The 1.1
SOLUTIONS
Notes: key exercises are 7 (or 11 or 12), 1922, 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.
1.
x1 + 5 x2 = 7 -2 x1 - 7 x2 = -5
1 -2
5 -7
7 -5
x1 + 5 x2 = 7
Replace R2 by R2 + (2)R1 and obtain:
3x2 = 9
x1 + 5 x2 = 7 x2 = 3 x1
1 0 1 0 1 0
5 3 5 1 0 1
7 9 7 3 -8 3
Scale R2 by 1/3: Replace R1 by R1 + (5)R2: The solution is (x1, x2) = (8, 3), or simply (8, 3).
2.
= -8
x2 = 3
2 x1 + 4 x2 = -4 5 x1 + 7 x2 = 11
2 5
4 7
-4 11
x1 + 2 x2 = -2
Scale R1 by 1/2 and obtain: Replace R2 by R2 + (5)R1: Scale R2 by 1/3: Replace R1 by R1 + (2)R2: The solution is (x1, x2) = (12, 7), or simply (12, 7).
5 x1 + 7 x2 = 11
x1 + 2 x2 = -2
1 5 1 0 1 0 1 0
2 7 2 -3 2 1 0 1
-2 11 -2 21 -2 -7 12 -7
-3x2 = 21
x1 + 2 x2 = -2 x2 = -7 x1
= 12
x2 = -7
1
2
CHAPTER 1
Linear Equations in Linear Algebra
3. The point of intersection satisfies the system of two linear equations: x1 + 5 x2 = 7 x1 - 2 x2 = -2
1 1
5 -2
7 -2
x1 + 5 x2 = 7
Replace R2 by R2 + (1)R1 and obtain: Scale R2 by 1/7: Replace R1 by R1 + (5)R2: The point of intersection is (x1, x2) = (4/7, 9/7).
-7 x2 = -9
x1 + 5 x2 = 7 x2 = 9/7 x1
1 0 1 0 1 0
5 -7 5 1 0 1
7 -9 7 9/7 4/7 9/7
= 4/7
x2 = 9/7
4. The point of intersection satisfies the system of two linear equations: x1 - 5 x2 = 1
3x1 - 7 x2 = 5
1 3
-5 -7
1 5
x1 - 5 x2 = 1
Replace R2 by R2 + (3)R1 and obtain: Scale R2 by 1/8: Replace R1 by R1 + (5)R2: The point of intersection is (x1, x2) = (9/4, 1/4).
8 x2 = 2
x1 - 5 x2 = 1 x2 = 1/4 x1
1 0 1 0 1 0
-5 8
1 2
-5 1 1 1/4 0 1 9/4 1/4
= 9/4
x2 = 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 4 0 -1 1 -6 0 2 -7 0 4 . After that, the next step is to scale the fourth row by 1/5. produces 0 0 1 2 -3 0 0 -5 15 0 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. The solution set is empty.
1.1
Solutions
3
8. The standard row operations are:
1 0 0
-4 1 0
9 7 2
0 1 0 ~ 0 0 0
-4 1 0
9 7 1
0 1 0 ~ 0 0 0
-4 1 0
0 0 1
0 1 0 ~ 0 0 0
0 1 0
0 0 1
0 0 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:
1 0 0 0 1 0 ~ 0 0
-1 1 0 0 -1 1 0 0
0 -3 1 0 0 0 1 0
0 0 -3 2 0 0 0 1
-4 1 -7 0 ~ -1 0 4 0 -4 1 8 0 ~ 5 0 2 0
-1 1 0 0 0 1 0 0 0 0 1 0
0 -3 1 0 0 0 0 1
0 0 -3 1 4 8 5 2
-4 1 7 0 ~ -1 0 2 0
-1 1 0 0
0 -3 1 0
0 0 0 1
-4 -7 5 2
Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2:
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 0 0 0
-2 1 0 0
0 0 1 0
3 -4 0 1
-2 1 7 0 ~ 6 0 -3 0
-2 1 0 0
0 0 1 0
0 0 0 1
7 1 -5 0 ~ 6 0 -3 0
0 1 0 0
0 0 1 0
0 0 0 1
-3 -5 6 -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.
3 5 -2 1 3 5 -2 0 1 4 -5 1 3 5 -2 1 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.
4 -4 1 -3 4 -4 1 -3 4 -4 1 -3 3 -7 ~ 0 ~ 0 7 -8 2 -5 4 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.
4
CHAPTER 1
Linear Equations in Linear Algebra
1 13. 2 0 1 ~ 0 0 1 14. -1 0 1 ~ 0 0
0 2 1 0 1 0 -3 1 1
-3 9 5 -3 5 1 0 5 1 0 1 1
8 1 7 ~ 0 -2 0 8 1 -2 ~ 0 -1 0 5 1 2 ~ 0 0 0 5 1 0 ~ 0 1 0
0 2 1 0 1 0 -3 -2 1 -3 1 0
-3 15 5 0 0 1 0 5 1 0 0 1
8 1 -9 ~ 0 -2 0
0 1 2
-3 5 15
8 1 -2 ~ 0 -9 0
0 1 0
-3 5 5
8 -2 -5
5 3 . The solution is (5, 3, 1). -1 5 1 7 ~ 0 0 0 5 1 -1 ~ 0 1 0 -3 1 -2 0 1 0 0 1 5 0 0 1 5 1 0 ~ 0 7 0 -3 1 0 0 1 7 5 0 7
-3 1 0
2 -1 . The solution is (2, 1, 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. 0 3 0 2 1 0 3 0 2 1 0 0 1 0 -3 3 1 0 -3 3 ~ 0 -2 3 2 1 0 -2 3 2 1 0 0 7 -5 0 0 -9 7 -11 3
1 0 ~ 0 0
0 1 0 0
3 0 3 -9
0 -3 -4 7
2 1 3 0 ~ 7 0 -11 0
0 1 0 0
3 0 3 0
0 -3 -4 -5
2 3 7 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 0 2 2 0 0 0 2 2 0 0 ~ 0 0 1 3 1 0 0 1 3 1 1 5 0 3 2 -3 -1 -2 3 2
1 0 ~ 0 0
0 2 0 0
0 2 1 -1
-2 0 3 -3
-3 1 0 0 ~ 1 0 -1 0
0 2 0 0
0 2 1 0
-2 0 3 0
-3 0 1 0
The system is now in triangular form and has a solution. The next section discusses how to continue with this type of system.
1.1
Solutions
5
17. Row reduce the augmented matrix corresponding to the given system of three equations: 1 1 -4 1 1 -4 1 1 -4 2 -1 -3 ~ 0 ~ 0 7 -5 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 4 1 2 1 4 1 2 1 4 1 2 0 1 -1 1 ~ 0 1 -1 ~ 0 1 -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. 4 h 1 h 4 1 19. ~ 0 6 - 3h -4 Write c for 6 3h. If c = 0, that is, if h = 2, then the system has no 3 6 8 solution, because 0 cannot equal 4. Otherwise, when h 2, the system has a solution.
h -3 1 h -3 1 20. ~ 0 4 + 2h 0 . Write c for 4 + 2h. Then the second equation cx2 = 0 has a solution 6 -2 4 for every value of c. So the system is consistent for all h.
3 -2 1 3 -2 1 21. ~ . Write c for h + 12. Then the second equation cx2 = 0 has a solution -4 h 8 0 h + 12 0 for every value of c. So the system is consistent for all h. 2 -3 h 2 22. ~ 9 5 0 -6 if h = 5/3.
-3 0
h . The system is consistent if and only if 5 + 3h = 0, that is, if and only 5 + 3h
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 applied 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. d. True. This definition of equivalent systems is in the second paragraph after equation (2).
6
CHAPTER 1
Linear Equations in Linear Algebra
7 g 1 -4 7 g 1 -4 7 g 1 -4 0 ~ 0 ~ 0 25. 3 -5 h 3 -5 h 3 -5 h -2 5 -9 k 0 -3 5 k + 2 g 0 0 0 k + 2 g + h Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution if and only if k + 2g + h = 0.
26. A basic principle of this section is that row operations do not affect the solution set of a linear system. Begin with a simple augmented matrix for which the solution is obviously (2, 1, 0), and then perform any elementary row operations to produce other augmented matrices. Here are three examples. The fact that they are all row equivalent proves that they all have the solution set (2, 1, 0).
1 0 0 1 c
0 1 0
0 0 1
-2 1 1 ~ 2 0 0
0 1 0
0 0 1
-2 1 -3 ~ 2 0 2
0 1 0
0 0 1
-2 -3 -4
27. Study the augmented matrix for the given system, replacing R2 by R2 + (c)R1:
3 d
f 1 ~ g 0
3 d - 3c
f g - cf
This shows that shows d 3c must be nonzero, since f and g are arbitrary. Otherwise, for some choices of f and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus d 3c.
28. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible since a is nonzero. Then replace R2 by R2 + (c)R1.
a c
b d
f 1 ~ g c
b/a d
f / a 1 ~ g 0
b/a d - c(b / a)
f /a g - c( f / a )
The quantity d c(b/a) must be nonzero, in order for the system to be consistent when the quantity g c( f /a) is nonzero (which can certainly happen). The condition that d c(b/a) 0 can also be written as ad bc 0, or ad bc.
29. Swap R1 and R2; swap R1 and R2. 30. Multiply R2 by 1/2; multiply R2 by 2. 31. Replace R3 by R3 + (4)R1; replace R3 by R3 + (4)R1. 32. Replace R3 by R3 + (3)R2; replace R3 by R3 + (3)R2. 33. The first equation was given. The others are: T2 = (T1 + 20 + 40 + T3 )/4, or 4T2 - T1 - T3 = 60
T3 = (T4 + T2 + 40 + 30)/4, T4 = (10 + T1 + T3 + 30)/4,
or or
4T3 - T4 - T2 = 70 4T4 - T1 - T3 = 40
1.1
Solutions
7
Rearranging, 4T1 -T1 -T1 - + T2 4T2 -T2 - - T3 + 4T3 - T3 T4 = = = = 30 60 70 40
- T4 + 4T4
34. Begin by interchanging R1 and R4, then create zeros in the first column:
4 -1 0 -1 1 0 ~ 0 0 1 0 ~ 0 0 1 0 ~ 0 0
-1 4 -1 0 0 1 -1 -1 0 1 0 0 0 1 0 0
0 -1 4 -1 1 0 4 -4 1 0 4 0 0 0 1 0
-1 0 -1 4 -4 -1 -1 15 -4 -1 -2 1 0 0 0 1
30 -1 60 -1 ~ 70 0 40 4 -40 1 5 0 ~ 70 0 190 0 -40 1 5 0 ~ 75 0 22.5 0
0 4 -1 -1 0 1 0 0 0 1 0 0 1 0 4 0
-1 -1 4 0 1 0 4 -4 0 0 0 1
4 0 -1 -1 -4 -1 -2 14
40 -1 60 0 ~ 70 0 30 0 -40 1 5 0 ~ 75 0 195 0 0 1 0 0
0 4 -1 -1 0 1 0 0 1 0 1 0
-1 0 4 -4 1 0 4 0 0 0 0 1
4 40 -4 20 -1 70 15 190 -4 -1 -2 12 50 27.5 30 22.5 -40 5 75 270
Scale R1 by 1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:
Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 50 1 27.5 0 ~ 120 0 22.5 0
The last step is to replace R1 by R1 + (1)R3: 20.0 27.5 . The solution is (20, 27.5, 30, 22.5). 30.0 22.5
Notes: The Study Guide includes a "Mathematical Note" about statements, "If ... , then ... ."
