"《线性代数极其应用第五版》习题答案解析"

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Linear algebra is a fundamental branch of mathematics that involves the study of vectors, matrices, and systems of linear equations. In "Linear Algebra and Its Applications Fifth Edition", authors David C. Lay, Steven R. Lay, and Judi J. McDonald provide a comprehensive overview of the subject, highlighting its relevance and practical applications in various fields. The Instructor's Solutions Manual offers detailed answers and explanations to key exercises throughout the textbook. With a focus on exercises 7, 11, 12, 19-22, and 25, instructors and students can gain a deeper understanding of important concepts and techniques in linear algebra. The manual also includes helpful notes and symbols to aid in the comprehension of complex problems. Linear algebra plays a crucial role in numerous disciplines, including physics, engineering, computer science, and economics. By studying vector spaces, linear transformations, and eigenvalues, students can develop problem-solving skills and analytical thinking abilities that are essential in today's data-driven world. Overall, "Linear Algebra and Its Applications Fifth Edition" and its accompanying Solutions Manual provide a comprehensive and insightful introduction to linear algebra, equipping students with the knowledge and tools needed to excel in their academic and professional pursuits. With a focus on practical applications and real-world examples, this textbook is an invaluable resource for anyone seeking to deepen their understanding of linear algebra.

1.9 • Solutions 1-15

13165

21224



  

   

  



  

. The intermediate step is often not written.

32 32 5

212(1)1



     

   

     



     

Using the definitions carefully,

323(2)(2)3(4)1

2(2)

212(2)(1)224



         

     

         



         

, or, more quickly,

32341

21224



  

   

  



  

. The intermediate step is often not written.

631

147

505



 

 

 

 



 

63 1

505















 





















63 1















 





















63 1



 



Usually the intermediate steps are not displayed.

123

28 10

35 60

xxx



  

 

  



  

312

635

xxx











 



















123

35 6

xxx

xx x

 











223

123

28 0

35 60

xxx

xx x

  



Usually the intermediate steps are not displayed.

u – 2v

– 2v

u – v

– v

u + v

u – v

u + v

– v

– 2v

u – 2v

1-16 CHAPTER 1 • Linear Equations in Linear Algebra

7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every

vector in

 can be written as a linear combination of u and v.

To write a vector

a as a linear combination of u and v, imagine walking from the origin to a along

the grid "streets" and keep track of how many "blocks" you travel in the

u-direction and how many in

the

v-direction.

a. To reach a from the origin, you might travel 1 unit in the u-direction and –2 units in the v-

direction (that is, 2 units in the negative

v-direction). Hence a = u – 2v.

b. To reach b from the origin, travel 2 units in the u-direction and –2 units in the v-direction. So

b = 2u – 2v. Or, use the fact that b is 1 unit in the u-direction from a, so that

b = a + u = (u – 2v) + u = 2u – 2v

. The vector c is –1.5 units from b in the v-direction, so

c = b – 1.5v = (2u – 2v) – 1.5v = 2u – 3.5v

. The “map” suggests that you can reach d if you travel 3 units in the u-direction and –4 units in

the

v-direction. If you prefer to stay on the paths displayed on the map, you might travel from the

origin to –3

v, then move 3 units in the u-direction, and finally move –1 unit in the v-direction. So

d = –3v + 3u – v = 3u – 4v

Another solution is

d = b – 2v + u = (2u – 2v) – 2v + u = 3u – 4v

Figure for Exercises 7 and 8

8. See the figure above. Since the grid can be extended in every direction, the figure suggests that every

vector in

can be written as a linear combination of u and v.

w. To reach w from the origin, travel –1 units in the u-direction (that is, 1 unit in the negative

u-direction) and travel 2 units in the v-direction. Thus, w = (–1)u + 2v, or w = 2v – u.

. To reach x from the origin, travel 2 units in the v-direction and –2 units in the u-direction. Thus,

x = –2u + 2v. Or, use the fact that x is –1 units in the u-direction from w, so that

x = w – u = (–u + 2v) – u = –2u + 2v

. The vector y is 1.5 units from x in the v-direction, so

y = x + 1.5v = (–2u + 2v) + 1.5v = –2u + 3.5v

z. The map suggests that you can reach z if you travel 4 units in the v-direction and –3 units in the

u-direction. So z = 4v – 3u = –3u + 4v. If you prefer to stay on the paths displayed on the “map,”

you might travel from the origin to –2

u, then 4 units in the v-direction, and finally move –1 unit

in the

u-direction. So

z = –2u + 4v – u = –3u + 4v

–2v

–u

–v

1.9 • Solutions 1-17

123

46 0

380

xxx





  

123

46 0

38 0

xxx

xx x















 













 







12 3

050

46 0

380

xx x

 

 



 



 

123

01 50

46 10

13 80

xxx

























Usually, the intermediate calculations are not displayed.

Note: The Study Guide says, “Check with your instructor whether you need to “show work” on a

problem such as Exercise 9.”

10.

123

439

72 2

86 515

xx x







12 3

123

439

72 2

865 15

xx x

xxx















 





















123

439

722

86 515

xxx



 





 



  



 



 







 



12 3

4139

1722

86515

xx x

























Usually, the intermediate calculations are not displayed.

11. The question

b a linear combination of a

, a

, and a

is equivalent to the question

Does the vector equation

+ x

= b have a solution?

The equation

123

10 52

21 61

02 86

xxx

  

  

  

  

  



aa ab

(*)

has the same solution set as the linear system whose augmented matrix is

10 5 2

21 6 1

02 8 6





  





Row reduce M until the pivot positions are visible:

1052 1052

~0 1 4 3~0 1 4 3

0286 0000







The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and

therefore b is a linear combination of a

, a

, and a

1.9 • Solutions 1-19

1·v

+ 0·v







, 0·v

+ 1·v









1·v

+ 1·v







, 1·v

– 1·v









17. [a

b] =

124 12 4 12 4 12 4

4 3 1~0 5 15 ~0 1 3 ~0 1 3

27 03 8 03 8 00 17hhhh



 

 



 



 

. The vector

b is in Span{a

, a

} when h + 17 is zero, that is, when h = –17.

18. [v

y] =

13 13 13

015~01 5~01 5

283 0232 0072

hh h

 

  

  

 

  

  

  

. The vector y is in

Span{v

, v

} when 7 + 2h is zero, that is, when h = –7/2.

19. By inspection, v

= (3/2)v

. Any linear combination of v

and v

is actually just a multiple of v

. For

instance,

+ bv

= av

+ b(3/2)v

= (a + 3b/2)v

So Span{v

, v

} is the set of points on the line through v

and 0.

Note: Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7.

20. Span{v

, v

} is a plane in

 through the origin, because neither vector in this problem is a multiple

of the other. Every vector in the set has 0 as its second entry and so lies in the xz-plane in ordinary

3-space. So Span{v

, v

} is the xz-plane.

21. Let y =







. Then [u v y] =

22 22

11 02 /2

kkh

 

 



 

. This augmented matrix

corresponds to a consistent system for all h and k. So y is in Span{u, v} for all h and k.

22. Construct any 3×4 matrix in echelon form that corresponds to an inconsistent system. Perform

sufficient row operations on the matrix to eliminate all zero entries in the first three columns.

23. a. False. The alternative notation for a (column) vector is (–4, 3), using parentheses and commas.

b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If









were on

the line through









and the origin, then















would have to be a multiple of









, which is

not the case.

c. True. See the line displayed just before Example 4.

d. True. See the box that discusses the matrix in (5).

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