 2

10

1 13 2

01

LNMOQPLNMOQP =LNMOQP


abcd

.

Solving each system, we get

ab

cd

LNMOQP =LNMOQPLNMOQP = −

−−LNMOQPLNMOQP =

−LNMOQP

LNMOQP =LNMOQPLNMOQP = −

−−LNMOQPLNMOQP = −LNMOQP

−

−

1 13 2

10

11

2 13 1

10

23

1 13 2

01

11

2 13 1

01

11

1

1

.

Because a and b are the elements in the first row of A−1, and c and d are the elements in thesecond row, we have

A−−

= LNMOQP =

−−

LNMOQP

111 3

1 22 31 1

.

!!!! Unique Inverse

16. We show that if B and C are both inverse of A, then B C= . Because B is an inverse of A, we canwrite BA I= . If we now multiply both sides on the right by C, we get

BA C IC C( ) = = .

But then we have

BA C B AC BI B( ) = ( ) = = , so B C= .

!!!! Invertible Matrix Method

17. Using the inverse found in Problem 6, yields

xx

1

2

1 5 32 1

410

5018

LNMOQP = =

−−

LNMOQP−LNMOQP = −LNMOQP

−A b .

!!!! Solution by Invertible Matrix

18. Using the inverse found in Problem 7, yields

xyz

L

NMMM

O

QPPP= = − −

−

L

N

MMMMMM

O

Q

PPPPPP

L

NMMM

O

QPPP= −L

NMMM

O

QPPP

−A b1

1 0 13

2 12

56

3 12

56

520

59

14.


!!!! Noninvertible 2 2× Matrices

19. From Problem 14, the inverse of

A = LNMOQP

a bc d

can be written as

A− =−

−−LNMOQP

1 1ad bc

d bc a

,

which does not exist when ad bc− = 0 or ad bc= . Also, if we reduce A to RREF we get

1

0

ba

ad bca−

L

N

MMM

O

Q

PPP,

which says that the matrix is invertible when ad bca−

≠ 0 , or equivalently when ad bc≠ .

!!!! Cancellation Works

20. Given that AB AC= and A are invertible, we premultiply by A−1, getting

A AB A AC− −=1 1

or

IB IC= or B C= .

!!!! An Inverse

21. If A is an invertible matrix and AB I= , then we can premultiply each side of the equation byA−1 getting

A AB A I− −( ) =1 1

or

AA B A− −=1 1b g .

Hence, B A= −1.

!!!! Invertible Product

22. If AB is an invertible matrix, then there exists a matrix X that satisfies

AB X IA BX I( ) =( ) = .


Hence, A has a right inverse BX. We also know

X AB IAB X

( ) =

= −1.

If we further assume that B is invertible then we can postmultiply this last equation by B−1,which yields

A X B BX= = ( )− − −1 1 1.

Now multiplying on the left by BX we have BX A I( ) = , which says that BX is also a left inverse

of A. Hence, the inverse of A exists and is BX.

!!!! Inconsistency

23. If Ax b= is inconsistent for some vector b, then A−1 does not exist because if A−1 did exist,then x A b= −1 exists for all b.

!!!! Elementary Matrices

24. (a) Eint =L

NMMM

O

QPPP

0 1 01 0 00 0 1

(b) Erepl =L

NMMM

O

QPPP

1 0 00 1 0

0 1k(c) Escale =

L

NMMM

O

QPPP

1 0 00 00 0 1

k

!!!! Invertibility of Elementary Matrices

25. Because the inverse of any elementary row operation is also an elementary row operation, andbecause elementary matrices are constructed from elementary row operations starting with theidentity matrix, we can convert any elementary row operation to the identity matrix byelementary row operations.

For example, the inverse of Eint can be found by performing the operation R R1 2↔ on

the augmented matrix

E Iint =L

NMMM

O

QPPP→L

NMMM

O

QPPP

0 1 0 1 0 01 0 0 0 1 00 0 1 0 0 1

1 0 0 0 1 00 1 0 1 0 00 0 1 0 0 1

.

Hence, Eint− =L

NMMM

O

QPPP

10 1 01 0 00 0 1

. In other words E Eint int= −1 . We leave finding Erepl−1 and Escale

−1 for the

reader.


!!!! Leontief Model

26. T = LNMOQP

05 00 05.

., d = LNM

OQP

1010

The basic equation is

Total Output = External Demand + Internal Demand,

so we have

xx

xx

1

2

1

2

1010

05 00 05

LNMOQP =LNMOQP +LNMOQPLNMOQP

..

.

Solving these equations yields x x1 2 20= = . This should be obvious because for every 20 units of

product each industry produces, 10 goes back into the industry to produce the other 10.

27. T = LNMOQP

0 010 2 0

..

, d = LNMOQP

1010



so we have

xx

xx

1

2

1

2

1010

0 0102 0

LNMOQP =LNMOQP +LNMOQPLNMOQP

..

.

Solving these equations yields x1 112= . , x2 12 2= . .

28. T = LNMOQP

02 0505 02. .. .

, d = LNMOQP

1010



so we have

xx

xx

1

2

1

2

1010

02 0505 02

LNMOQP =LNMOQP +LNM

OQPLNMOQP

. .. .

.

Solving these equations yields x1 3313

= , x2 3313

= .

29. T = LNMOQP

05 0201 03. .. .

, d = LNMOQP

5050




so we have

xx

xx

1

2

1

2

5050

05 0201 03

LNMOQP =LNMOQP +LNM

OQPLNMOQP

. .

. ..

Solving these equations yields x1 1364= . , x2 909= . .

!!!! How Much Is Left Over?

30. The basic demand equation is


so we have

150250

03 0 405 03

150250

1

2

LNMOQP =LNMOQP +LNM

OQPLNMOQP

dd

. .

. ..

Solving for d1, d2 yields d1 5= , d2 100= .

!!!! Israeli Economy

31. (a) I T− = − −− −

L

NMMM

O

QPPP

0 70 000 0 00010 080 0 20005 001 0 98

. . .. . .. . .

(b) I T− =L

NMMM

O

QPPP

−b g 1143 0 00 0 000 20 125 0 260 07 0 01 102

. . .

. . .

. . .

(c) x I T d= −( ) =L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

−1143 000 000020 125 026007 0 01 102

140 00020 0002,000

200520040

. . .

. . .

. . .

,,

$200,$53,$12,


32. Student Project

SECTION 3.4 Determinants and Cramer’s Rule 209

3.4 Determinants and Cramer’s Rule

!!!! Calculating Determinants

1. Expanding by cofactors down the first column we get0 7 92 1 15 6 2

01 16 2

27 96 2

57 91 1

0− =−

− +−

= .

2. Expanding by cofactors across the middle row we get1 2 30 1 01 0 3

02 30 3

11 31 3

01 21 0

6−

=−

+−

− = − .

3. Expanding by cofactors down the third column we get

1 3 0 20 1 1 51 2 1 71 1 0 6

11 3 21 2 71 1 6

1 3 20 1 51 1 6

6 6 12

−−

− −−

=−

− −−

+−

−= + = .

4. Expanding by cofactors across the third row we get

1 4 2 24 7 3 53 0 8 05 1 6 9

34 2 27 3 51 6 9

81 4 24 7 55 1 9

3 14 8 250 2042

− −−

− −

=− −

−−

+− −

− −= ( ) + ( ) = .

!!!! Find the Properties

5. Subtract the first row from the second row in the matrix in the first determinant to get the matrixin the second determinant.

6. Factor out 3 from the second row of the matrix in the first determinant to get the matrix in thesecond determinant.

7. Interchange the two rows of the matrix.

!!!! Basketweave for 3 3×

8. Direct computation as in Problems 1–4.


9.0 7 92 1 15 6 2

0 35 108 45 0 28 0− = − + − − − =

10.1 2 30 1 01 0 3

3 0 0 3 0 0 6−

= − + + − − − = −

11. By an extended basketweave hypothesis,

0 1 1 01 1 0 10 0 0 10 1 1 0

0 0 0 0 0 0 1 0 1= + + + − − − − = − .

However, the determinant is clearly 0 (because rows 1 equals row 4), so the basketweave methoddoes not generalize to dimensions higher than 3.

!!!! Triangular Determinants

12. We verify this for 4 4× matrices. Higher-order matrices follow along the same lines. Given theupper-triangular matrix

A =

a a a aa a a

a aa

11 12 13 14

22 23 24

33 34

44

00 00 0 0

,

we expand down the first column, getting

a a a aa a a

a aa

aa a a

a aa

a aa a

aa a a a

11 12 13 14

22 23 24

33 34

44

11

22 23 24

33 34

44

11 2233 34

4411 22 33 44

00 00 0 0

00 0

0= = = .