This early in the course, students typically use single row operations to reduce a matrix. As a result, even the small grid for Exercise 34 leads to about 25 multiplications or additions (not counting operations with zero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization. For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/86/89 and HP-48G calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section 1.1 describes how to access the data that is available for all numerical exercises in the text. This feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale. These commands are included in the text data sets, available from the text web site, www.laylinalgebra.com.
8
CHAPTER 1
Linear Equations in Linear Algebra
1.2
SOLUTIONS
Notes: The key exercises are 120 and 2328. (Students should work at least four or five from Exercises
714, in preparation for Section 1.5.)
1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c. 2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.
1 3. 4 6
2 5 7
3 6 8
4 1 7 ~ 0 9 0 2 1 0 3 2 0
2 -3 -5 4 1 3 ~ 0 0 0 3 -4 -8 7 1 3 ~ 0 1 0
0
3 0
3 -6 -10 0 1 0 5 -8 -16 3 1 0
4 1 -9 ~ 0 -15 0 -1 2 0
2 1 -5
3 2 -10
4 3 -15
1 4 6
1 ~ 0 0 1 4. 3 5
-2 3 . Pivot cols 1 and 2. 0 3 1 -8 0 1 0 5 2 -16 -1 2 0
2 5 7 3 1 0
3 6 8 5 2 0
1 3 5
4 7 9 7 3 -10
3 5 7 5 7 9 7 9 1
3 5 7
5 7 9
7 1 9 ~ 0 1 0 3 1 0 5 2 0
7 1 -12 ~ 0 -34 0 5 2 0 0 1 0 ~ 0 1 0
7 1 3 ~ 0 -34 0
1 ~ 0 0 5. 0 1 7. 3
* , 0 3 9 4 7
0 Pivot cols 0 . 1, 2, and 4 1
* 0 0 , 0 0 0
* 0 , 0 0 7 1 ~ 6 0
6. 0 0
4 -5 7 1 ~ -15 0 3 0
* , 0 0 0 4 1
0 0
3 0 0 1
7 1 ~ 3 0
-5 3
Corresponding system of equations:
x1 + 3x2 x3
= -5 = 3
The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is free. Solve for the basic variables in terms of the free variable. The general solution is
x1 = -5 - 3x2 x2 is free x = 3 3
Note: Exercise 7 is paired with Exercise 10.
1.2
Solutions
9
1 8. 2
4 7
0 0
7 1 ~ 10 0
4 -1
0 0
7 1 ~ -4 0
4 1
0 0
7 1 ~ 4 0
0 1
0 0
-9 4
Corresponding system of equations:
x1 x2
= -9 = 4
The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic variables do not depend on the value of the free variable.
x1 = -9 General solution: x2 = 4 x is free 3
Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)
0 9. 1
1 -2 -6 7 5 1 ~ -6 0 x1 x2 -2 1 7 -6 -6 1 ~ 5 0 = 4 = 5 0 1 -5 -6 4 5
Corresponding system:
- 5 x3 - 6 x3
x1 = 4 + 5 x3 Basic variables: x1, x2; free variable: x3. General solution: x2 = 5 + 6 x3 x is free 3 1 10. 3
-2 -6 -1 -2 3 1 ~ 2 0 x1 -2 0 -1 1 3 1 ~ -7 0 x3 -2 0 0 1 -4 -7
Corresponding system:
- 2 x2
= -4 = -7
x1 = -4 + 2 x2 Basic variables: x1, x3; free variable: x2. General solution: x2 is free x = -7 3 3 11. -9 -6
-4 12 8 2 -6 -4 0 3 0 ~ 0 0 0 x1 Corresponding system: -4 0 0 - 2 0 0 4 x2 3 0 1 0 ~ 0 0 0 + 2 x3 3 0 0 -4 / 3 0 0 = 0 = 0 = 0 2/3 0 0 0 0 0
10
CHAPTER 1
Linear Equations in Linear Algebra
4 2 x1 = 3 x2 - 3 x3 Basic variable: x1; free variables x2, x3. General solution: x2 is free x is free 3
1 12. 0 -1
-7 0 7
0 1 -4
6 -2 2
5 1 -3 ~ 0 7 0
-7 0 0
0 1 -4
6 -2 8
5 1 -3 ~ 0 12 0
-7 0 0
0 1 0
6 -2 0
5 -3 0
x1
Corresponding system:
- 7 x2 x3
+ 6 x4 = 5 - 2 x4 = -3 0 = 0
x1 = 5 + 7 x2 - 6 x4 x is free Basic variables: x1 and x3; free variables: x2, x4. General solution: 2 x3 = -3 + 2 x4 x4 is free 1 0 13. 0 0
-3 1 0 0
0 0 0 0
-1 0 1 0
0 -4 9 0
x1
-2 1 1 0 ~ 4 0 0 0
x2
-3 1 0 0
0 0 0 0
- 3x5 - 4 x5 + 9 x5
0 0 1 0
9 -4 9 0
= 5 = 1 = 4
2 1 1 0 ~ 4 0 0 0
0 1 0 0
0 0 0 0
0 0 1 0
-3 -4 9 0
5 1 4 0
Corresponding system:
x4
0 = 0
x1 = 5 + 3x5 x = 1 + 4x 5 2 Basic variables: x1, x2, x4; free variables: x3, x5. General solution: x3 is free x = 4 - 9x 5 4 x5 is free
Note: The Study Guide discusses the common mistake x3 = 0.
1 0 14. 0 0
2 1 0 0 -5 -6 0 0 -6 -3 0 0 0 0 1 0 -5 1 2 0 ~ 0 0 0 0 0 1 0 0 7 -6 0 0 0 -3 0 0 0 0 1 0 -9 2 0 0
1.2
Solutions
11
x1 Corresponding system: x2
+ 7 x3 - 6 x3 - 3 x4 x5
= -9 = = 2 0 0
0 =
x1 = -9 - 7 x3 x = 2 + 6 x + 3x 3 4 2 Basic variables: x1, x2, x5; free variables: x3, x4. General solution: x3 is free x is free 4 x5 = 0
15. a. The system is consistent, with a unique solution. b. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column). 16. a. The system is consistent, with a unique solution. b. The system is consistent. There are many solutions because x2 is a free variable.
2 17. 4
3 6
h 2 ~ 7 0
3 0
h The system has a solution only if 7 2h = 0, that is, if h = 7/2. 7 - 2h
-3 -2 1 -3 -2 1 18. If h +15 is zero, that is, if h = 15, then the system has no solution, ~ 0 h + 15 h -7 3 5 because 0 cannot equal 3. Otherwise, when h -15, the system has a solution.
h 2 1 h 2 1 19. ~ 0 8 - 4h k - 8 4 8 k a. When h = 2 and k 8, the augmented column is a pivot column, and the system is inconsistent.
b. When h 2, the system is consistent and has a unique solution. There are no free variables. c. When h = 2 and k = 8, the system is consistent and has many solutions.
3 2 1 3 2 1 20. ~ 0 h - 9 k - 6 3 h k a. When h = 9 and k 6, the system is inconsistent, because the augmented column is a pivot column.
b. When h 9, the system is consistent and has a unique solution. There are no free variables. c. When h = 9 and k = 6, the system is consistent and has many solutions. 21. a. b. c. d. e.
False. See Theorem 1. False. See the second paragraph of the section. True. Basic variables are defined after equation (4). True. This statement is at the beginning of Parametric Descriptions of Solution Sets. False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a contradiction. So the system might be consistent or it might be inconsistent.
12
CHAPTER 1
Linear Equations in Linear Algebra
22. a. False. See the statement preceding Theorem 1. Only the reduced echelon form is unique. b. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are determined completely by the positions of the leading entries in the nonzero rows of any echelon form obtained from the matrix. c. True. See the paragraph after Example 3. d. False. The existence of at least one solution is not related to the presence or absence of free variables. If the system is inconsistent, the solution set is empty. See the solution of Practice Problem 2. e. True. See the paragraph just before Example 4. 23. Yes. The system is consistent because with three pivots, there must be a pivot in the third (bottom) row of the coefficient matrix. The reduced echelon form cannot contain a row of the form [0 0 0 0 0 1]. 24. The system is inconsistent because the pivot in column 5 means that there is a row of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows that the system has no solution. 25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom row, and there is no room for a pivot in the augmented column. So, the system is consistent, by Theorem 2. 26. Since there are three pivots (one in each row), the augmented matrix must reduce to the form
x1 = a 1 0 0 a 0 1 0 b and so = b x2 0 0 1 c x3 = c No matter what the values of a, b, and c, the solution exists and is unique.
