Solution Manual for Linear Algebra and Its Applications| 9780136880929| All Chapters| LATEST 2026

SOLUTION MANUAL
LINEAR ALGEBRA AND ITS APPLICATIONS

, INSTRUCTOR‘S SOLUTIONS
MANUAL
JUDI J. MCDONALD
Washington State University




LINEAR ALGEBRA
AND ITS APPLICATIONS
SIXTH EDITION
GLOBAL EDITION


David C. Lay
University Of Maryland


Steven R. Lay
Lee University


Judi J. Mcdonald
Washington State University




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, Contents



Chapter 1 1-1

Chapter 2 2-1

Chapter 3 3-1

Chapter 4 4-1

Chapter 5 5-1

Chapter 6 6-1

Chapter 7 7-1

Chapter 8 8-1




Iii
Copyright © 2016 Pearson Education, Ltd.

,1.1 Solutions

Notes: The Key Exercises Are 7 (Or 11 Or 12), 19–22, And 25. For Brevity, The Symbols R1, R2,…,
Stand For Row 1 (Or Equation 1), Row 2 (Or Equation 2), And So On. Additional Notes Are At The End
Of The Section.

X  5x  7  1 7
1. 2x 1 7x 2  5 5
5
2 7
1 2  
X1  5x2  7 1 5 7
Replace R2 By R2 + (2)R1 And
3x2  9 0 3 9
Obtain:
Scale R2 By 1/3: X1  5x2  1 5 7
7
X2  3 0 1 3
Replace R1 By R1 + (–5)R2: X1  8 1 0 8
X2  0 1 3
The Solution Is (X1, X2) = (–8, 3), Or Simply (–8, 3
3).
2x  4x  4 4
2. 5x1  7x 2  11 52 4
7 11
1 2  
Scale R1 By 1/2 And X1  2x2  1 2 2
Obtain:
2 5
5x  7x  11 7 11
1 2  
Replace R2 By R2 + (–5)R1: X1  2x2  1 2 2
2 0
3x  21 3 21
2  
Scale R2 By –1/3: X1  2x2  1 2 2
2 0
X  7
Replace 1 7
By R1 + R1
2)R2: (– 2


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, X1  12  
1 0 12 0
X  7 1 7
2  
The Solution Is (X1, X2) = (12, –7), Or Simply (12, –7).




Copyright © 2016 Pearson Education, Ltd. 1-1

,1-2 CHAPTER 1 • Linear Equations in Linear Algebra


3. The Point Of Intersection Satisfies The System Of Two Linear Equations:
X1  5x2  7 1 7
1 2 2
5
2  2x 
X
1 2  
X1  5x2   1 7
9
5
Replace R2 By R2 + (–1)R1 And 0 7
7
Obtain: 7x  9
2  
X1  5x2  7 1 5 7 
Scale R2 By –1/7: 0
X  9/7
2 1 9/7
X1  4/7 1 0 4/7
Replace R1 By R1 + (–5)R2:
X2  9/7 0 1 9/7
The Point Of Intersection Is (X1, X2) = (4/7,
9/7).

4. The Point Of Intersection Satisfies The System Of Two Linear Equations:
X1  5x2  1 5 1
1 3 7 5
3x  7x  5
1 2  
X1  5x2  1 1 5 1
Replace R2 By R2 + (–3)R1 And
8x2  2 0 8 2
Obtain:
X1  5x2  1 1 5 1
Scale R2 By 1/8:
X2  1/4 0 1 1/4
X1  9/4 1 0 9/4
Replace R1 By R1 + (5)R2:
X2  1/4 0 1 1/4
The Point Of Intersection Is (X1, X2) = (9/4,
1/4).

5. The System Is Already In ―Triangular‖ Form. The Fourth Equation Is X4 = –5, And The Other
Equations Do Not Contain The Variable X4. The Next Two Steps Should Be To Use The Variable X3
In The Third Equation To Eliminate That Variable From The First Two Equations. In Matrix
Notation, That Means To Replace R2 By Its Sum With 3 Times R3, And Then Replace R1 By Its
Sum With –5 Times R3.

6. One More Step Will Put The System In Triangular Form. Replace R4 By Its Sum With –3 Times R3, Which
 1 6 4 0 1
0 2 7 0 4
Produces   . After That, The Next Step Is To Scale The Fourth Row By –1/5.
0
 0 1 2 3

0 0 0 5 15
 

7. Ordinarily, The Next Step Would Be To Interchange R3 And R4, To Put A 1 In The Third Row
And Third Column. But In This Case, The Third Row Of The Augmented Matrix Corresponds To
The Equation 0 X1 + 0 X2 + 0 X3 = 1, Or Simply, 0 = 1. A System Containing This Condition Has
No Solution. Further Row Operations Are Unnecessary Once An Equation Such As 0 = 1 Is
Evident.
Copyright © 2016 Pearson Education, Ltd.