!!!! Think Diagonal

13. The matrix is upper triangular, hence the determinant is the product of the diagonal elements−

= −( )( )( ) = −3 4 00 7 60 0 5

3 7 5 105.


14. The matrix is a diagonal matrix, hence the determinant is the product of the diagonal elements.

4 0 00 3 00 0 1

2

4 3 12

6− = ( ) −( ) = − .

15. The matrix is lower triangular, hence the determinant is the product of the diagonal elements.

1 0 0 03 4 0 00 5 1 0

11 0 2 2

1 4 1 2 8−

−−

= ( )( ) −( )( ) = − .

16. The matrix is upper triangular, hence the determinant is the product of the diagonal elements.

6 22 0 30 1 0 40 0 13 00 0 0 4

6 1 13 4 312

−−

= ( ) −( )( )( ) = − .

!!!! Invertibility Test

17. The matrix does not have an inverse because its determinant is zero.

18. The matrix has an inverse because its determinant is nonzero.




!!!! Product Verification

21. A

B

AB

A 2

B

AB

= LNMOQP

= LNMOQP

= LNMOQP

= LNMOQP = −

= LNMOQP =

= LNMOQP = −

1 23 41 01 13 27 41 23 41 01 1

1

3 27 4

2

Hence AB A B= .

22. A A

B B

AB AB

=L

NMMM

O

QPPP⇒ = −

= −−

L

NMMM

O

QPPP⇒ = −

=−

−

L

NMMM

O

QPPP⇒ =

0 1 01 0 01 2 2

2

1 2 31 2 00 1 1

7

1 2 01 2 31 8 1

14

Hence AB A B= .

!!!! Determinant of an Inverse

23. We have

1 1 1= = =− −I AA A A

and hence AA

− =1 1 .

!!!! Do Determinants Commute?

24. Because

AB A B B A BA= = = ,

then A B is a product of real or complex numbers.

!!!! Determinant of Similar Matrices

25. The key to the proof lies in the determinant of a product of matrices. If

A P BP= −1 ,

we use the general properties

AA

− =1 1 , AB A B=


shown in Problems 23 and 24, and write

A P BP P B PP

B P BP

P B= = = = =− −1 1 1 1 .

!!!! Determinant of An

26. (a) If An = 0 for some integer n, we have

A An n= = 0

or

A = 0.

Hence, A is noninvertible.

(b) If An ≠ 0 for some integer n, then

A An n= ≠ 0

for some integer n. This implies A ≠ 0 , so A is invertible. In other words, for everymatrix A either An = 0 for all positive integers n or it is never zero.

!!!! Determinants of Sums

27. An example is

A = LNMOQP

1 00 1

, B =−

−LNMOQP

1 00 1

,

so

A B+ = LNMOQP

0 00 0

,

which has the determinant

A B+ = 0,

whereas

A B= = 1,

so A B+ = 2 . Hence,

A B A B+ ≠ + .


!!!! Determinants of Sums Again

28. Letting

A = LNMOQP

1 10 0

, B =− −LNMOQP

1 10 0

,

we get

A B+ = LNMOQP

0 00 0

.

Thus A B+ = 0. Also, we have A = 0, B = 0 , so

A B+ = 0.

Hence,

A B A B+ = + .

!!!! Scalar Multiplication

29. For a 2 2× matrix, we see

ka kaka ka

k a a k a a ka aa a

11 12

21 22

211 22

221 12

2 11 12

21 22= − = .

For an n n× matrix, A, we can factor a k out of each row getting k knA A= .

!!!! Inversion by Determinants

30. Given the matrix

A =L

NMMM

O

QPPP

1 0 22 2 31 1 1

the matrix of minors can easily be computed and is

M =− −− −− −

L

NMMM

O

QPPP

1 1 02 1 14 1 2

.

The matrix of cofactors ~A , which we get by multiplying the minors by −( ) +1 i j , is given by

~A M= −( ) =−

− −−

L

NMMM

O

QPPP

+11 1 02 1 14 1 2

i j .


Taking the transpose of this matrix gives

~AT =− −

−−

L

NMMM

O

QPPP

1 2 41 1 10 1 2

.

Computing the determinant of A, we get A = −1. Hence, we have the inverse

AA

A− = =−

− −−

L

NMMM

O

QPPP

1 11 2 41 1 10 1 2

~ T .

!!!! Determinants of Elementary Matrices

31. (a) If we interchange the rows of the 2 2× identity matrix, we change the sign of thedeterminant because

1 00 1

1= , 0 11 0

1= − .

For a 3 3× matrix if we interchange the first and second rows, we get0 1 01 0 00 0 1

1= − .

You can verify yourself that if any two rows of the 3 3× identity matrix areinterchanged, the determinant is –1.

For a 4 4× matrix suppose the ith and jth rows are interchanged and that wecompute the determinant by expanding by minors across one of the rows that was notinterchanged. (We can always do this.) The determinant is then

A M M M M= − + −a a a a11 11 12 12 13 13 14 14 .

But the minors M11, M12 , M13 , M14 are 3 3× matrices, and we know each of these

determinants is –1 because each of these matrices is a 3 3× elementary matrix with tworows changed from the identity matrix. Hence, we know 4 4× matrices with two rowsinterchanged from the identity matrix have determinant –1. The idea is to proceedinductively from 4 4× matrices to 5 5× matrices and so on.

(b) The matrix1 0 0

1 00 0 1kL

NMMM

O

QPPP


shows what happens to the 3 3× identity matrix if we add k times the 1st row to the 2ndrow. If we expand this matrix by minors across any row we see that the determinant isthe product of the diagonal elements and hence 1. For the general n n× matrix adding ktimes the ith row to the jth row places a k in the jith position of the matrix with all otherentries looking like the identity matrix. This matrix is an upper-triangular matrix, and itsdeterminant is the product of elements on the diagonal or 1.

(c) Multiplying a row, say the first row, by k of a 3 3× matrixk 0 00 1 00 0 1

L

NMMM

O

QPPP

and expanding by minors across any row will give a determinant of k. Higher-ordermatrices give the same result.

!!!! Determinant of a Product

32. (a) If A is not invertible then A = 0. If A is not invertible then neither is AB, so AB = 0.Hence, it yields AB A B= because both sides of the equation are zero.

(b) We first show that EA E A= for elementary matrices E. An elementary matrix is one

that results in changing the identity matrix using one of the three elementary operations.There are three kinds of elementary matrices. In the case when E results in multiplying arow of the identity matrix I by a constant k, we have:

EA

A E A

=

L

N

MMMM

O

Q

PPPP⋅

L

N

MMMM

O

Q

PPPP=

L

N

MMMM

O

Q

PPPP= =

1 0 00 0

0 0 1

11 12 1

21 22 2

1 2

11 12 1

21 22 2

1 2

!!

! ! ! !!

!!

! ! ! !!

!!

! ! ! !!

ka a aa a a

a a a

a a aka ka ka

a a ak

n

n

n n nn

n

n

n n nn

In those cases when E is a result of interchanging two rows of the identity or by adding amultiple of one row to another row, the verification follows along the same lines.

Now if A is invertible it can be written as the product of elementary matrices

A E E E= −p p 1 1 … .

If we postmultiply this equation by B, we get

AB E E E B= −p p 1 1 … ,

so

AB E E E B E E E B A B= = =− −p p p p1 1 1 1 … … .


!!!! Cramer’s Rule

33. x yx y+ =+ =

2 22 5 0

To solve this system we write it in matrix form as

1 22 5

20


xy

.

Using Cramer’s rule, we compute the determinants

A

A

A

= =

= =

= = −

1 22 5

1

2 20 5

10

1 22 0

4

1

2 .

Hence, the solution is

x

y

= = =

= = − = −

AAAA

1

2

101

10

41

4.

34. x yx y+ =+ =

λ2 1

To solve this system we write it in matrix form as

1 11 2 1LNMOQPLNMOQP =LNMOQP

xy

λ.


A

A

A

= =

= = −

= = −

1 11 2

1

11 2

2 1

11 1

1

1

2

λλ

λλ .


x

y

= =−

= −

= =−

= −

AAAA

1

2

2 11

2 1

11

1

λ λ

λ λ .


35. x y zy z

x z

+ + =+ =+ =

3 52 5 7

2 3

To solve this system, we write it in matrix form as1 1 30 2 51 0 2

573

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

xyz

.


A

A

A

A

= =

= =

= =

= =

1 1 30 2 51 0 2

3

5 1 37 2 53 0 2

3

1 5 30 7 51 3 2

3

1 1 50 2 71 0 3

3

1

2

3 .