27. "If a linear system is consistent, then the solution is unique if and only if every column in the coefficient matrix is a pivot column; otherwise there are infinitely many solutions. " This statement is true because the free variables correspond to nonpivot columns of the coefficient matrix. The columns are all pivot columns if and only if there are no free variables. And there are no free variables if and only if the solution is unique, by Theorem 2. 28. Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column. 29. An underdetermined system always has more variables than equations. There cannot be more basic variables than there are equations, so there must be at least one free variable. Such a variable may be assigned infinitely many different values. If the system is consistent, each different value of a free variable will produce a different solution. 30. Example:
x1 2 x1
+
x2
+
x3
= 4 = 5
+ 2 x2
+ 2 x3
31. Yes, a system of linear equations with more equations than unknowns can be consistent. x1 + x2 = 2 Example (in which x1 = x2 = 1): x1 - x2 = 0
3x1
+ 2 x2
= 5
1.2
Solutions
13
32. According to the numerical note in Section 1.2, when n = 30 the reduction to echelon form takes about 2(30)3/3 = 18,000 flops, while further reduction to reduced echelon form needs at most (30)2 = 900 flops. Of the total flops, the "backward phase" is about 900/18900 = .048 or about 5%. When n = 300, the estimates are 2(300)3/3 = 18,000,000 phase for the reduction to echelon form and (300)2 = 90,000 flops for the backward phase. The fraction associated with the backward phase is about (9104) /(18106) = .005, or about .5%. 33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 12), (2, 15), and (3, 16), the coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the augmented matrix:
1 1 1
1 2 3
1 12 1 4 15 ~ 0 9 16 0
1 1 2
1 12 1 3 3 ~ 0 8 4 0 7 6 -1
1 1 0
1 3 2
12 1 3 ~ 0 -2 0
1 1 0
1 3 1
12 3 -1
1 1 0 13 1 0 0 ~ 0 1 0 6 ~ 0 1 0 0 0 1 -1 0 0 1 The polynomial is p(t) = 7 + 6t t2.
34. [M] The system of equations to be solved is:
a0 a0 a0 a0 a0 a0
+ + + + +
a1 0 a1 2 a1 4 a1 6 a1 8
+ + + + +
a2 0 2 a2 2 a2 4 a2 6 a2 8
2 2 2
+ + + + +
a3 03 a3 2 a3 4 a3 6 a3 8
3 3
+ + + + +
a4 04 a4 2 a4 4 a4 6 a4 8
4 4 4 4 4
+ + + + +
a5 05 a5 2 a5 4 a5 6 a5 8
5 5
=
0
= 2.90 = 14.8 = 39.6 = 74.3 = 119
3
5
2 2
3 3
5 5
+ a1 10 + a2 10
+ a3 10
+ a4 10
+ a5 10
The unknowns are a0, a1, ..., a5. Use technology to compute the reduced echelon of the augmented matrix:
1 0 1 2 1 4 1 6 1 8 1 10
0 4 16 36 64 10
2
0 8 64 216 512 10
3
0 16 256 1296 4096 10
4
0 32 1024 7776 32768 105
0 1 2.9 0 14.8 0 ~ 39.6 0 74.3 0 119 0
0 2 0 0 0 0
0 4 8 24 48 80
0 8 48 192 480 960
0 16 224 1248 4032 9920
0 32 960 7680 32640 99840
0 2.9 9 30.9 62.7 104.5
1 0 0 ~ 0 0 0
0 2 0 0 0 0
0 4 8 0 0 0
0 8 48 48 192 480
0 16 224 576 2688 7680
0 32 960 4800 26880 90240
0 1 2.9 0 9 0 ~ 3.9 0 8.7 0 14.5 0
0 2 0 0 0 0
0 4 8 0 0 0
0 8 48 48 0 0
0 16 224 576 384 1920
0 32 960 4800 7680 42240
0 2.9 9 3.9 -6.9 -24.5
14
CHAPTER 1
Linear Equations in Linear Algebra
1 0 0 ~ 0 0 0 1 0 0 ~ 0 0 0
0 2 0 0 0 0 0 2 0 0 0 0
0 4 8 0 0 0 0 4 8 0 0 0
0 8 48 48 0 0 0 8 48 48 0 0
0 16 224 576 384 0 0 16 224 576 384 0
0 32 960 4800 7680 3840 0 0 0 0 0 1
0 1 2.9 0 9 0 ~ 3.9 0 -6.9 0 10 0
0 2 0 0 0 0
0 4 8 0 0 0 0 1 0 0 0 0
0 8 48 48 0 0 0 0 1 0 0 0
0 16 224 576 384 0 0 0 0 1 0 0 0 0 0 0 1 0
0 0 32 2.9 960 9 4800 3.9 7680 -6.9 1 .0026 0 0 0 0 0 1 0 1.7125 -1.1948 .6615 -.0701 .0026
0 1 0 2.8167 0 6.5000 ~"~ -8.6000 0 0 -26.900 .002604 0
Thus p(t) = 1.7125t 1.1948t2 + .6615t3 .0701t4 + .0026t5, and p(7.5) = 64.6 hundred lb.
Notes: In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) = 64.8.
If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists. Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2. The Study Guide includes a "Mathematical Note" about the phrase, "If and only if," used in Theorem 2.
1.3
SOLUTIONS
Notes: The key exercises are 1114, 1722, 25, and 26. A discussion of Exercise 25 will help students understand the notation [a1 a2 a3], {a1, a2, a3}, and Span{a1, a2, a3}.
-1 -3 -1 + ( -3) -4 1. u + v = + = = . 2 -1 2 + (-1) 1 Using the definitions carefully, -1 -3 -1 (-2)(-3) -1 + 6 5 u - 2 v = + (-2) = + = = , or, more quickly, 2 -1 2 (-2)(-1) 2 + 2 4 -1 -3 -1 + 6 5 u - 2v = - 2 = = . The intermediate step is often not written. 2 -1 2 + 2 4 3 2 3 + 2 5 2. u + v = + = = . 2 -1 2 + (-1) 1 Using the definitions carefully,
1.3
Solutions
15
3 2 3 (-2)(2) 3 + (-4) -1 u - 2 v = + (-2) = + = = , or, more quickly, 2 -1 2 (-2)(-1) 2 + 2 4 3 2 3 - 4 -1 u - 2v = - 2 = = . The intermediate step is often not written. 2 -1 2 + 2 4
3.
x2 uv u u+v v x1 v 2v u 2v
4.
u 2v x2 uv u 2v v x1 v u+v
6 x1 -3 x2 1 6 -3 1 -1 + x 4 = -7 , - x + 4 x = -7 , 5. x1 1 2 2 5 x1 0 -5 5 0 -5 6 x1 - 3x2 = 1 - x1 + 4 x2 = -7 5 x1 = -5 Usually the intermediate steps are not displayed. -2 8 1 0 6. x1 + x2 + x3 = , 3 5 -6 0 -2 x2 + 8 x2 + x3 = 3x1 + 5 x2 - 6 x3 =
6 x1 - 3x2 1 - x + 4 x = -7 2 1 5 x1 -5
-2 x1 8 x2 x3 0 = , + + 3x1 5 x2 -6 x3 0 0 0
-2 x1 + 8 x2 + x3 0 3 x + 5 x - 6 x = 0 2 3 1
Usually the intermediate steps are not displayed.
7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every vector in R2 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.
16
CHAPTER 1
Linear Equations in Linear Algebra
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 c. 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 d. 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 3v, 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
d c a
b
u v 0
2v w y x
v 2v u
z
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 R2 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. x. 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 y. 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 2u, then 4 units in the v-direction, and finally move 1 unit in the u-direction. So z = 2u + 4v u = 3u + 4v
x2 + 6 x2
9. 4 x1 - x1
+ 3x2
+ 5 x3 - x3 - 8 x3
= 0 = 0, = 0
x2 + 5 x3 0 4 x + 6 x - x = 0 2 3 1 - x1 + 3 x2 - 8 x3 0 0 1 5 0 4 + x 6 + x -1 = 0 x1 2 3 -1 3 -8 0
0 x2 5 x3 0 4 x + 6 x + - x = 0 , 3 1 2 - x1 3 x2 -8 x3 0
Usually, the intermediate calculations are not displayed.
1.3
Solutions
17
Note: The Study Guide says, "Check with your instructor whether you need to "show work" on a problem such as Exercise 9."
4 x1 x1 8 x1 + x2 - 7 x2 + 6 x2 + 3x3 - 2 x3 - 5 x3 = 9 = 2 , = 15 4 x1 + x2 + 3 x3 9 x - 7x - 2x = 2 2 3 1 8 x1 + 6 x2 - 5 x3 15 4 1 3 9 1 + x -7 + x -2 = 2 x1 2 3 8 6 -5 15
10.
4 x1 x2 3x3 9 x1 + -7 x2 + -2 x3 = 2 , 8 x1 6 x2 -5 x3 15
Usually, the intermediate calculations are not displayed.
11. The question Is b a linear combination of a1, a2, and a3? is equivalent to the question Does the vector equation x1a1 + x2a2 + x3a3 = b have a solution? The equation
1 0 5 2 x1 -2 + x2 1 + x3 -6 = -1 0 2 8 6 a1 a2 a3 b (*)
has the same solution set as the linear system whose augmented matrix is 5 2 1 0 -2 1 -6 -1 M = 0 2 8 6 Row reduce M until the pivot positions are visible: 1 0 5 2 1 0 5 2 M ~ 0 1 4 3 ~ 0 1 4 3 0 2 8 6 0 0 0 0 The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and therefore b is a linear combination of a1, a2, and a3. 12. The equation 1 0 2 -5 -2 + x 5 + x 0 = 11 x1 2 3 2 5 8 -7 a1 a2 a3 b (*)
has the same solution set as the linear system whose augmented matrix is
18
CHAPTER 1
Linear Equations in Linear Algebra
-5 5 0 11 5 8 -7 Row reduce M until the pivot positions are visible: 1 M = -2 2 0 0 2 -5 5 4 5 4 1 5 4 0 0 2 The linear system corresponding to M has no solution, so the vector equation (*) has no solution, and therefore b is not a linear combination of a1, a2, and a3. 1 M ~ 0 0 2 0 2 13. Denote the columns of A by a1, a2, a3. To determine if b is a linear combination of these columns, use the boxed fact on page 34. Row reduced the augmented matrix until you reach echelon form: 1 0 -2 -4 3 8 2 5 -4 3 1 -7 ~ 0 -3 0 -4 3 0 2 5 0 3 -7 3 -5 1 1 ~ 0 3 0
The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns of A. 1 -2 -6 11 1 -2 -6 11 14. [a1 a2 a3 b] = 0 3 7 -5 ~ 0 3 7 -5 . The linear system corresponding to this 1 -2 5 9 0 0 11 -2 matrix has a solution, so b is a linear combination of the columns of A. 15. Noninteger weights are acceptable, of course, but some simple choices are 0v1 + 0v2 = 0, and 7 1v1 + 0v2 = 1 , 0v1 + 1v2 = -6 2 1v1 + 1v2 = 4 , 1v1 1v2 = -6 3 1v1 + 0v2 = 0 , 0v1 + 1v2 = 2 1 1v1 + 1v2 = 0 , 1v1 1v2 = 5 -2 0 3 5 0 -1 -5 3 0 12 -2 -6
16. Some likely choices are 0v1 + 0v2 = 0, and
1.3
Solutions
19
4 1 -2 1 -2 4 1 -2 4 -3 1 ~ 0 17. [a1 a2 b] = 5 -15 ~ 0 1 -2 0 0 7 h 3 h + 8 3 in Span{a1, a2} when h + 17 is zero, that is, when h = 17.