, 1.1 • Solutions 1-3
The Solution Set Is Empty.




Copyright © 2016 Pearson Education, Ltd. 1-1

,1-4 CHAPTER 1 • Linear Equations in Linear Algebra


8. The Standard Row Operations Are:
 1 4 9 0  1 4 9 0  1 4 0 0 1 0 0 0
0 1 7 0 ~ 0 1 7 0 ~ 0 1 0 0 ~ 0 1 0 0
       
0 0 2 0 0 0 1 0 0 0 1 0 0 0 1 0
The Solution Set Contains One Solution: (0, 0, 0).

9. The System Has Already Been Reduced To Triangular Form. Begin By Scaling The Fourth Row
By 1/2 And Then Replacing R3 By R3 + (3)R4:
01 1 0 0 4  1 1 0 0 4  1 1 0 0 4
7

0 7  0   
 1 3 ~ 1 3 0 7  0
~ 1 3 0 
0 0 1 3 1 0 0 1 3 1 0 0 1 0 5
     
0 0 0 2 4 0 0 0 1 2 0 0 0 1 2
     
Next, Replace R2 By R2 + (3)R3. Finally, Replace R1 By R1 + R2:
 1 1 0 0 4  1 0 0 0 4
0 1 0 0 8 0 1 0 0 8
~ ~ 
0 0 1 0 5 0 0 1 0 5
   
0 0 0 1 2  0 0 0 1 2 
The Solution Set Contains One Solution: (4, 8, 5, 2).

10. The System Has Already Been Reduced To Triangular Form. Use The 1 In The Fourth Row To Change The
–4 And 3 Above It To Zeros. That Is, Replace R2 By R2 + (4)R4 And Replace R1 By R1 + (–3)R4.
For The Final Step, Replace R1 By R1 + (2)R2.
 1 2 0 3 2  1 2 0 0 7  1 0 0 0 3
0 1 0 4 7 0 1 0 0 5 0 1 0 0 5
 ~ ~ 
0
 0 1 0 6 0
 0 1 0 6 0
 0 1 0 6
0 0 0 1 3 0 0 0 1 3 0 0 0 1 3
     
The Solution Set Contains One Solution: (–3, –5, 6, –3).

11. First, Swap R1 And R2. Then Replace R3 By R3 + (–3)R1. Finally, Replace R3 By R3 + (2)R2.
0 1 4 5  1 3 5 2  1 3 5 2  1 3 5 2
1 3 5 2 ~ 0 1 4 5 ~ 0 1 4 5 ~ 0 1 4 5
       
3 7 7 6 3 7 7 6 0 2 8 12 0 0 0 2
The System Is Inconsistent, Because The Last Row Would Require That 0 = 2 If There Were A
Solution. The Solution Set Is Empty.

12. Replace R2 By R2 + (–3)R1 And Replace R3 By R3 + (4)R1. Finally, Replace R3 By R3 + (3)R2.
 1 3 4 4  1 3 4 4  1 3 4 4
 3 7 7 8 ~ 0 2 5 4 ~ 0 2 5 4
     
4 6 1 
7  0 6 15 9   0 0 0 3
The System Is Inconsistent, Because The Last Row Would Require That 0 = 3 If There Were A
Solution. The Solution Set Is Empty.


Copyright © 2016 Pearson Education, Ltd.

, 1.1 • Solutions 1-5


13. Replace R2 By R2 + (–2)R1. Then Interchange R2 And R3. Next Replace R3 By R3 + (–2)R2.
Then Divide R3 By 5. Finally, Replace R1 By R1 + (–2)R3.
 1 0 3 8  1 0 3 8  1 0 3 8  1 0 3 8
       2
2 2 9 7 ~ 0 2 15 9 ~ 0 1 5 2 ~ 0 1 5 
      
0 1 5 2   0 1 5 2    0 2 15 9   0 0 5 5
 1 0 3 8 1 0 0 5
~ 0 1 5 2 ~ 0 1 0 3 . The Solution Is (5, 3, –1).
   
0 0 1 1 0 0 1 1

14. Replace R2 By R2 + R1. Then Interchange R2 And R3. Next Replace R3 By R3 + 2r2. Then
Divide R3 By 7. Next Replace R2 By R2 + (–1)R3. Finally, Replace R1 By R1 + 3r2.