All determinants are 3, so

x

y

z

= = =

= = =

= = =

AAAAAA

1

2

3

33

1

33

1

33

1.

36. x x xx x xx x x

1 2 3

1 2 3

1 2 3

2 63 8 9 102 2 2

+ − =+ + =− + = −

To solve this system, we write it in matrix form as1 2 13 8 92 1 2

6102

1

2

3

−

−

L

NMMM

O

QPPP

L

NMMM

O

QPPP=−

L

NMMM

O

QPPP

xxx

.



A

A

A

A

=−

−=

=−

− −=

=−

−=

=− −

= −

1 2 13 8 92 1 2

68

6 2 110 8 92 1 2

68

1 6 13 10 92 2 2

136

1 2 63 8 102 1 2

68

1

2

3 .


x

x

x

11

22

33

6868

1

13668

2

6868

1

= = =

= = =

= = − = −

AAAAAA

.

!!!! The Wheatstone Bridge

37. (a) Each equation represents the fact that the sum of the currents into the respective nodes A,B, C, and D is zero. For example

node :node :node :node :

ABCD

I I I I I II I I I I II I I I I I

I I I I I I

g x g x

x x

g g

− − = ⇒ = +− − = ⇒ = +

− + + = ⇒ = ++ − = ⇒ = +

1 2 1 2

1 1

3 3

2 3 3 2

0000 .

(b) If a current I flows through a resistance R, then the voltage drop across the resistance isRI. Applying Kirchhoff’s voltage law, the sum of the voltage drops around each of thethree circuits is set to zero giving the desired three equations:

voltage drop around the large circuit E R I R Ix x0 1 1 0− − = ,

voltage drop around the upper-left circuit R I R I R Ig g1 1 2 2 0+ − = ,

voltage drop around the upper-right circuit R I R I R Ix x g g− − =3 3 0 .


(c) Using the results from part (a) and writing the three currents I3, Ix , and I in terms of I1,I2 , Ig . gives

I I II I I

I I I

g

x g

3 2

1

1 2

= +

= −

= + .

We substitute these into the three given equations to obtain the 3 3× linear system forthe currents I1, I2 , Ig :

R R R R RR R R

R R R

III E

x x g

g

x x g

− − − −−

+ −

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

3 3

1 2

1

1

2

00

00 .

Solving for Ig (we only need to solve for one of the three unknowns) using Cramer’s

rule, we find

Ig =AA

1

where

A1

3

1 2

1 0

0 2 1 3

00

0=

−−

+

L

NMMM

O

QPPP= − +

R RR R

R R EE R R R R

x

x

xb g .

Hence, Ig = 0 if R R R Rx2 1 3= . Note: The proof of this result is much easier if we assumethe resistance Rg is negligible, and we take it as zero.

!!!! Least Squares Derivation

38. Starting with

F m k y k mxi ii

n, ( ) = − +

=∑ a f 2

1,

we compute the equations

∂∂

=Fk

0 , ∂∂

=Fm

0

yielding

∂∂

=∂∂

− + = − + −( ) =

∂∂

=∂∂

− + = − + − =

= =

= =

∑ ∑

∑ ∑

Fk k

y k mx y k mx

Fm m

y k mx y k mx x

i ii

n

i ii

n

i ii

n

i ii

n

i

a f a f

a f a f a f

2

1 1

2

1 1

2 1 0

2 0.


Carrying out a little algebra, we get

kn m x y

k x m x x y

ii

n

ii

n

ii

n

ii

n

i ii

n

+ =

+ =

= =

= = =

∑ ∑

∑ ∑ ∑

1 1

1

2

1 1

or in matrix form

n x

x x

km

y

x y

ii

n

ii

n

ii

n

ii

n

i ii

n=

= =

=

=

∑

∑ ∑

∑

∑

L

N

MMMM

O

Q

PPPPLNMOQP =L

N

MMMM

O

Q

PPPP1

1

2

1

1

1

.

!!!! Alternative Derivation of Least Squares Equations

39. (a) Equation (9) in the textk mk mk mk m

+ =+ =+ =+ =

17 112 3 3131 2 34 0 38

. .. .. .. .

can be written in matrix form

1 171 2 31 311 4 0

11312 338

....

.

...

L

N

MMMM

O

Q

PPPPLNMOQP =L

N

MMMM

O

Q

PPPPkm

which is the form of Ax b= .

(b) Given the matrix equation Ax b= , where

A =

L

N

MMMM

O

Q

PPPP

1111

1

2

3

4

xxxx

, x = LNMOQP

km

, b =

L

N

MMMM

O

Q

PPPP

yyyy

1

2

3

4

if we premultiply each side of the equation by AT, we get A Ax A bT T= , or

1 1 1 11111

1 1 1 1

1 2 3 4

1

2

3

4

1 2 3 4

1

2

3

4

x x x x

xxxx

km x x x x

yyyy

LNM

OQPL

N

MMMM

O

Q

PPPPLNMOQP =LNM

OQPL

N

MMMM

O

Q

PPPP


or

41

4

1

42

1

41

4

1

4

x

x x

km

y

x y

ii

ii

ii

ii

i ii

=

= =

=

=

∑

∑ ∑

∑

∑

L

N

MMMM

O

Q

PPPPLNMOQP =L

N

MMMM

O

Q

PPPP.

!!!! Least Squares Calculation

40. Here we are given the data points

x y

0 1

1 1

2 3

3 3

so

x

x

y

x y

ii

ii

ii

i ii

=

=

=

=

∑

∑

∑

∑

=

=

=

=

1

4

2

1

4

1

4

1

4

6

14

8

16.

The constants m, k in the least squares line

y mx k= +

satisfy the equations

4 66 14

816


km

,

which yields k m= = 080. . The least squares lineis y x= +08 08. . . 42

2

1

3

4

1

3

y x= +08 08. .

x

y

!!!! Computer or Calculator

41. To find the least-squares approximation of the form y k mx= + , we solve to a set of data pointsx y i ni i, : , , a fk p= 1 2, … , to get the system


n x

x x

km

y

x y

ii

n

ii

n

ii

n

ii

n

i ii

n=

= =

=

=

∑

∑ ∑

∑

∑

L

N

MMMM

O

Q

PPPPLNMOQP =L

N

MMMM

O

Q

PPPP1

1

2

1

1

1

.

Using a spreadsheet to compute the element of the coefficient matrix and the right-hand-sidevector, we get

Spreadsheet to compute least squares

x y x^2 xy

1.6 1.7 2.56 2.72

3.2 5.3 10.24 16.96

6.9 5.1 47.61 35.19

8.4 6.5 70.56 54.60

9.1 8.0 82.81 72.80

sum x sum y sum x^2 sum xy

29.2 26.6 213.78 182.27

We must solve the system

50 29 2029 2 21378

26 60182 27

. .

. ...

LNM

OQPLNMOQP =LNMOQP

km

getting k = 168. , m = 0 62. . Hence, we have theleast squares line

y x= +0 62 168. . ,

whose graph is shown next. 84

4

2

6

8

2

6x

y

10

10

y x= +062 168. .

42. To find the least-square approximation of the form y k mx= + , we solve to a set of data pointsx y i ni i, : , , a fk p= 1 2, … to get the system

n x

x x

km

y

x y

ii

n

ii

n

ii

n

ii

n

i ii

n=

= =

=

=

∑

∑ ∑

∑

∑

L

N

MMMM

O

Q

PPPPLNMOQP =L

N

MMMM

O

Q

PPPP1

1

2

1

1

1

.

Using a spreadsheet to compute the elements of the coefficient matrix and the right-hand-sidevector, we get


Spreadsheet to compute least squares

x y x^2 xy

0.91 1.35 0.8281 1.2285

1.07 1.96 1.1449 2.0972

2.56 3.13 6.5536 8.0128

4.11 5.72 16.8921 23.5092

5.34 7.08 28.5156 37.8072

6.25 8.14 39.0625 50.8750

sum x sum y sum x^2 sum xy

20.24 27.38 92.9968 123.5299

We must solve the system

600 20240020 24 92 9968

27 38001235299

. .

. ...

LNM

OQPLNMOQP =LNM

OQP

km

getting k = 0 309. , m = 126. . Hence, the least-squares line is y x= +0 309 126. . .

5

5

x

y

10

10

y x= +126 0309. .

!!!! Least Squares in Another Dimension

43. We seek the constants α, β1, and β2 that minimize

F y T Pi i ii

nα β β α β β, , 1 2 1 2

2

1a f a f= − + +

=∑ .