4 1 -3 ~ 0 h + 8 0
-2 1 0
4 -3 . The vector b is h + 17
h 1 -3 h 1 -3 1 -3 18. [v1 v2 y] = 0 -5 ~ 0 1 -5 ~ 0 1 1 -2 8 -3 0 2 -3 + 2h 0 0 Span{v1, v2} when 7 + 2h is zero, that is, when h = 7/2.
h -5 . The vector y is in 7 + 2h
19. By inspection, v2 = (3/2)v1. Any linear combination of v1 and v2 is actually just a multiple of v1. For instance, av1 + bv2 = av1 + b(3/2)v2 = (a + 3b/2)v1 So Span{v1, v2} is the set of points on the line through v1 and 0.
Note: Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7.
20. Span{v1, v2} is a plane in R3 through the origin, because the 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{v1, v2} is the xz-plane. h h 2 2 h 2 2 21. Let y = . Then [u v y] = ~ 0 2 k + h / 2 . This augmented matrix corresponds to k -1 1 k a consistent system for all h and k. So y is in Span{u, v} for all h and k. 22. Construct any 34 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. -5 b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If were on 2 -2 -2 -5 the line through and the origin, then would have to be a multiple of , which is not 5 5 2 the case. c. True. See the line displayed just before Example 4. d. True. See the box that discusses the matrix in (5). e. False. The statement is often true, but Span{u, v} is not a plane when v is a multiple of u, or when u is the zero vector. 24. a. b. c. d. e. True. See the beginning of the subsection Vectors in Rn. True. Use Fig. 7 to draw the parallelogram determined by u v and v. False. See the first paragraph of the subsection Linear Combinations. True. See the statement that refers to Fig. 11. True. See the paragraph following the definition of Span{v1, ..., vp}.
20
CHAPTER 1
Linear Equations in Linear Algebra
25. a. There are only three vectors in the set {a1, a2, a3}, and b is not one of them. b. There are infinitely many vectors in W = Span{a1, a2, a3}. To determine if b is in W, use the method of Exercise 13.
1 0 -2 a1
0 3 6
-4 -2 3
4 1 1 ~ 0 -4 0
0 3 6
-4 -2 -5
4 1 1 ~ 0 4 0
0 3 0
-4 -2 -1
4 1 2
a 2 a3 b
The system for this augmented matrix is consistent, so b is in W. c. a1 = 1a1 + 0a2 + 0a3. See the discussion in the text following the definition of Span{v1, ..., vp}. 2 b] = -1 1
0 8 -2 6 5 1 10
26. a. [a1 a2 a3
1 ~ -1 3 3 1
0 8 -2
3 5 1
5
1 ~ 0 3 3 0
0 8 -2
3 8 -2
1 ~ 0 8 -2 0
5
0 8 0
3 8 0
8 0
5
Yes, b is a linear combination of the columns of A, that is, b is in W. b. The third column of A is in W because a3 = 0a1 + 0a2 + 1a3. 27. a. 5v1 is the output of 5 days' operation of mine #1. 150 b. The total output is x1v1 + x2v2, so x1 and x2 should satisfy x1 v1 + x2 v 2 = . 2825 20 c. [M] Reduce the augmented matrix 550 30 500 150 1 ~ 2825 0 0 1 1.5 . 4.0
Operate mine #1 for 1.5 days and mine #2 for 4 days. (This is the exact solution.) 28. a. The amount of heat produced when the steam plant burns x1 tons of anthracite and x2 tons of bituminous coal is 27.6x1 + 30.2x2 million Btu. b. The total output produced by x1 tons of anthracite and x2 tons of bituminous coal is given by the 27.6 30.2 3100 + x 6400 . vector x1 2 250 360 27.6 30.2 162 3100 + x 6400 = 23,610 . c. [M] The appropriate values for x1 and x2 satisfy x1 2 250 360 1,623 To solve, row reduce the augmented matrix: 27.6 3100 250 30.2 6400 360 162 1.000 23610 ~ 0 1623 0 0 1.000 0 3.900 1.800 0
The steam plant burned 3.9 tons of anthracite coal and 1.8 tons of bituminous coal.
1.3
Solutions
21
29. The total mass is 2 + 5 + 2 + 1 = 10. So v = (2v1 +5v2 + 2v3 + v4)/10. That is,
5 4 -4 -9 10 + 20 - 8 - 9 1.3 1 3 + 2 -3 + 8 = 1 -8 + 15 - 6 + 8 = .9 2 -4 + 5 v= 10 10 3 -2 -1 6 6 - 10 - 2 + 6 0 30. Let m be the total mass of the system. By definition, m m 1 v = (m1 v1 + " + mk v k ) = 1 v1 + " + k v k m m m The second expression displays v as a linear combination of v1, ..., vk, which shows that v is in Span{v1, ..., vk}.
31. a. The center of mass is
8 2 10 / 3 1 0 1 + 1 + 1 = . 3 1 1 4 2
b. The total mass of the new system is 9 grams. The three masses added, w1, w2, and w3, satisfy the equation
0 8 2 2 1 ( w1 + 1) + ( w2 + 1) + ( w3 + 1) = 1 1 9 4 2
which can be rearranged to
( w1 + 1) 1 + ( w2 + 1) 1 + ( w3 + 1) 4 = 18
and
0
8
2
18
0 8 2 8 w1 + w2 + w3 = 1 1 4 12
The condition w1 + w2 + w3 = 6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown below, along with a sequence of row operations:
1 0 1
1 8 1
1 2 4
6 1 8 ~ 0 12 0
1 8 0 1 8 0
1 2 3 0 0 1
6 1 8 ~ 0 6 0 4 1 4 ~ 0 2 0
1 8 0 0 8 0
1 2 1 0 0 1
6 8 2 3.5 1 4 ~ 0 2 0 0 1 0 0 0 1 3.5 .5 2
1 ~ 0 0
Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4).
Extra problem: Ignore the mass of the plate, and distribute 6 gm at the three vertices to make the center of mass at (2, 2). Answer: Place 3 g at (0, 1), 1 g at (8, 1), and 2 g at (2, 4).
32. See the parallelograms drawn on Fig. 15 from the text. Here c1, c2, c3, and c4 are suitable scalars. The darker parallelogram shows that b is a linear combination of v1 and v2, that is c1v1 + c2v2 + 0v3 = b
22
CHAPTER 1
Linear Equations in Linear Algebra
The larger parallelogram shows that b is a linear combination of v1 and v3, that is, c4v1 + 0v2 + c3v3 = b So the equation x1v1 + x2v2 + x3v3 = b has at least two solutions, not just one solution. (In fact, the equation has infinitely many solutions.)
v3 c3v3 c2v2 v2 b
0
v1
c1v1
c4v1
33. a. For j = 1,..., n, the jth entry of (u + v) + w is (uj + vj) + wj. By associativity of addition in R, this entry equals uj + (vj + wj), which is the jth entry of u + (v + w). By definition of equality of vectors, (u + v) + w = u + (v + w). b. For any scalar c, the jth entry of c(u + v) is c(uj + vj), and the jth entry of cu + cv is cuj + cvj (by definition of scalar multiplication and vector addition). These entries are equal, by a distributive law in R. So c(u + v) = cu + cv. 34. a. For j = 1,..., n, uj + (1)uj = (1)uj + uj = 0, by properties of R. By vector equality, u + (1)u = (1)u + u = 0. b. For scalars c and d, the jth entries of c(du) and (cd )u are c(duj) and (cd )uj, respectively. These entries in R are equal, so the vectors c(du) and (cd)u are equal.
Note: When an exercise in this section involves a vector equation, the corresponding technology data (in the data files on the web) is usually presented as a set of (column) vectors. To use MATLAB or other technology, a student must first construct an augmented matrix from these vectors. The MATLAB note in the Study Guide describes how to do this. The appendices in the Study Guide give corresponding information about Maple, Mathematica, and the TI and HP calculators.
1.4
SOLUTIONS
Notes: Key exercises are 120, 27, 28, 31 and 32. Exercises 29, 30, 33, and 34 are harder. Exercise 34 anticipates the Invertible Matrix Theorem but is not used in the proof of that theorem.
1. The matrix-vector product Ax product is not defined because the number of columns (2) in the 32 -4 2 3 1 6 does not match the number of entries (3) in the vector -2 . matrix 0 1 7
1.4
Solutions
23
2. The matrix-vector product Ax product is not defined because the number of columns (1) in the 31 2 5 matrix 6 does not match the number of entries (2) in the vector . -1 -1
6 3. Ax = -4 7 6 Ax = -4 7 8 4. Ax = 5 8 Ax = 5
3 1 3 1
5 6 5 12 -15 -3 2 = 2 -4 - 3 -3 = -8 + 9 = 1 , and -3 -3 6 7 6 14 -18 -4 5 6 2 + 5 (-3) -3 2 = (-4) 2 + (-3) (-3) = 1 -3 -3 7 2 + 6 (-3) -4 6
1 -4 8 3 -4 8 + 3 - 4 7 1 = 1 5 + 1 1 + 1 2 = 5 + 1 + 2 = 8 , and 2 1 1 -4 8 1 + 3 1 + (-4) 1 7 = 1 = 2 5 1 + 1 1 + 2 1 8 1
5. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A:
5 1 -8 4 -8 5 - 1 + 3 - 2 = -2 -7 3 -5 16
6. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A:
7 -3 1 2 1 -9 -2 - 5 = 9 -6 12 -3 2 -4
7. The left side of the equation is a linear combination of three vectors. Write the matrix A whose columns are those three vectors, and create a variable vector x with three entries:
4 -1 A = 7 -4 4 -1 7 -4
-5 3 -5 1
-5 3 -5 1
7 4 -8 -1 = 0 7 2 -4
-5 3 -5 1
7 x1 -8 and x = x . Thus the equation Ax = b is 2 0 x3 2
7 6 x1 -8 -8 x2 = 0 0 x 2 3 -7
24
CHAPTER 1
Linear Equations in Linear Algebra
For your information: The unique solution of this equation is (5, 7, 3). Finding the solution by hand would be time-consuming.