 1 3 0 5  1 3 0 5  1 3 0 5  1 3 0 5  1 3 0 5
1 1 5 2 ~ 0 2 5 7 ~ 0 1 1 0 ~ 0 1 1 0 ~ 0 1 1 0
         
 0 1 1 0 0 1 1 0 0 2 5 7 0 0 7 7 0 0 1 1
 1 3 0 5  1 0 0 2
~ 0 1 0 1 ~ 0 1 0 1 . The Solution Is (2, –1, 1).
   
0 0 1 1 0 0 1 1

15. First, Replace R4 By R4 + (–3)R1, Then Replace R3 By R3 + (2)R2, And Finally Replace
R4 By R4 + (3)R3.
1 0 3 0 2  1 0 3 0 2  1 0 3 0 2
0  0 1 3 3
3  0 3
3 0
 1 0 3 ~ 1 0  ~ 
0
 2 3 2 1 0 2 3 2 1 0
 0 3 4 7
3    0 0 9 7 11
 0 0 7 5   0 0 9 7 11   
0 01 03 3
1 0 2
3
~ .
0 0 3 4 7
0 0 0 5 10
 
The Resulting Triangular System Indicates That A Solution Exists. In Fact, Using The Argument From
Example 2, One Can See That The Solution Is Unique.

16. First Replace R4 By R4 + (2)R1 And Replace R4 By R4 + (–3/2)R2. (One Could Also Scale R2
Before Adding To R4, But The Arithmetic Is Rather Easy Keeping R2 Unchanged.) Finally,
Replace R4 By R4 + R3.
 1 0 0 2 3  1 0 0 2 3  1 0 0 2 3  1 0 0 2 3
    0 0 2 2 0 0
 0 2 2 0 0
~
0 2 2 0 0 ~ 0 2 2 0
~ 
 0 0 1 3 
1 0
 0 1 3 
1 0
 0 1 3 1 0
 0 1 3 
1
 5 0 3 2 3  0 0 1 3  0 0 0 0 0
 2 3 2 1   1   1   
The System Is Now In Triangular Form And Has A Solution. The Next Section Discusses How To
Continue With This Type Of System.


Copyright © 2016 Pearson Education, Ltd.

, 1-6 CHAPTER 1 • Linear Equations in Linear Algebra


17. Row Reduce The Augmented Matrix Corresponding To The Given System Of Three Equations:
 1 4 1  1 4 1  1 4 1
 2 1 3 ~ 0 7 5 ~ 0 7 5
     
1 3 4 0 7 5 0 0 0
The System Is Consistent, And Using The Argument From Example 2, There Is Only One Solution.
So The Three Lines Have Only One Point In Common.

18. Row Reduce The Augmented Matrix Corresponding To The Given System Of Three Equations:
1 2 1 4  1 2 1 4  1 2 1 4
0 1 1 1 ~ 0 1 1 1 ~ 0 1 1 1
     
1 3 0 0 0 1 1 4 0 0 0 5
The Third Equation, 0 = –5, Shows That The System Is Inconsistent, So The Three Planes Have No
Point In Common.
1 4
19. 4 1 Write C For 6 – 3h. If C = 0, That Is, If H = 2, Then The System Has No
~
H
3 6 H 8 0 6  3h 4
   
Solution, Because 0 Cannot Equal –4. Otherwise, When H  2, The System Has A Solution.
 1 3
20. 3 1 . Write C For 4 + 2h. Then The Second Equation = 0 Has A
~ Cx
H
2 4 H6 0 4  2h 0
2

   
Solution For Every Value Of C. So The System Is Consistent For All H.
2
21.  1 3 2 1 3 . Write C For H + 12. Then The Second = 0 Has A
~ Equation Cx
4  2
8 0 H  12 0
H
   
Solution For Every Value Of C. So The System Is Consistent For All H.
 2 H 
22. 3 H 2 3 . The System Is Consistent If And Only If 5 + 3h = 0, That Is, If And
~
6 9 5  0 0 5  3h 

Only If H = –5/3.

23. A. True. See The Remarks Following The Box Titled ―Elementary Row Operations‖.
b. False. A 5 × 6 Matrix Has Five Rows.
c. False. The Description Given Applies To A Single Solution. The Solution Set Consists Of All
Possible Solutions. Only In Special Cases Does The Solution Set Consist Of Exactly One
Solution. Mark A Statement True Only If The Statement Is Always True.
d. True. See The Box Before Example 2.

24. A. True. See The Box Preceding The Subsection Titled ―Existence And Uniqueness Questions‖.
b. False. The Definition Of Row Equivalent Requires That There Exist A Sequence Of Row
Operations That Transforms One Matrix Into The Other.
c. False. By Definition, An Inconsistent System Has No Solution.
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