We write the equations

∂∂

= − + + −( ) =

∂∂

= − + + − =

∂∂

= − + + − =

=

=

=

∑

∑

∑

F y T P

F y T P T

F y T P P

i i ii

n

i i ii

n

i

i i ii

n

i

αα β β

βα β β

βα β β

2 1 0

2 0

2 0

1 21

11 2

1

21 2

1

a f

a f a f

a f a f .


Simplifying, we get

n T P

T T T P

P T P P

y

T y

Py

ii

n

ii

n

ii

n

ii

n

i ii

n

ii

n

i ii

n

ii

n

ii

n

i ii

n

i ii

n

= =

= = =

= = =

=

=

=

∑ ∑

∑ ∑ ∑

∑ ∑ ∑

∑

∑

∑

L

N

MMMMMMM

O

Q

PPPPPPP

L

NMMM

O

QPPP=

L

N

MMMMMMM

O

Q

PPPPPPP

1 1

1

2

1 1

1 1

2

1

1

2

1

1

1

αββ

Solving for α, β1, and β2 , we get the least-squares plane

y T P= + +α β β1 2 .

!!!! Least Squares System Solution

44. Premultiplying each side of the system Ax b= by AT gives A Ax A bT T= , or

1 0 11 1 1

1 10 11 1

1 0 11 1 1

121

−LNM

OQP −

L

NMMM

O

QPPPLNMOQP =

−LNM

OQPL

NMMM

O

QPPP

xy

or simply

2 00 3

04


xy

.

Solving this 2 2× system, gives

x = 0, y = 43

,

which is the least squares approximation to theoriginal system.

2–1

–3

1

3

2

–2

–2

least squaressolution

y = 2

0 4 3,a f− + =x y 1

x y+ = 1

y

x


45. Student Project


3.5 Vector Spaces and Subspaces

!!!! They Don’t All Look Like Vectors

1. A typical vector is 1 2, ( ) , the zero vector is 0 0, ( ) and the negative of x y, ( ) is − −( )x y, .

2. A typical vector is 1 3 2, , ( ) , the zero vector is 0 0 0, , ( ) and the negative of x y z, , ( ) is− − −( )x y z, , .

3. A typical vector is 1 4, 3 0, , ( ) , the zero vector is 0 0 0 0, , , ( ) and the negative of a b c d, , , ( ) is− − − −( )a b c d, , , .

4. A typical vector is 1 3 9, , ( ) , the zero vector is 0 0 0, , ( ) and the negative of a b c, , ( ) is− − −( )a b c, , .

5. A typical vector is 1 9 02 3 0−LNM

OQP , the zero vector is

0 0 00 0 0LNMOQP and the negative of

a b cd e fLNM

OQP is

− − −− − −LNM

OQP

a b cd e f

.

6. A typical vector is 1 5 43 4 32 6 7

L

NMMM

O

QPPP, the zero vector is

0 0 00 0 00 0 0

L

NMMM

O

QPPP and the negative of

a b cd e fg h i

L

NMMM

O

QPPP

is

− − −− − −− − −

L

NMMM

O

QPPP

a b cd e fg h i

.

7. A typical vector is a linear function p t at b( ) = + , the zero vector is p ≡ 0 and the negative ofp t( ) is − ( )p t .

8. A typical vector is a linear function

p t at bt cb g = + +2 ,

the zero vector is p ≡ 0 and the negative of p t( )is − ( )p t .

4–4

–2

2

y

t

4

–4

–2 2

p t1( ) p t( )

p t3( )

p t2( )

Segments of typical vectors in P2

SECTION 3.5 Vector Spaces and Subspaces 227

9. A typical vector is a continuous and differenti-able function, such as f t t( ) = sin , the zero vectoris f t( ) ≡ 0 and the negative of f t( ) is − ( )f t .

4–4

–2

2

y

x

4

–4

–2 2

g t( )h t( )

f t( )

10. C 2 0 1, : Typical vectors are continuous and

twice differentiable functions such as

f t t( ) = sin , g t t t( ) = + −2 2 ,

and so on. The zero vector is the zero functionf t( ) ≡ 0, and the negative of a typical vector, sayf t e tt( ) = sin , is − ( ) = −f t e tt sin .

4

y

t

8

–4

–8

g t( )

f t( )

–4 4

f t( )

!!!! Are They Vector Spaces?

11. Not a vector space; there is no additive inverse.

12. First octant of space: No, the vectors have no negatives. For example, 1 3 3, , ( ) belongs to the setbut − − −( )1 3 3, , does not.

13. Not a vector space; e.g., the negative of (2, 1) does not lie in the set.

14. Not a vector space; e.g., x x2 + and −( )1 2x each belongs, but their sum x x x x2 21+ + −( ) = does

not.

15. Not a vector space, polynomial of degree 2.

16. Yes, a vector space, diagonal 2 × 2 matrices.

17. Not a vector space; the set is not closed under vector addition. 2 × 2 singular matrices: The set isnot closed under addition as indicated by

1 01 0

0 10 3

1 11 3

LNMOQP +LNMOQP =LNMOQP .

18. All invertible matrices: No, you can’t add matrices of different orders.

19. Yes, a vector space, 3 × 3 upper-triangular matrices.


20. Not a vector space, does not contain the zero function. Continuous functions satisfying f 0 1( ) = .

21. Not a vector space; not closed under scalar multiplication; no additive inverse.

22. Yes, a vector space. All differentiable functions on −∞ ∞( ), .

23. Yes, a vector space. All integrable functions on 0 1, .

!!!! A Familiar Vector Space

24. Yes a vector space. Straightforward verification of the 10 commandments of a vector space; thatis, the sum of two vectors (real numbers in this case) is a vector (another real number), theproduct of a real number by a scalar (another real number) is a real number. The zero vector isthe number 0. Every number has a negative. The distributivity and associatively properties aresimply properties of the real numbers, and so on.

!!!! Not a Vector Space

25. Not a vector space; not closed under scalar multiplication.

!!!! DE Solution Space

26. Properties A3, A4, S1, S2, S3, and S4 are basic properties that hold for all functions; inparticular, solutions of a differential equation.

!!!! Another Solution Space

27. Yes, the solution space of the linear homogeneous DE

−( )′′ + ( ) − ′( ) + ( ) −( ) = − ′′ + ( ) ′ + ( ) =y p t y q t y y p t y q t y 0

is indeed a vector space; the linearity properties are sufficient to prove all the vector spaceproperties.

!!!! The Space C −∞ ∞( ),

28. This result follows from basic properties of continuous functions; the sum of continuous func-tions is continuous, scalar multiples of continuous functions are continuous, the zero function iscontinuous, the negative of a continuous function is continuous, the distributive properties holdfor all functions, and so on.

!!!! Vector Space Properties

29. Unique Zero: We prove that if a vector Z satisfies v Z v+ = for any vector v, then Z 0= . We canwrite

Z Z 0 Z v v Z v v v v 0= + = + + −( )( ) = +( ) + −( ) = + −( ) = .


30. Unique Negative: We show that if v is an arbitrary vector in some vector space, then there is onlyone vector N (which we call –v) in that space that satisfies v N 0+ = . Suppose another vector ′Nalso satisfies v N 0+ ′ = . Then

N N 0 N v v N v v 0 v v N v v v N0 N N

= + = + + −( )( ) = +( ) + −( ) = + −( ) = + ′( ) + −( ) = + −( ) + ′= + ′ = ′ .

31. Zero as Multiplier: We can write

v v v v v v v+ = + = +( ) = =0 1 0 1 0 1 .

Hence, by the result of Problem 30, we can conclude that 0v 0= .

32. Negatives as Multiples: From Problem 30, we know that –v is the only vector that satisfiesv v 0+ −( ) = . Hence, if we write

v v v v v v 0+ −( ) = × + −( ) = + −( )( ) = × =1 1 1 1 1 0 .

Hence, we conclude that −( ) = −1 v v .

!!!! A Vector Space Equation

33. Let v be an arbitrary vector and c an arbitrary scalar. Set cv 0= . Then either c = 0 or v 0= . Forc ≠ 0,

v v v 0 0= = ( ) = ( ) =1 1 1c

cc

,

which proves the result.

!!!! Nonstandard Definitions

34. x y x y x x1 1 2 2 1 2 0, , , a f a f a f+ ≡ + , c x y cx y, , ( ) ≡ ( )

All vector space properties clearly hold for these operations. The set R2 with indicated vectoraddition and scalar multiplication is a vector space.