Note: The skill of writing a vector equation as a matrix equation will be important for both theory and application throughout the text. See also Exercises 27 and 28.
8. The left side of the equation is a linear combination of four vectors. Write the matrix A whose columns are those four vectors, and create a variable vector with four entries:
4 A = -2
-4 5
-5 4
3 4 = 0 -2
-4 5
-5 4
z1 z 3 2 , and z = . Then the equation Az = b z 3 0 z4
4 is -2
-4 5
-5 4
z1 3 z 2 4 . = 0 z3 13 z4
For your information: One solution is (7, 3, 3, 1). The general solution is z1 = 6 + .75z3 1.25z4, z2 = 5 .5z3 .5z4, with z3 and z4 free. 9. The system has the same solution set as the vector equation 3 1 -5 9 x1 + x2 + x3 = 0 1 4 0 and this equation has the same solution set as the matrix equation 3 1 0 1 x1 -5 9 x = 4 2 0 x3
10. The system has the same solution set as the vector equation 8 -1 4 x1 5 + x2 4 = 1 1 -3 2 and this equation has the same solution set as the matrix equation 8 5 1 1 0 -2 -1 4 x1 4 = 1 x2 2 -3 -2 1 2 ~ 0 9 0 -2 1 2 ~ 0 5 0 -2 1 2 ~ 0 1 0 -6 1 -3 ~ 0 1 0 0 -3 1
11. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system: 2 1 -4 4 5 -3 2 1 0 4 5 5 2 1 0 4 5 1 2 1 0 0 0 1 0 1 0 0 0 1
1.4
Solutions
25
x1 The solution is x2 x 3
=
x1 0 = -3 . As a vector, the solution is x = x2 = -3 . x3 1 = 1 0
12. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system: 1 -3 0 1 ~ 0 0 2 -1 5 2 5 0 1 2 3 0 0 1 0 1 1 ~ 0 -1 0 -1 1 -4 ~ 0 1 0 = 3/ 5 2 5 5 2 1 0 1 5 3 0 0 1 0 1 1 ~ 0 -1 0 -1 1 -4 / 5 ~ 0 1 0 2 5 0 0 1 0 1 5 -2 0 0 1 0 1 1 ~ 0 -2 0 3/ 5 -4 / 5 1 2 5 0 1 5 1 0 1 1
x1 The solution is x2 x 3
x1 3/ 5 = -4 / 5 . As a vector, the solution is x = x2 = -4 / 5 . x3 1 1 =
13. The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of the columns of A. This happens if and only if the equation Ax = u has a solution. (See the box preceding Example 3 in Section 1.4.) To study this equation, reduce the augmented matrix [A u] 3 -2 1 -5 6 1 0 1 4 ~ -2 4 3 1 6 -5 4 1 4 ~ 0 0 0 1 8 -8 4 1 12 ~ 0 -12 0 1 8 0 4 12 0
The equation Ax = u has a solution, so u is in the plane spanned by the columns of A. For your information: The unique solution of Ax = u is (5/2, 3/2). 14. Reduce the augmented matrix [A u] to echelon form: 5 0 1 8 1 3 7 -1 0 2 1 -3 ~ 0 2 5 3 1 8 0 -1 7 2 1 -3 ~ 0 2 0 3 1 -7 0 -1 7 2 1 -3 ~ 0 -8 0 3 1 0 0 -1 0 2 -3 -29
The equation Ax = u has no solution, so u is not in the subset spanned by the columns of A. b1 2 -1 b1 2 -1 . 15. The augmented matrix for Ax = b is , which is row equivalent to 0 3 b2 0 b2 + 3b1 -6 This shows that the equation Ax = b is not consistent when 3b1 + b2 is nonzero. The set of b for which the equation is consistent is a line through the originthe set of all points (b1, b2) satisfying b2 = 3b1. 1 16. Row reduce the augmented matrix [A b]: A = -3 5 1 -3 5 -3 2 -1 -4 6 -8 b1 1 b2 ~ 0 b3 0 -3 -7 14 -4 -6 12 b1 b2 + 3b1 b3 - 5b1 -3 2 -1 -4 b1 , b = b . 6 2 b3 -8
26
CHAPTER 1
Linear Equations in Linear Algebra
1 ~ 0 0
-3 -7 0
-4
1 = 0 -6 b2 + 3b1 0 b3 - 5b1 + 2(b2 + 3b1 ) 0 b1
-3 -7 0
-4
-6 b2 + 3b1 0 b1 + 2b2 + b3 b1
The equation Ax = b is consistent if and only if b1 + 2b2 + b3 = 0. The set of such b is a plane through the origin in R3. 17. Row reduction shows that only three rows of A contain a pivot position: 1 -1 A= 0 2 3 -1 -4 0 0 -1 2 3 3 1 1 0 ~ -8 0 -1 0 3 2 -4 -6 0 -1 2 3 3 1 4 0 ~ -8 0 -7 0 3 2 0 0 0 -1 0 0 3 1 4 0 ~ 0 0 5 0 3 2 0 0 0 -1 0 0 3 4 5 0
Because not every row of A contains a pivot position, Theorem 4 in Section 1.4 shows that the equation Ax = b does not have a solution for each b in R4. 18. Row reduction shows that only three rows of B contain a pivot position: 1 0 B= 1 -2 3 1 2 -8 -2 1 -3 2 2 1 -5 0 ~ 7 0 -1 0 3 1 -1 -2 -2 1 -1 -2 2 1 -5 0 ~ 5 0 3 0 3 1 0 0 -2 1 0 0 2 1 -5 0 ~ 0 0 -7 0 3 1 0 0 -2 1 0 0 2 -5 -7 0
Because not every row of B contains a pivot position, Theorem 4 in Section 1.4 shows that the equation Bx = y does not have a solution for each y in R4. 19. The work in Exercise 17 shows that statement (d) in Theorem 4 is false. So all four statements in Theorem 4 are false. Thus, not all vectors in R4 can be written as a linear combination of the columns of A. Also, the columns of A do not span R4. 20. The work in Exercise 18 shows that statement (d) in Theorem 4 is false. So all four statements in Theorem 4 are false. Thus, not all vectors in R4 can be written as a linear combination of the columns of B. The columns of B certainly do not span R3, because each column of B is in R4, not R3. (This question was asked to alert students to a fairly common misconception among students who are just learning about spanning.) 21. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row. 1 0 -1 0 0 -1 0 1 1 1 0 0 ~ 0 0 -1 0 0 -1 0 1 1 1 0 0 ~ 1 0 -1 0 0 -1 0 0 1 1 0 0 ~ 1 0 -1 0 0 1 0 0 1 0 . 1 0
The matrix [v1 v2 v3] does not have a pivot in each row, so the columns of the matrix do not span R4, by Theorem 4. That is, {v1, v2, v3} does not span R4.
Note: Some students may realize that row operations are not needed, and thereby discover the principle
covered in Exercises 31 and 32.
1.4
Solutions
27
22. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row. 0 4 -2 8 -5 0 0 -3 -1 ~ 0 -3 -1 -2 8 -5 0 0 4 The matrix [v1 v2 v3] has a pivot in each row, so the columns of the matrix span R4, by Theorem 4. That is, {v1, v2, v3} spans R4. 23. a. b. c. d. e. f. False. See the paragraph following equation (3). The text calls Ax = b a matrix equation. True. See the box before Example 3. False. See the warning following Theorem 4. True. See Example 4. True. See parts (c) and (a) in Theorem 4. True. In Theorem 4, statement (a) is false if and only if statement (d) is also false.
24. a. True. This statement is in Theorem 3. However, the statement is true without any "proof" because, by definition, Ax is simply a notation for x1a1 + + xnan, where a1, ..., an are the columns of A. b. True. See Example 2. c. True, by Theorem 3. d. True. See the box before Example 2. Saying that b is not in the set spanned by the columns of A is the same a saying that b is not a linear combination of the columns of A. e. False. See the warning that follows Theorem 4. f. True. In Theorem 4, statement (c) is false if and only if statement (a) is also false. 25. By definition, the matrix-vector product on the left is a linear combination of the columns of the matrix, in this case using weights 3, 1, and 2. So c1 = 3, c2 = 1, and c3 = 2. 26. The equation in x1 and x2 involves the vectors u, v, and w, and it may be viewed as x1 v ] = w. By definition of a matrix-vector product, x1u + x2v = w. The stated fact that x2 3u 5v w = 0 can be rewritten as 3u 5v = w. So, a solution is x1 = 3, x2 = 5.
[u
27. Place the vectors q1, q2, and q3 into the columns of a matrix, say, Q and place the weights x1, x2, and x3 into a vector, say, x. Then the vector equation becomes Qx = v, where Q = [q1 q2 x1 q3] and x = x2 x3
Note: If your answer is the equation Ax = b, you need to specify what A and b are. 28. The matrix equation can be written as c1v1 + c2v2 + c3v3 + c4v4 + c5v5 = v6, where c1 = 3, c2 = 2, c3 = 4, c4 = 1, c5 = 2, and -3 5 -4 9 7 8 v1 = , v 2 = , v 3 = , v 4 = , v 5 = , v 6 = 5 8 1 -2 -4 -1
28
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Linear Equations in Linear Algebra
29. Start with any 33 matrix B in echelon form that has three pivot positions. Perform a row operation (a row interchange or a row replacement) that creates a matrix A that is not in echelon form. Then A has the desired property. The justification is given by row reducing A to B, in order to display the pivot positions. Since A has a pivot position in every row, the columns of A span R3, by Theorem 4. 30. Start with any nonzero 33 matrix B in echelon form that has fewer than three pivot positions. Perform a row operation that creates a matrix A that is not in echelon form. Then A has the desired property. Since A does not have a pivot position in every row, the columns of A do not span R3, by Theorem 4. 31. A 32 matrix has three rows and two columns. With only two columns, A can have at most two pivot columns, and so A has at most two pivot positions, which is not enough to fill all three rows. By Theorem 4, the equation Ax = b cannot be consistent for all b in R3. Generally, if A is an mn matrix with m > n, then A can have at most n pivot positions, which is not enough to fill all m rows. Thus, the equation Ax = b cannot be consistent for all b in R3. 32. A set of three vectors in cannot span R4. Reason: the matrix A whose columns are these three vectors has four rows. To have a pivot in each row, A would have to have at least four columns (one for each pivot), which is not the case. Since A does not have a pivot in every row, its columns do not span R4, by Theorem 4. In general, a set of n vectors in Rm cannot span Rm when n is less than m. 33. If the equation Ax = b has a unique solution, then the associated system of equations does not have any free variables. If every variable is a basic variable, then each column of A is a pivot column. So the 1 0 0 0 1 0 . reduced echelon form of A must be 0 0 1 0 0 0
Note: Exercises 33 and 34 are difficult in the context of this section because the focus in Section 1.4 is on existence of solutions, not uniqueness. However, these exercises serve to review ideas from Section 1.2, and they anticipate ideas that will come later.
34. If the equation Ax = b has a unique solution, then the associated system of equations does not have any free variables. If every variable is a basic variable, then each column of A is a pivot column. So the 1 0 0 reduced echelon form of A must be 0 1 0 . Now it is clear that A has a pivot position in each row. 0 0 1 By Theorem 4, the columns of A span R3. 35. Given Ax1 = y1 and Ax2 = y2, you are asked to show that the equation Ax = w has a solution, where w = y1 + y2. Observe that w = Ax1 + Ax2 and use Theorem 5(a) with x1 and x2 in place of u and v, respectively. That is, w = Ax1 + Ax2 = A(x1 + x2). So the vector x = x1 + x2 is a solution of w = Ax. 36. Suppose that y and z satisfy Ay = z. Then 4z = 4Ay. By Theorem 5(b), 4Ay = A(4y). So 4z = A(4y), which shows that 4y is a solution of Ax = 4z. Thus, the equation Ax = 4z is consistent.
2 8 7 2 8 5 5 7 2 5 8 7 5 3 4 9 0 11/ 7 3/ 7 23/ 7 0 11/ 7 3/ 7 23 / 7 ~ ~ 37. [M] 6 10 2 7 0 58 / 7 16 / 7 1/ 7 0 0 50 /11 189 /11 11 23 0 0 0 0 3 7 9 2 15 0
1.4
Solutions
29
2 8 -5 7 0 -1.57 .429 -3.29 , to three significant figures. The original matrix does not or, approximately 0 0 4.55 -17.2 0 0 0 0 have a pivot in every row, so its columns do not span R4, by Theorem 4. 5 6 38. [M] 4 -9 -7 -8 -4 11 -4 -7 -9 16 9 5 5 0 ~ -9 0 7 0 -7 2/5 8/5 -8 / 5 -4 -11/ 5 -29 / 5 44 / 5 9 5 -29 / 5 0 ~ -81/ 5 0 116 / 5 0 -7 2/5 0 0 -4 -11/ 5 3 * 9 -29 / 5 7 *
MATLAB shows starred entries for numbers that are essentially zero (to many decimal places). So, with pivots only in the first three rows, the original matrix has columns that do not span R4, by Theorem 4. 12 -9 39. [M] -6 4 12 0 ~ 0 0 -7 4 11 -6 11 -8 -7 10 11 1/ 4 0 28 / 5 -9 7 3 -5 5 12 -3 0 ~ -9 0 12 0 -9 1/ 4 0 -41/15 -7 -5 / 4 15 / 2 -11/ 3 11 1/ 4 -3/ 2 19 / 3 -7 -5 / 4 0 0 -9 1/ 4 -3/ 2 -2 11 1/ 4 28 / 5 0 5 3/ 4 -13/ 2 31/ 3 -9 1/ 4 -41/15 0 5 3/ 4 122 /15 -2
-7 -5 / 4 0 0
5 12 3/ 4 0 ~ -2 0 122 /15 0
The original matrix has a pivot in every row, so its columns span R4, by Theorem 4. 8 -7 40. [M] 11 -3 8 0 ~ 0 0 11 -8 7 4 -6 5 -7 1 -6 -1/ 4 0 0 -7 6 -9 8 -7 -1/ 8 0 6 13 8 -9 0 ~ -6 0 7 0 11 13/ 8 -65 / 8 65 / 8 -6 -1/ 4 5/ 4 -5 / 4 -7 -1/ 8 5/8 43/ 8 13 19 / 8 -191/ 8 95 / 8 13 19 / 8 0 -12
11 13/ 8 0 0
13 8 19 / 8 0 ~ -12 0 0 0
11 13/ 8 0 0
-6 -1/ 4 0 0
-7 -1/ 8 6 0
The original matrix has a pivot in every row, so its columns span R4, by Theorem 4. 41. [M] Examine the calculations in Exercise 39. Notice that the fourth column of the original matrix, say A, is not a pivot column. Let Ao be the matrix formed by deleting column 4 of A, let B be the echelon form obtained from A, and let Bo be the matrix obtained by deleting column 4 of B. The sequence of row operations that reduces A to B also reduces Ao to Bo. Since Bo is in echelon form, it shows that Ao has a pivot position in each row. Therefore, the columns of Ao span R4. It is possible to delete column 3 of A instead of column 4. In this case, the fourth column of A becomes a pivot column of Ao, as you can see by looking at what happens when column 3 of B is deleted. For later work, it is desirable to delete a nonpivot column.
30
CHAPTER 1
Linear Equations in Linear Algebra
Note: Exercises 41 and 42 help to prepare for later work on the column space of a matrix. (See Section 2.9 or
4.6.) The Study Guide points out that these exercises depend on the following idea, not explicitly mentioned in the text: when a row operation is performed on a matrix A, the calculations for each new entry depend only on the other entries in the same column. If a column of A is removed, forming a new matrix, the absence of this column has no affect on any row-operation calculations for entries in the other columns of A. (The absence of a column might affect the particular choice of row operations performed for some purpose, but that is not being considered here.) 42. [M] Examine the calculations in Exercise 40. The third column of the original matrix, say A, is not a pivot column. Let Ao be the matrix formed by deleting column 3 of A, let B be the echelon form obtained from A, and let Bo be the matrix obtained by deleting column 3 of B. The sequence of row operations that reduces A to B also reduces Ao to Bo. Since Bo is in echelon form, it shows that Ao has a pivot position in each row. Therefore, the columns of Ao span R4. It is possible to delete column 2 of A instead of column 3. (See the remark for Exercise 41.) However, only one column can be deleted. If two or more columns were deleted from A, the resulting matrix would have fewer than four columns, so it would have fewer than four pivot positions. In such a case, not every row could contain a pivot position, and the columns of the matrix would not span R4, by Theorem 4.
Notes: At the end of Section 1.4, the Study Guide gives students a method for learning and mastering linear
algebra concepts. Specific directions are given for constructing a review sheet that connects the basic definition of "span" with related ideas: equivalent descriptions, theorems, geometric interpretations, special cases, algorithms, and typical computations. I require my students to prepare such a sheet that reflects their choices of material connected with "span", and I make comments on their sheets to help them refine their review. Later, the students use these sheets when studying for exams. The MATLAB box for Section 1.4 introduces two useful commands gauss and bgauss that allow a student to speed up row reduction while still visualizing all the steps involved. The command B = gauss(A,1) causes MATLAB to find the left-most nonzero entry in row 1 of matrix A, and use that entry as a pivot to create zeros in the entries below, using row replacement operations. The result is a matrix that a student might write next to A as the first stage of row reduction, since there is no need to write a new matrix after each separate row replacement. I use the gauss command frequently in lectures to obtain an echelon form that provides data for solving various problems. For instance, if a matrix has 5 rows, and if row swaps are not needed, the following commands produce an echelon form of A:
B = gauss(A,1), B = gauss(B,2), B = gauss(B,3), B = gauss(B,4)
If an interchange is required, I can insert a command such as B = swap(B,2,5) . The command bgauss uses the left-most nonzero entry in a row to produce zeros above that entry. This command, together with scale, can change an echelon form into reduced echelon form. The use of gauss and bgauss creates an environment in which students use their computer program the same way they work a problem by hand on an exam. Unless you are able to conduct your exams in a computer laboratory, it may be unwise to give students too early the power to obtain reduced echelon forms with one command--they may have difficulty performing row reduction by hand during an exam. Instructors whose students use a graphic calculator in class each day do not face this problem. In such a case, you may wish to introduce rref earlier in the course than Chapter 4 (or Section 2.8), which is where I finally allow students to use that command.
1.5
SOLUTIONS
Notes: The geometry helps students understand Span{u, v}, in preparation for later discussions of subspaces. The parametric vector form of a solution set will be used throughout the text. Figure 6 will appear again in Sections 2.9 and 4.8.