35. x y x y x1 1 2 2 20, , , a f a f a f+ ≡ , c x y cx cy, , ( ) ≡ ( )

Not a vector space because, for example, the new vector addition is not commutative:

2, 3 4, 0 4 5 ( ) + ( ) = ( ), , 4, 2, 3 0 2 5 ( ) + ( ) = ( ), .

36. x y x y x x y y1 1 2 2 1 2 1 2, , , a f a f a f+ ≡ + + , c x y cx cy, , ( ) ≡ d hNot a vector space, for example,

c d c d+( ) ≠ +x x x .


For c = 4, d = 9 and vector x = x x1 2, a f, we have

c d x x x x

c d x x x x x x x x x x

+( ) = =

+ = + = + =

x

x x

13 13 13

4 9 2 2 3 3 5 51 2 1 2

1 2 1 2 1 2 1 2 1 2

, ,

, , , , , .

a f d ha f a f a f a f a f

!!!! Sifting Subsets for Subspaces

37. W = =x y y,a fk p0 is a subspace of R2.

38. W = + =x y x y,a fl q2 2 1 is not a subspace of R2 because it does not contain the zero vector

(0, 0). It is also not closed under vector addition and scalar multiplication.

39. W = =x x x x1 2 3 3 0, ,a fk p is a subspace of R3.

40. W = ( ) ( ) ={ }p t p tdegree 2 is not a subspace of P2 because it does not contain the zero vectorp t( ) ≡ 0 .

41. W = ( ) ( ) ={ }p t p 0 0 is a subspace of P3.

42. W = ( ) ( ) ={ }f t f 0 0 is a subspace of C 0 1, .

43. W = ( ) ( ) = ( ) ={ }f t f f0 1 0 is a subspace of C 0 1, .

44. W = ( ) ( ) =zf t f t dta

b0{ } is a subspace of C a b, .

45. W = ( ) ′′ + ={ }f t f f 0 is a subspace of C2 0 1, .

46. W = ′′ + =f t f f( ) 1m r is not a subspace of C2 0 1, . It does not contain the zero vector y t( ) ≡ 0 .

It is also not closed under vector addition and scalar multiplication because the sum of twosolutions is not necessarily a solution. For example, y t1 1= + sin and y t2 1= + cos are both

solutions, but the sum

y y t t1 2 2+ = + +sin cos

is not a solution. Likewise 2 2 21y t= + sin is not a solution.

!!!! Hyperplanes as Subspaces

47. We select two arbitrary vectors

u = x y z w1 1 1 1, , ,a f , v = x y z w2 2 2 2, , ,a ffrom the subset W. Hence, we have

ax by cz dwax by cz dw

1 1 1 1

2 2 2 2

00

+ + + =+ + + = .


Adding, we get

a x x b y y c z z d w w ax by cz dw ax by cz dw1 2 1 2 1 2 1 2 1 1 1 1 2 2 2 2

0+ + + + + + + = + + + + + + +

=a f a f a f a f

which says that u v+ ∈W . To show ku ∈W , we must show that the scalar multiple

k kx ky kz kwu = 1 1 1 1, , ,a fsatisfies

akx bky ckz dkw1 1 1 1 0+ + + = .

But this follows from

akx bky ckz dkw k ax by cz dw1 1 1 1 1 1 1 1 0+ + + = + + + =a f .

!!!! Differentiable Subspaces

48. f t f( ) ′ ={ }0 . It is a subspace.

49. f t f( ) ′ ={ }1 . It is not a subspace, because it does not contain the zero vector and is not closedunder vector addition. Hence, f t t( ) = , g t t( ) = + 2 belongs to the subset but f g t+( )( ) does notbelong. It is also not closed under scalar multiplication. For example f t t( ) = belongs to thesubset, but 2 2f t t( ) = does not.

50. f t f f( ) ′ ={ } . It is a subspace.

51. f t f f( ) ′ = 2l q . It is not a subspace; e.g., not closed under scalar multiplication. ( f may satisfy

equation ′ =f f 2 , but 2f will not, since 2 4 2′ ≠f f .

!!!! Property Failures

52. The first quadrant (including the coordinate axes) is closed under vector addition, but not scalarmultiplications.

53. An example of a set in R2 that is closed under scalar multiplication but not under vector additionis that of two different lines passing through the origin.

54. The unit circle is not closed under either vector addition or scalar multiplication.

!!!! Nonlinear Differential Equations

55. ′ =y y2 . Writing the equation in differential form, we have y dy dt− =2 . We get the general

solution yc t

=−1 . Hence, from c = 0 and 1, we have two solutions

y tt11

( ) = − , y tt2

11

( ) =−

.


But, if we compute

y t y tt t1 21 1

1( ) + ( ) = − +

−

it would not be a solution of the DE. So the solution set of this nonlinear DE is not a vectorspace.

56. ′′ + =y ysin 0 . Assume that y is a solution of the equation. Hence, we have the equation

′′ + =y ysin 0 .

But cy does not satisfy the equation because we have

cy cy c y y( )′′ + ( ) ≠ ′′ +( ) =sin sin 0 .

57. ′′ + =yy1 0 . From the DE we can see that the zero vector is not a solution, so the solution space

of this nonlinear DE is not a vector space.

!!!! Do They or Don’t They?

58. ′ + =y y et2 . Not a vector space, it doesn’t contain the zero vector.

59. ′ + =y y2 0 . The solutions are yt c

=−1 , and the sum of two solutions is not a solution, so the

general solution space of this nonlinear DE is not a vector space.

60. ′′ + =y ty 0 . If y1 , y2 satisfy the equation, then

′′+ =′′+ =

y tyy ty

1 1

2 2

00.

By adding, we obtain

′′+ + ′′ + =y ty y ty1 1 2 2 0a f a f ,

which from properties of the derivative is equivalent to

c y c y t c y c y1 1 2 2 1 1 2 2 0+ ′′ + + =a f a f .

This shows the set of solutions is a vector space.

61. ′′ + +( ) =y t y1 0sin . If y1 , y2 satisfy the equation then

′′+ +( ) =′′+ +( ) =

y t yy t y

1 1

2 2

1 01 0

sinsin .

By adding, we have

′′+ +( ) + ′′+ +( ) =y t y y t y1 1 1 11 1 0sin sin ,


which from properties of the derivative is equivalent to

c y c y t c y c y1 1 2 2 1 1 2 21 0+ ′′ + +( ) + =a f a fsin ,

which shows the set of solutions is a vector space. This is true for the solution set of any linearhomogeneous DE.

!!!! Line of Solutions

62. (a) x p h= + = + = +t t t t0 1 2, 3 2 1 3, ,a f a f a fHence, calling x1 , x2 the coordinates of the vector x = x x1 2,a f , we have x t1 2= ,x t2 1 3= + .

(b) x = + −2, 1 3 2, 3 0, ,a f a ft

(c) Showing that solutions of ′ + =y y 0 are closed under vector addition is a result of the

fact that the sum of two solutions is a solution. The fact that solutions are closed underscalar multiplication is a result of the fact that scalar multiples of solutions are alsosolutions. The zero vector (zero function) is also a solution because the negative of asolution is a solution. Computing the solution of the equation gives y t ce t( ) = − , which is

scalar multiple of e t− . We will later discuss that this collection of solutions is a one-dimensional vector space.

(d) The solutions of ′ + =y y t are given by y t t ce t( ) = −( ) + −1 . The abstract point of view is

a line through the vector t −1 (remember functions are vectors here) in the direction ofthe vector e t− .

(e) The solution of any linear equation Ly f= can be interpreted as a line passing throughany particular solution yp in the direction of any homogeneous solution yh ; that is,y y cyp h= + .


63. Student Project


3.6 Bases and Dimension

!!!! The Spin on Spans

1. V R= 2 . Let

x y c cc c

c

, , ,,,

a f a f a fa fa f

= +==

1 2

2 2

2

0 0 1 1

1 1.

The given vectors do not span R2 , although they span the one-dimensional subspacek k1 1,a fk p ∈R .

2. V R= 3 . Letting

x y z c c c, , , , , , ,a f a f a f a f= + +1 2 31 0 0 0 1 0 2, 3 1

yields the system of equations

c c xc c y

c z

1 3

2 3

3

23

+ =+ =

=

or

c zc y c y zc x c x z

3

2 3

1 3

3 32 2

== − = −= − = − .

Hence, W spans R3.

3. V R= 3 . Letting

x y z c c c c, , , , , , , , ,a f a f a f a f a f= − + + − +1 2 3 41 0 1 2, 0 4 5 0 2 0 0 1

yieldsx c c cyz c c c c

= + −== − + + +

1 2 3

1 2 3 4

2 50

4 2 .

These vectors do not span R3 because they cannot give any vector with y ≠ 0 .