1.5
Solutions
31
For solving homogeneous systems, the text recommends working with the augmented matrix, although no calculations take place in the augmented column. See the Study Guide comments on Exercise 7 that illustrate two common student errors. All students need the practice of Exercises 114. (Assign all odd, all even, or a mixture. If you do not assign Exercise 7, be sure to assign both 8 and 10.) Otherwise, a few students may be unable later to find a basis for a null space or an eigenspace. Exercises 2934 are important. Exercises 33 and 34 help students later understand how solutions of Ax = 0 encode linear dependence relations among the columns of A. Exercises 3538 are more challenging. Exercise 37 will help students avoid the standard mistake of forgetting that Theorem 6 applies only to a consistent equation Ax = b. 1. Reduce the augmented matrix to echelon form and circle the pivot positions. If a column of the coefficient matrix is not a pivot column, the corresponding variable is free and the system of equations has a nontrivial solution. Otherwise, the system has only the trivial solution. 2 -2 4 -5 -7 2 8 1 7 0 2 0 ~ 0 0 0 -5 -12 12 8 9 -9 0 2 0 ~ 0 0 0 -5 -12 0 8 9 0 0 0 0
The variable x3 is free, so the system has a nontrivial solution. 1 2. -2 1 -3 1 2 7 -4 9 0 1 0 ~ 0 0 0 -3 -5 5 7 10 2 0 1 0 ~ 0 0 0 -3 -5 0 7 10 12 0 0 0
There is no free variable; the system has only the trivial solution. 5 -7 0 -3 5 -7 0 -3 3. ~ 0 -3 15 0 . The variable x3 is free; the system has nontrivial solutions. 1 0 -6 7 An alert student will realize that row operations are unnecessary. With only two equations, there can be at most two basic variables. One variable must be free. Refer to Exercise 31 in Section 1.2. 7 9 0 1 -2 6 0 1 -2 6 0 -5 4. ~ -5 ~ 0 -3 39 0 . x3 is a free variable; the system has 7 9 0 1 -2 6 0 nontrivial solutions. As in Exercise 3, row operations are unnecessary. 1 5. -4 0 3 -9 -3 1 2 -6 0 1 0 ~ 0 0 0 3 3 -3 1 6 -6 0 1 0 ~ 0 0 0 0 3 0 -5 6 0 0 1 0 ~ 0 0 0 0 1 0 -5 2 0 0 0 0
x1
- 5 x3 = 0 x2 + 2 x 3 = 0 . The variable x3 is free, x1 = 5x3, and x2 = 2x3. 0 = 0
x1 5 x3 5 x = -2 x = x -2 . In parametric vector form, the general solution is x = 2 3 3 x3 x3 1
32
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Linear Equations in Linear Algebra
1 6. 1 -3
3 4 -7
-5 -8 9
0 1 0 ~ 0 0 0
3 1 2
-5 -3 -6
0 1 0 ~ 0 0 0
3 1 0
-5 -3 0
0 1 0 ~ 0 0 0
0 1 0
4 -3 0
0 0 0
x1
+ 4 x3 = 0 x2 - 3 x 3 = 0 . The variable x3 is free, x1 = 4x3, and x2 = 3x3. 0 = 0
x1 -4 x3 -4 x = 3 x = x 3 . In parametric vector form, the general solution is x = 2 3 3 x3 x3 1 1 7. 0 3 1 -3 -4 7 5 0 1 ~ 0 0 0 1 9 -4 -8 5
+ 9 x3 - 8 x4 = 0 0 x1 . 0 x2 - 4 x3 + 5 x4 = 0
The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = 9x3 + 8x4, and x2 = 4x3 5x4. The general solution is x1 -9 x3 + 8 x4 -9 x3 8 x4 -9 8 x 4 x - 5 x 4 x -5 x 4 -5 4 4 x = 2 = 3 = 3 + = x3 + x4 x3 x3 0 1 0 x3 x4 0 1 x4 0 x4 1 8. 0 -2 1 -9 2 5 -6 0 1 ~ 0 0 0 1 -5 2 -7 -6
- 5 x3 - 7 x4 = 0 0 x1 . 0 x2 + 2 x3 - 6 x4 = 0
The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = 5x3 + 7x4 and x2 = 2x3 + 6x4. The general solution in parametric vector form is x1 5 x3 + 7 x4 5 x3 7 x4 5 7 x -2 x + 6 x -2 x 6 x -2 6 3 4 3 = + 4 = x3 + x4 x = 2 = x3 x3 0 1 0 x3 x4 0 1 x4 0 x4 3 9. -1 -9 3 6 -2 0 1 ~ 0 3 -3 -9 2 6 0 1 ~ 0 0 -3 0 2 0 0 0
x1 - 3x2 + 2 x3 = 0 . 0 = 0
The solution is x1 = 3x2 2x3, with x2 and x3 free. In parametric vector form, 3 x2 - 2 x3 3 x2 -2 x3 3 -2 x = x + 0 = x 1 + x 0 . x= 2 2 3 2 x3 0 x3 0 1
- 4 x4 = 0 1 3 0 -4 0 1 3 0 -4 0 x1 - 3 x2 10. . ~ 0 0 0 0 0 0 = 0 2 6 0 -8 0 The only basic variable is x1, so x2, x3, and x4 are free. (Note that x3 is not zero.) Also, x1 = 3x2 + 4x4. The general solution is
1.5
Solutions
33
x1 3 x2 + 4 x4 3x2 0 4 x4 3 0 4 x x 0 0 1 0 x2 = 2 + + = x + x + x 0 x = 2 = 2 3 4 x3 0 x3 0 0 1 0 x3 x4 0 0 1 x4 0 0 x4 1 0 11. 0 0 -4 0 0 0 -2 1 0 0 0 0 0 0 3 0 1 0 -5 -1 -4 0 0 1 0 0 ~ 0 0 0 0 -4 0 0 0 -2 1 0 0 0 0 0 0 0 0 1 0 7 -1 -4 0 0 1 0 0 ~ 0 0 0 0 -4 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 5 -1 -4 0 0 0 0 0
x1 - 4 x2 x3
+ 5 x6 = - x6 = x5 - 4 x6 = 0 =
0 0 . The basic variables are x1, x3, and x5. The remaining variables are free. 0 0
In particular, x4 is free (and not zero as some may assume). The solution is x1 = 4x2 5x6, x3 = x6, x5 = 4x6, with x2, x4, and x6 free. In parametric vector form, x1 4 x2 - 5 x6 4 x2 0 -5 x6 0 -5 4 x x 0 0 0 0 1 x2 2 2 x3 0 0 x6 0 1 0 x6 x= = = = x2 + x4 + x6 + + x4 x4 0 x4 0 1 0 0 x5 4 x6 0 0 4 x6 0 4 0 x6 0 1 0 x6 0 0 x6
u
v
w
Note: The Study Guide discusses two mistakes that students often make on this type of problem.
1 0 12. 0 0 5 0 0 0 2 1 0 0 -6 -7 0 0 9 4 0 0 0 -8 1 0 0 1 0 0 ~ 0 0 0 0 5 0 0 0 2 1 0 0 -6 -7 0 0 9 4 0 0 0 0 1 0 0 1 0 0 ~ 0 0 0 0 5 0 0 0 0 1 0 0 8 -7 0 0 1 4 0 0 0 0 1 0 0 0 0 0
x1 + 5 x2
+ 8 x4 + x5 x3 - 7 x4 + 4 x5
= = x6 = 0 =
0 0 . 0 0
The basic variables are x1, x3, and x6; the free variables are x2, x4, and x5. The general solution is x1 = 5x2 8x4 x5, x3 = 7x4 4x5, and x6 = 0. In parametric vector form, the solution is
34
CHAPTER 1
Linear Equations in Linear Algebra
x1 -5 x2 - 8 x4 - x5 -5 x2 -8 x4 - x5 -5 -8 -1 x 0 x 0 0 1 x2 2 0 2 x 7 x4 - 4 x5 0 7 x4 -4 x5 7 -4 0 x = 3 = = + + = x2 + x4 + x5 x4 x4 1 0 0 x4 0 0 x5 0 1 0 0 x5 0 x5 0 0 0 0 0 0 0 x6
13. To write the general solution in parametric vector form, pull out the constant terms that do not involve the free variable:
x1 5 + 4 x3 5 4 x3 5 4 x = -2 - 7 x = -2 + -7 x = -2 + x -7 = p + x q. x = 2 3 3 3 3 x3 x3 0 x3 0 1
p
q
4 -7 . 1
5 Geometrically, the solution set is the line through -2 in the direction of 0
14. To write the general solution in parametric vector form, pull out the constant terms that do not involve the free variable:
x1 3 x4 0 3x4 0 3 x 8 + x 8 x 8 1 2 4 4 x= = = + = + x4 = p + x4q x3 2 - 5 x4 2 -5 x4 2 -5 x4 x4 0 x4 0 1 p q
The solution set is the line through p in the direction of q.
15. Row reduce the augmented matrix for the system: 1 -4 0 1 ~ 0 0
3 -9 -3 3 1 0 1 2 -6 1 2 0 1 1 -1 ~ 0 -3 0 1 1 1 ~ 0 0 0 0 1 0 3 3 -3 -5 2 0 1 6 -6 1 1 3 ~ 0 -3 0 3 3 0 1 6 0 1 3 0
- 5 x3 = -2 -2 x1 . x2 + 2 x3 = 1 . 1 0 0 = 0
Thus x1 = 2 + 5x3, x2 = 1 2x3, and x3 is free. In parametric vector form,
x1 -2 + 5 x3 -2 5 x3 -2 5 x = 1 - 2 x = 1 + -2 x = 1 + x -2 x = 2 3 3 3 x3 x3 0 x3 0 1
1.5
Solutions
35
-2 The solution set is the line through 1 , parallel to the line that is the solution set of the homogeneous 0 system in Exercise 5. 16. Row reduce the augmented matrix for the system: 1 1 -3
3 4 -7 -5 -8 9 4 1 7 ~ 0 -6 0 3 1 2 -5 -3 -6 4 1 3 ~ 0 6 0 3 1 0 -5 -3 0 4 1 3 ~ 0 0 0 0 1 0 4 -3 0 -5 3 0
x1
+ 4 x3 = -5 x2 - 3 x3 = 3 . Thus x1 = 5 4x3, x2 = 3 + 3x3, and x3 is free. In parametric vector form, 0 = 0
x1 -5 - 4 x3 -5 -4 x3 -5 -4 x = 3 + 3 x = 3 + 3x = 3 + x 3 x = 2 3 3 3 x3 x3 0 x3 0 1 -5 The solution set is the line through 3 , parallel to the line that is the solution set of the homogeneous 0 system in Exercise 6. 17. Solve x1 + 9x2 4x3 = 2 for the basic variable: x1 = 2 9x2 + 4x3, with x2 and x3 free. In vector form, the solution is x1 -2 - 9 x2 + 4 x3 -2 -9 x2 4 x3 -2 -9 4 = 0 + x + 0 = 0 + x 1 + x 0 x = x2 = x2 2 2 3 x3 0 0 x3 0 0 1 x3
The solution of x1 + 9x2 4x3 = 0 is x1 = 9x2 + 4x3, with x2 and x3 free. In vector form,
x1 -9 x2 + 4 x3 -9 x2 4 x3 -9 4 x = = x + 0 = x 1 + x 0 = x u + x v x = 2 x2 2 3 2 2 3 x3 0 x3 0 1 x3
The solution set of the homogeneous equation is the plane through the origin in R3 spanned by u and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through the -2 point p = 0 . 0
18. Solve x1 3x2 + 5x3 = 4 for the basic variable: x1 = 4 + 3x2 5x3, with x2 and x3 free. In vector form, the solution is x1 4 + 3x2 - 5 x3 = x = x2 = x2 x3 x3
4 3x2 -5 x3 0 + x2 + 0 = 0 0 x3 4 3 -5 + x 1 + x 0 0 2 3 0 1 0
36
CHAPTER 1
Linear Equations in Linear Algebra
The solution of x1 3x2 + 5x3 = 0 is x1 = 3x2 5x3, with x2 and x3 free. In vector form, x1 3 x2 - 5 x3 3x2 -5 x3 3 -5 x = x = x + 0 = x 1 + x 0 = x u + x v x = 2 2 3 2 2 3 2 x3 x3 0 x3 0 1 The solution set of the homogeneous equation is the plane through the origin in R3 spanned by u and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through the 4 point p = 0 . 0
19. The line through a parallel to b can be written as x = a + t b, where t represents a parameter:
x1 = -2 - 5t x -2 -5 x = 1 = + t , or x2 0 3 x2 = 3t
20. The line through a parallel to b can be written as x = a + tb, where t represents a parameter:
x1 = 3 - 7t x 3 -7 x = 1 = + t , or x2 -4 8 x2 = -4 + 8t
2 -3 21. The line through p and q is parallel to q p. So, given p = and q = , form -5 1 -3 - 2 -5 2 -5 q -p = = 6 , and write the line as x = p + t(q p) = -5 + t 6 . 1 - (-5) -6 0 22. The line through p and q is parallel to q p. So, given p = and q = , form 3 -4 0 - (-6) 6 -6 6 q -p = = -7 , and write the line as x = p + t(q p) = 3 + t -7 -4 - 3
Note: Exercises 21 and 22 prepare for Exercise 27 in Section 1.8.