4. V P= 2 . Let

at bt c c c t c t t21 2 3

21 1 2 3+ + = ( ) + +( ) + − +b g .

SECTION 3.6 Bases and Dimension 235

Setting the coefficients of t2 , t, and 1 equal to each other gives

t c at c c b

c c c c

23

2 3

1 2 3

21 3

:::

=− =

+ + =

which has the solution c c b a1 5= − − , c b a2 2= + , c a3 = . Any vector in V can be written as a

linear combination of vectors in W. Hence, the vectors in W span V.

5. V P= 2 . Let

at bt c c t c t c t t21 2

23

21 1+ + = +( ) + + + −b g b g .Setting the coefficients of t2 , t, and 1 equal to each other gives

t c c at c c b

c c c

22 3

1 3

1 21

::: .

+ =− =

+ =

If we add the first and second equations, we get

c c a bc c c

1 2

1 2

+ = ++ =

This means we have a solution only if c a b= + . In other words, the given vectors do not spanP2 ; they only span a one-dimensional vector space of R3.

6. V M= 22 . Letting

a bc d

c c c cLNMOQP =LNMOQP +LNMOQP +LNMOQP +LNMOQP1 2 3 4

1 10 0

0 01 1

1 01 0

0 10 1

we have the equations

c c ac c bc c cc c d

1 3

1 4

2 3

2 4

+ =+ =+ =+ = .

If we add the first and last equation, and then the second and third equations, we obtain theequations

c c c c a dc c c c b c1 2 3 4

1 2 3 4

+ + + = ++ + + = + .

Hence, we have a solution if and only if a d b c+ = + . This means we can solve for c1 , c2 , c3 ,

c4 for only a subset of vectors in V. Hence, W does not span R4.


!!!! Independence Day

7. V R= 2 . Setting

c c1 21 1 1 1 0 0, , ,− + − =a f a f a fwe get

c cc c1 2

1 2

00

− =− + =

which does not imply c c1 2 0= = . Hence, the vectors in W are linearly dependent.

8. V R= 2 . Setting

c c1 21 1 1 1 0 0, , ,a f a f a f+ − =

we get

c cc c1 2

1 2

00

+ =− =

which implies c c1 2 0= = . Hence, the vectors in W are linearly independent.

9. V R= 3 . Setting

c c c1 2 31 0 0 1 1 0 1 1 1 0 0 0, , , , , , , ,a f a f a f a f+ + =

we get

c c cc c

c

1 2 3

2 3

3

000

+ + =+ =

=

which implies c c c1 2 3 0= = = . Hence vectors in W are linearly independent.

10. V R= 3 . Setting

c c1 22, 1 4 4, 2, 8 0 0 0− + − =, , ,a f a f a fwe get

2 4 02 0

4 8 0

1 2

1 2

1 2

c cc cc c

+ =− − =

+ =

which (the equations are all the same) has a nonzero solution c1 2= − , c2 1= . Hence, the vectors

in W are linearly dependent.


11. V R= 3 . Setting

c c c1 2 31 1 8 3 4, 2 7, 1 3 0 0 0, , , , , ,a f a f a f a f+ − + − =

we get

c c cc c cc c c

1 2 3

1 2 3

1 2 3

3 7 04 0

8 2 3 0

− + =+ − =+ + =

which has only the solution c c c1 2 3 0= = = . Hence, the vectors in W are linearly independent.

12. V P= 1. Setting

c c t1 2 0+ = ,

we get c1 0= , c2 0= . Hence, the vectors in W are linearly independent.

13. V P= 1. Setting

c t c t1 21 1 0+( ) + −( ) =

we get

c cc c1 2

1 2

00

+ =− =

which has a unique solution c c1 2 0= = . Hence, the vectors in W are linearly independent.

14. V P= 2 . Setting

c t c t1 2 1 0+ −( ) = ,

we get

c cc

1 2

2

00

− ==

which implies c c1 2 0= = . Hence, the vectors in W are linearly independent.


c t c t c t1 2 321 1 0+( ) + −( ) + =

we get

c cc c

c

1 2

1 2

3

000

+ =− =

=

which implies c c c1 2 3 0= = = . Hence, the vectors in W are linearly independent.



c t c t c t t1 22

323 1 2 5 0+( ) + − + − − =b g b g

we get

3 5 00

2 0

1 2 3

1 3

2 3

c c cc c

c c

− − =− =+ =

which has the nonzero solution c1 1= − , c2 2= , c3 1= − . Hence, the vectors in W are linearly

dependent.

17. V D= 22 . Setting

ab

c c0

01 00 0

0 00 11 2

LNMOQP =LNMOQP +LNMOQP

we get c a1 = , c b2 = . Hence, these vectors are linearly independent and span D22 .

18. V D= 22 . Setting

ab

c c0

01 00 1

1 00 11 2

LNMOQP =LNMOQP + −LNMOQP

we get c c a1 2+ = , c c b1 2− = . We can solve these equations for c1 , c2 , and hence these vectors

are linearly independent and span D22 .

!!!! Function Space Dependence

19. S = −e et t,l q . We set

c e c et t1 2 0+ =− .

Because we assume this holds for all t, it holds in particular for t = 0 , 1, so

c cec e c

1 2

11

2

00

+ =+ =−

which has only the zero solution c c1 2 0= = . Hence, the functions are linearly independent.

20. S = e te t et t t, , 2l q . We assume

c e c te c t et t t1 2 3

2 0+ + =

for all t. We let t = 0 , 1, 2, so


cec ec ec

e c e c e c

1

1 2 32

12

22

3

00

2 4 0

=+ + =+ + =

which has only the zero solution c c c1 2 3 0= = = . Hence, these vectors are linearly independent.

21. S = sin , sin , sint t t2 3k p . Letting

c t c t c t1 2 32 3 0sin sin sin+ + =

for all t. In particular if we choose three values of t, say π6

, π4

, π2

, we obtain three equations to

solve for c1 , c2 , c3 , namely,

c c c

c c c

c c

1 2 3

1 2 3

1 3

12

32

0

22

22

0

0

FHIK +FHGIKJ + =

FHGIKJ + +

FHGIKJ =

− = .

We used Maple to compute the determinant of this coefficient matrix and found it to be

− +32

12

6 . Hence, the system is nonsingular, and the only solution is c c c1 2 3 0= = = . Thus,

sin t , sin2t , and sin3t are linearly independent.

22. S = 1 2 2, sin , cost tl q . Because

1 02 2− − =sin cost t

the vectors are linearly dependent.

23. S = − −( )1 1 1 2, ,t tn s . Setting

c c t c t1 2 321 1 0+ −( ) + −( ) =

we get for the coefficients of 1, t, t2 the system of equations

c c cc c

c

1 2 3

2 3

3

02 0

0

− + =− =

=

which has the only zero solution c c c1 2 3 0= = = . Hence, these vectors are linearly independent.


24. S = −e e tt t, , coshl q . Because

cosh t e et t= + −12b g

we have that 2 0cosh t e et t− − =− is a nontrivial linear combination that is identically zero forall t. Hence, the vectors are linearly dependent.

25. S = sin , cos2 4, 2t tl q . Recall the trigonometric identity

sin cos2 12

1 2t t= −( ) ,

which can be rewritten as

2 14

4 2 02sin cost t− ( ) + = .

Hence, we have found a nontrivial linear combination of the three vectors that is identically zero.Hence, the three vectors are linearly dependent.

!!!! Independence Testing

26. We will show the only values for which

cee

cee

t

t

t

t1 2

22 00

LNMOQP +LNMOQP =LNMOQP

for all t are c c1 2 0= = and, hence, conclude that the vectors are linearly independent. If it is true

for all t, then it must be true for t = 0 (which is the easiest place to test), which yields the twolinear equations

c cc c1 2

1 2

2 00

+ =+ =

whose only solution is c c1 2 0= = . Hence, the vectors are linearly independent.

Another approach is to say the vectors are linearly independent because clearly there isno constant k such that one vector is k times the other vector for all t.

27. We will show that

ctt

ctt1 2

00

sincos

cossin

LNMOQP + −LNMOQP =LNMOQP

for all t implies c c1 2 0= = , and hence, the vectors are linearly independent. If it is true for all t,then it must be true for t = 0 , which gives the two equations c2 0= , c1 0= . This proves the

vectors are linearly independent.


Another approach is to say that the vectors are linearly independent because clearly thereis no constant k such that one vector is k times the other vector for all t.

28. We write

ceee

ceee

ceee

t

t

t

t

t

t

t

t

t1 2 3

2

2

22 2 3

000

L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP=L

NMMM

O

QPPP

−

−

for all t and see if there are nonzero solutions for c1 , c2 , and c3 . Because the preceding equation

is assumed true for all t, it must be true for t = 1 (testing at t = 0 won’t work because there AREnonzero solutions there), or

c e c e c ec e c e c ec e c e c e

1 21

32

1 21

32

1 2 32

02 2 3 0

0

+ + =+ + =+ + =

−

−

.

Writing this in matrix form gives

e e ee e e

e e e

ccc

−

−

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

1 2

1 2

2

1

2

3

2 2 3000

.

Evaluating the determinant of this matrix with Maple (do it yourself if you like) we obtain anonzero value (the exact value is unimportant), so the vectors are linearly independent.

Another approach is to say that the vectors are linearly independent simply by looking atthem because clearly it is impossible to write any one vector as a linear combination of the othertwo due to the nature of the exponents.

29. We write

ceee

ce

ec

eee

t

t

t

t

t

t

t

t1 2 3

8

8

84 0

2

2

000

−

−

−

−

−

−L

NMMM

O

QPPP+

−

L

NMMM

O

QPPP+L

NMMM

O

QPPP=L

NMMM

O

QPPP

for all t and see if there are nonzero solutions for c1 , c2 , c3 . Because the above equation is

assumed true for all t, it must be true for t = 0 , or

c c cc cc c c

1 2 3

1 3

1 2 3

2 04 0

2 0

+ + =− + =

− + = .


Writing this in matrix form gives

1 1 24 0 11 1 2

000

1

2

3

−−

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

ccc

.

The determinant of the coefficient matrix is 18, so the only solution of this linear system isc c c1 2 3 0= = = , and thus the vectors are linearly independent.

!!!! Twins?

30. We have

span

span

cos sin , cos sin cos sin cos sincos sin

cos sinsin , cos .

t t t t c t t c t tc c t c c t

C t C tt t

+ − = +( ) + −( )

= + + −= +=

k p k pa f a fk pk pk p

1 2

1 2 1 2

1 2

!!!! Wronskian

31. We assume that the Wronskian function

W f g t f t g t f t g t, ( ) = ( ) ′( ) − ′( ) ( ) ≠ 0

for every t ∈ 0 1, . To show f and g are linearly independent on [0, 1], we assume thatc f t c g t1 2 0( ) + ( ) = for all t in the interval [0, 1]. Differentiating, we have

c f t c g t1 2 0′( ) + ′( ) =

on [0, 1]. Hence, we have the two equations

f t g tf t g t

cc

( ) ( )

′( ) ′( )LNM

OQPLNMOQP =LNMOQP

1

2

00

.

The determinant of the coefficient matrix is the Wronskian of f and g, which is assumed to benonzero on [0, 1]. Since c c1 2 0= = , the vectors are linearly independent.

!!!! Linearly Independent Exponentials

32. We compute the Wronskian of f and g:

W f g tf t g tf t g t

e eae be

be ae e b aat bt

at bta b t a b t a b t, b g b g b gb g b g b gb g b g b g=

′ ′= = − = − ≠+ + + 0

for any t. Hence, f and g are linearly independent.


!!!! Looking Ahead

33. The Wronksian is

We tee e t

e t te et t

t tt t t=

+( )= +( ) − =

112 2 2 .

Hence, the vectors are linearly independent.

!!!! Revisiting Linear Independence

34. The Wronskian is

We e ee e ee e e

ee e

e ee

e ee e

ee ee e

e e

t t t

t t t

t t t

tt t

t tt

t t

t tt

t t

t t

t t

= − =−

− +−

= − − − − + + = −

−

−

−

−

−

−

−

−

−

55 3

5 9

5 35 9

55 9

55 3

45 15 45 5 15 5 80

3

3

3

3

3

3

3

3

3

3 3b g b g b gHence, the vectors are linearly independent.

!!!! Getting on Base in R2

35. Not a basis because 1 1,a fk p does not span R2.

36. Is a basis because 1 2 2, 1, ,a f a fk p are linearly independent and span R2.

37. − −1 1 1 1, , ,a f a fk p Not a basis because − −1 1 1 1, , ,a f a fk pand are linearly dependent.

38. 1 0 1 1, , ,a f a fk p Is a basis because the vectors are linearly independent and span R2.

39. 1 0 0 1 1 1, , , , ,a f a f a fk p Not a basis because the vectors are linearly dependent.

40. 0 0 1 1 2, 2, , , ,a f a f a fk p Not a basis because the vectors are linearly dependent.

!!!! The Base for the Space

41. V R= 3 : Not a basis because the two vectors are not enough to span R3.

42. V R= 3 : Yes, these vectors are linearly independent and span R3.

43. V R= 3 : Not a basis because four vectors must be linearly dependent in R3.

44. V P= 2 : Clearly the two vectors t t2 3 1+ + and t t2 2 4− + are linearly independent because theyare not constant multiples of one another. To verify they do not span P2 , however, we set

c t t c t t at bt c12

22 23 1 2 4+ + + − + = + +b g b g .


Comparing coefficients we get

c c ac c bc c c

1 2

1 2

1 2

3 24

+ =− =+ =

which can easily be seen to have no solution for arbitrary a, b, c. The two vectors are not a basisfor P2 . (Two vectors can never form a basis for a three-dimensional space.)

45. V P= 3: We assume that

c t c t c t c t c t t12

2 33

4 523 4 1 5 1 0+ +( ) + + + −( ) + − + =b g b g

and compare coefficients of t3 , t2 , t, 1. Not a basis because the five vectors must be linearlydependent in t3 , t2 , t, 1.

46. V P= 4 : We assume that

c t c t c t c t c t t14

2 33

4 523 4 1 5 1 0+ +( ) + + + −( ) + − + =b g b g

and compare coefficients. We find a homogeneous system of equations which has only the zerosolution c c c c c1 2 3 4 5 0= = = = = . Hence, the vectors are linearly independent. To show the

vectors span P4 , we set the above linear combination equal to an arbitrary vector

at bt ct dt e4 3 2+ + + + , and comparing coefficients arrive at a system equations, which can besolved for c1 , c2 , c3 , c4 , and c5 in terms of a, b, c, d, e. Hence, the vectors span P4 and

combined with linear independence are a basis for P4 .

47. V M= 22 . We assume that

c c c c1 2 3 41 00 0

0 10 0

0 01 0

1 11 1

0 00 0

LNMOQP +LNMOQP +LNMOQP +LNMOQP =LNMOQP

yields the equations

c cc c

c cc

1 4

2 4

3 4

4

0000

+ =+ =+ =

=

which has the zero solution c c c c1 2 3 4 0= = = = . Hence, the vectors are linearly independent. If

we replace the zero vector on the right of the preceding equation by an arbitrary vector

a bc dLNMOQP ,


we get the four equations

c c ac c b

c c cc d

1 4

2 4

3 4

4

+ =+ =+ =

=

This yields the solutionc dc c dc b dc a d

4

3

2

1

== −= −= −

Hence, the four given vectors span M22 . Because they are linearly independent and span M22 ,

they are a basis.

48. V M= 23: If we set a linear combination of these vectors to an arbitrary vector, like

c c c c ca b cd e f1 2 3 4 5

1 0 10 0 0

1 1 00 0 0

0 0 01 0 1

0 0 01 1 0

0 0 01 1 1

LNMOQP +LNMOQP +LNMOQP +LNMOQP +LNMOQP =LNM

OQP

we arrive at the algebraic equations

c c ac b

c cc c c d

c c ec c f

1 2

2

1

3 4 5

4 5

3 5

+ ===

+ + =+ =+ = .

Looking at the first three equations gives c a b1 = − , c c1 = . If we pick an arbitrary matrix such

that a b c− ≠ , we have no solution. Hence, the vectors do not span M22 and do not form a basis.

(They are linearly independent however.)

!!!! Sizing Them Up

49. W = + + =x x x x x x1 2 3 1 2 3 0, ,b gl qLetting x2 = α , x3 = β , we can write x1 = − −α β . Any vector in W can be written as

xxx

1

2

3

110

101

L

NMMM

O

QPPP=

− −L

NMMM

O

QPPP=

−L

NMMM

O

QPPP+

−L

NMMM

O

QPPP

α βαβ

α β

where α and β are arbitrary real numbers. Hence, The dimension of W is 2; a basis is

{ −1 1 0, ,a f , −1 0 1, ,a f}.


50. W = + = =x x x x x x x x1 2 3 4 1 3 2 40, , , ,b gl qLetting x3 = α , x4 = β , we have

xxxx

1

2

3

4

= −===

αβαβ .

Any vector in W can be written as

xxxx

1

2

3

4

1010

0101

L

N

MMMM

O

Q

PPPP=

−L

N

MMMM

O

Q

PPPP=

−L

N

MMMM

O

Q

PPPP+

L

N

MMMM

O

Q

PPPP

αβαβ

α β

where α and β are arbitrary real numbers. Hence, the two vectors −1 0 1 0, , , and 0 1 0 1, , ,

form a basis of W, which is only two-dimensional.

!!!! Polynomial Dimensions

51. t t, −1k p . We write

at b c t c t+ = + −( )1 2 1

yielding the equations

t c c ac b

:: .

1 2

21+ =− =

We can represent any vector at b+ as some linear combination of t and t −1 . Hence, we havethat t t, −1k p spans a two-dimensional vector space.

52. t t t, ,− +1 12l q . We write

at bt c c t c t c t21 2 3

21 1+ + = + −( ) + +b gyielding the equations

t c at c c b

c c c

23

1 2

2 31

::: .

=+ =− + =

Because we can solve this system for c1 , c2 , c3 in terms of a, b, c getting

c a c bc a cc a

1

2

3

= − + += −=

the subspace spans the entire three-dimensional vector space P2 .


53. t t t2 2 1 1, ,− +n s . We write

at bt c c t c t c t21

22

231 1+ + = + − + +( )b g

getting the equations

t c c at c b

c c c

21 2

3

2 31

::: .

+ ==

− + =

Because we can solve this system for c1 , c2 , c3 in terms of a, b, c getting

c a b c1 = − + , c b c2 = − , c b3 = ,

the subspace spans the entire three-dimensional vector space P2 .

!!!! Basis for P2

54. We first show the vectors span P2 by selecting an arbitrary vector from P2 and show it can be

written as a linear combination of the three given vectors. We set

at bt c c t t c t c21

22 31 1+ + = + + + +( ) +b g

and try to solve for c1 , c2 , c3 in terms of a, b, c. Setting the coefficients of t2 , t, and 1 equal to

each other yields

t c at c c b

c c c c

21

1 2

1 2 31

::: ,

=+ =+ + =

giving the solutionc ac a bc b c

1

2

3

== − += − + .

Hence, the set spans P2 . We also know that the vectors

t t t2 1 1 1+ + +, ,l qare independent because setting

c t t c t c12

2 31 1 0+ + + +( ) + =b gwe get

cc cc c c

1

1 2

1 2 3

000

=+ =+ + =


which has only the solution c c c1 2 3 0= = = . Hence, the vectors are a basis for P2 , for example,

3 2 1 3 1 1 1 1 12 2t t t t t+ + = + + − +( ) − ( )b g .

!!!! Two-by-Two Basis

55. Setting

c c c1 2 31 00 0

0 11 0

0 01 1

0 00 0

LNMOQP +LNMOQP +LNMOQP =LNMOQP

gives c1 0= , c2 0= , and c3 0= . Hence, the given vectors are linearly independent. If we add the

vector

0 01 0LNMOQP ,

then the new vectors are still linearly independent (similar proof), and an arbitrary 2 2× matrixcan be written as

a bc d

c c c cLNMOQP =LNMOQP +LNMOQP +LNMOQP +LNMOQP1 2 3 4

1 00 1

0 11 0

0 01 1

0 01 0

because it reduces to

c ac b

c dc c c c

1

2

3

2 3 4

===

+ + = .

This yields

c ac b

c dc b d

1

2

3

4

==== − −

in terms of a, b, c, and d and form a basis for M22 , which is four-dimensional.

!!!! Basis for Trace Zero Matrices

56. Letting

c c ca bc d1 2 3

1 00 1

0 10 0

0 01 0−

LNMOQP +LNMOQP +LNMOQP =LNMOQP

we find a c= 1 , b c= 2 , c c= 3 , d c= − 1 . Given a b c d= = = = 0 implies c c c c1 2 3 4 0= = = = ,

which shows the vectors (matrices) are linearly independent. It also shows they span the set of


2 2× matrices with trace zero because if a d+ = 0 , we can solve for c a d1 = = − , c b2 = , c c3 = .

In other words we can write any zero trace 2 2× matrix as follows as a linear combination of thethree given vectors (matrices):

a bc a

a b c−

LNMOQP = −LNMOQP +LNMOQP +LNMOQP

1 00 1

0 10 0

0 01 0

.

Hence, the vectors (matrices) form a basis for the 2 2× zero trace matrices.

!!!! Hyperplane Basis

57. Solving the equation

x y z w+ − + =3 2 6 0

for x we get

x y z w= − + −3 2 6 .

Letting y = α , z = β and w = γ , we can write

x = − + −3 2 6α β γ .

Hence, an arbitrary vector x y z w, , ,a f in the hyperplane can be written

xyzw

L

N

MMMM

O

Q

PPPP=

− + −L

N

MMMM

O

Q

PPPP=

−L

N

MMMM

O

Q

PPPP+

L

N

MMMM

O

Q

PPPP+

−L

N

MMMM

O

Q

PPPP

3 2 6 3100

2010

6001

α β γαβγ

α β γ .

The set of four-dimensional vectors

−L

N

MMMM

O

Q

PPPP

L

N

MMMM

O

Q

PPPP

−L

N

MMMM

O

Q

PPPP

RS||

T||

UV||

W||

3100

2010

6001

, ,

is a basis for the hyperplane.

!!!! Solution Basis

58. Letting z = α we solve for x and y, obtaining x = −4α , y = 5α . An arbitrary solution of the

system can be expressed asxyz

L

NMMM

O

QPPP=

−L

NMMM

O

QPPP=

−L

NMMM

O

QPPP

45

451

ααα

α .

Hence, the vector [–4, 5, 1] is a basis for the solutions.


!!!! Cosets in R3

59. W = + + =x x x x x x1 2 3 1 2 3 0, ,a fk p, v = 0 0 1, ,a fWe want to write W in parametric form, so we solve the equation

x x x1 2 3 0+ + =

by letting x2 = β , x3 = γ and solving for x1 = − −β γ . These solutions can be written as

β γ β γ− + − ∈1 1 0 1 0 1, , , , : ,b g b gm r R ,

so the coset of (0, 0, 1) in W is the collection of vectors

0 0 1 1 1 0 1 0 1, , , , , , ,a f a f a fk p+ − + − ∈β γ β γ R .

Geometrically, this describes a plane passing through (0, 0, 1) and parallel to x x x1 2 3 0+ + = .

60. W = =x x x x1 2 3 3 0, ,a fk p , v = 1 1 1, ,a fHere a coset through the point (1, 1, 1) is given by the points

1 1 1 1 0 0 0 1 0, , , , , ,a f a f a fk p+ +β γ

where β and γ are arbitrary real numbers. This describes the plane through (1, 1, 1) parallel to

the x1 – x2 plane (i.e., the subspace W).

!!!! More Cosets

61. The coset through the point (1, –2, 1) is given by the points

1 2, 1 1 3 2, , ,− +a f a fk pt ;

t is an arbitrary number. This describes a line through (1, –2, 1) parallel to the line t 1 3 2, ,a f .!!!! Antiderivative Space

62. Because the general solution of the nth order equation d ydt

n

n = 0 is

y t c t c t c t cnn

nn( ) = + + + +−

−−

−1

12

21 0…

where c0 , c1 , … cn−1 are arbitrary constants, we have that the basis for the vector space of

solutions is given by

1 2 1, , ,t t tn… −l q .


!!!! DE Solution Space

63. The solutions of

′ − =y y2 0

are

y t ce t( ) = 2 ;

that is, all solutions are multiples of the continuous and differentiably-continuous function e t2 .Hence, the function e t2k p forms a basis for the solution set, which is a vector subspace of

C CR R( ) ( ), 1 , and so on.

!!!! Another Solution Space

64. If we solve the separable equation

′ − =y yt2 0

by separating variables, we get

dyy

tdt= 2 .

Hence,

ln y t c

y e e

y t e e c e

c t

c t t

= +

=

( ) = ± =

2

1

2

2 2

where c1 is an arbitrary constant. Hence, a basis for the solution set is et2n s . This vector space is

a subspace of many larger vector spaces including C −∞ ∞,a f and C2 −∞ ∞,a f .!!!! Line in Function Space

65. The general solution of ′ + = −y y e t2 2 is

y t ce tet t( ) = +− −2 2 .

We could say the solution is a “line” in the vector space of solutions, passing through te t−2 in thedirection of e t−2 .

!!!! Suggested Journal Entry I

66. Student Project


!!!! Suggested Journal Entry II

67. Student Project

!!!! Suggested Journal Entry III

68. Student Project

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