23. a. True. See the first paragraph of the subsection titled Homogeneous Linear Systems. b. False. The equation Ax = 0 gives an implicit description of its solution set. See the subsection entitled Parametric Vector Form. c. False. The equation Ax = 0 always has the trivial solution. The box before Example 1 uses the word nontrivial instead of trivial. d. False. The line goes through p parallel to v. See the paragraph that precedes Fig. 5. e. False. The solution set could be empty! The statement (from Theorem 6) is true only when there exists a vector p such that Ap = b. 24. a. False. A nontrivial solution of Ax = 0 is any nonzero x that satisfies the equation. See the sentence before Example 2. b. True. See Example 2 and the paragraph following it.
1.5
Solutions
37
c. True. If the zero vector is a solution, then b = Ax = A0 = 0. d. True. See the paragraph following Example 3. e. False. The statement is true only when the solution set of Ax = 0 is nonempty. Theorem 6 applies only to a consistent system. 25. Suppose p satisfies Ax = b. Then Ap = b. Theorem 6 says that the solution set of Ax = b equals the set S ={w : w = p + vh for some vh such that Avh = 0}. There are two things to prove: (a) every vector in S satisfies Ax = b, (b) every vector that satisfies Ax = b is in S. a. Let w have the form w = p + vh, where Avh = 0. Then Aw = A(p + vh) = Ap + Avh. By Theorem 5(a) in section 1.4 =b+0=b So every vector of the form p + vh satisfies Ax = b. b. Now let w be any solution of Ax = b, and set vh = w - p. Then
Avh = A(w p) = Aw Ap = b b = 0 So vh satisfies Ax = 0. Thus every solution of Ax = b has the form w = p + vh.
26. (Geometric argument using Theorem 6.) Since Ax = b is consistent, its solution set is obtained by translating the solution set of Ax = 0, by Theorem 6. So the solution set of Ax = b is a single vector if and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has only the trivial solution. (Proof using free variables.) If Ax = b has a solution, then the solution is unique if and only if there are no free variables in the corresponding system of equations, that is, if and only if every column of A is a pivot column. This happens if and only if the equation Ax = 0 has only the trivial solution. 27. When A is the 33 zero matrix, every x in R3 satisfies Ax = 0. So the solution set is all vectors in R3. 28. No. If the solution set of Ax = b contained the origin, then 0 would satisfy A0= b, which is not true since b is not the zero vector. 29. a. When A is a 33 matrix with three pivot positions, the equation Ax = 0 has no free variables and hence has no nontrivial solution. b. With three pivot positions, A has a pivot position in each of its three rows. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for every possible b. The term "possible" in the exercise means that the only vectors considered in this case are those in R3, because A has three rows. 30. a. When A is a 33 matrix with two pivot positions, the equation Ax = 0 has two basic variables and one free variable. So Ax = 0 has a nontrivial solution. b. With only two pivot positions, A cannot have a pivot in every row, so by Theorem 4 in Section 1.4, the equation Ax = b cannot have a solution for every possible b (in R3). 31. a. When A is a 32 matrix with two pivot positions, each column is a pivot column. So the equation Ax = 0 has no free variables and hence no nontrivial solution. b. With two pivot positions and three rows, A cannot have a pivot in every row. So the equation Ax = b cannot have a solution for every possible b (in R3), by Theorem 4 in Section 1.4. 32. a. When A is a 24 matrix with two pivot positions, the equation Ax = 0 has two basic variables and two free variables. So Ax = 0 has a nontrivial solution. b. With two pivot positions and only two rows, A has a pivot position in every row. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for every possible b (in R2).
38
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Linear Equations in Linear Algebra
-2 -6 7 + x 21 and notice that the second column is 3 times the first. So suitable values for 33. Look at x1 2 -3 -9 3 x1 and x2 would be 3 and 1 respectively. (Another pair would be 6 and 2, etc.) Thus x = -1 satisfies Ax = 0. 34. Inspect how the columns a1 and a2 of A are related. The second column is 3/2 times the first. Put 3 another way, 3a1 + 2a2 = 0. Thus satisfies Ax = 0. 2
Note: Exercises 33 and 34 set the stage for the concept of linear dependence.
35. Look for A = [a1 a2 a3] such that 1a1 + 1a2 + 1a3 = 0. That is, construct A so that each row sum (the sum of the entries in a row) is zero. 36. Look for A = [a1 a2 a3] such that 1a1 2a2 + 1a3 = 0. That is, construct A so that the sum of the first and third columns is twice the second column. 37. Since the solution set of Ax = 0 contains the point (4,1), the vector x = (4,1) satisfies Ax = 0. Write this equation as a vector equation, using a1 and a2 for the columns of A: 4a1 + 1a2 = 0 Then a2 = 4a1. So choose any nonzero vector for the first column of A and multiply that column by 4 1 -4 to get the second column of A. For example, set A = . 1 -4
Finally, the only way the solution set of Ax = b could not be parallel to the line through (1,4) and the origin is for the solution set of Ax = b to be empty. This does not contradict Theorem 6, because that theorem applies only to the case when the equation Ax = b has a nonempty solution set. For b, take any vector that is not a multiple of the columns of A.
Note: In the Study Guide, a "Checkpoint" for Section 1.5 will help students with Exercise 37.
38. No. If Ax = y has no solution, then A cannot have a pivot in each row. Since A is 33, it has at most two pivot positions. So the equation Ax = z for any z has at most two basic variables and at least one free variable. Thus, the solution set for Ax = z is either empty or has infinitely many elements. 39. If u satisfies Ax = 0, then Au = 0. For any scalar c, Theorem 5(b) in Section 1.4 shows that A(cu) = cAu = c0 = 0. 40. Suppose Au = 0 and Av = 0. Then, since A(u + v) = Au + Av by Theorem 5(a) in Section 1.4, A(u + v) = Au + Av = 0 + 0 = 0. Now, let c and d be scalars. Using both parts of Theorem 5, A(cu + dv) = A(cu) + A(dv) = cAu + dAv = c0 + d0 = 0.
Note: The MATLAB box in the Study Guide introduces the zeros command, in order to augment a matrix
with a column of zeros.
1.6
Solutions
39
1.6
SOLUTIONS
1. Fill in the exchange table one column at a time. The entries in a column describe where a sector's output goes. The decimal fractions in each column sum to 1. Distribution of Output From:
Goods output .2 .8 Services .7 .3 Purchased by: input Goods Services
Denote the total annual output (in dollars) of the sectors by pG and pS. From the first row, the total input to the Goods sector is .2 pG + .7 pS. The Goods sector must pay for that. So the equilibrium prices must satisfy income expenses pG = .2pG + .7 pS From the second row, the input (that is, the expense) of the Services sector is .8 pG + .3 pS. The equilibrium equation for the Services sector is income expenses pS = .8pG + .3 pS Move all variables to the left side and combine like terms: .8 pG - .7 pS = 0 -.8 pG + .7 pS = 0 Row reduce the augmented matrix:
.8 -.8
-.7 .7
0 .8 ~ 0 0
-.7 0
0 1 ~ 0 0
-.875 0
0 0
The general solution is pG = .875 pS, with pS free. One equilibrium solution is pS = 1000 and pG = 875. If one uses fractions instead of decimals in the calculations, the general solution would be written pG = (7/8) pS, and a natural choice of prices might be pS = 80 and pG = 70. Only the ratio of the prices is important: pG = .875 pS. The economic equilibrium is unaffected by a proportional change in prices.
2. Take some other value for pS, say 200 million dollars. The other equilibrium prices are then pC = 188 million, pE = 170 million. Any constant nonnegative multiple of these prices is a set of equilibrium prices, because the solution set of the system of equations consists of all multiples of one vector. Changing the unit of measurement to, say, European euros has the same effect as multiplying all equilibrium prices by a constant. The ratios of the prices remain the same, no matter what currency is used. 3. a. Fill in the exchange table one column at a time. The entries in a column describe where a sector's output goes. The decimal fractions in each column sum to 1.
40
CHAPTER 1
Linear Equations in Linear Algebra
Distribution of Output From: output Chemicals .2 .3 .5 Fuels .8 .1 .1 Machinery .4 .4 .2 input
Purchased by: Chemicals Fuels Machinery
b. Denote the total annual output (in dollars) of the sectors by pC, pF, and pM. From the first row of the table, the total input to the Chemical & Metals sector is .2 pC + .8 pF + .4 pM. So the equillibrium prices must satisfy income expenses
pC
=
.2pC + .8 pF + .4 pM
From the second and third rows of the table, the income/expense requirements for the Fuels & Power sector and the Machinery sector are, respectively, pF = .3 pC + .1 pF + .4 pM
pM = .5 pC + .1 pF + .2 pM
Move all variables to the left side and combine like terms: .8 pC .8 pF .4 pM = 0 .3 pC + .9 pF .4 pM = 0 .5 pC .1 pF + .8 pM = 0
c. [M] You can obtain the reduced echelon form with a matrix program. Actually, hand calculations are not too messy. To simplify the calculations, first scale each row of the augmented matrix by 10, then continue as usual. 8 -3 -5 1 ~ 0 0
-8 9 -1 -1 1 0 -4 -4 0 1 0 ~ -3 8 0 -5 0 1 0 ~ 0 0 0 -1 9 -1 0 1 0 -.5 -4 8 0 1 0 ~ 0 0 0 -1 6 -6 -.5 0 -5.5 0 5.5 0
-.5 -.917 0
-1.417 -.917 0
0 The number of decimal 0 places displayed is 0 somewhat arbitrary.
The general solution is pC = 1.417 pM, pF = .917 pM, with pM free. If pM is assigned the value 100, then pC = 141.7 and pF = 91.7. Note that only the ratios of the prices are determined. This makes sense, for if the were converted from, say, dollars to yen or Euros, the inputs and outputs of each sector would still balance. The economic equilibrium is not affected by a proportional change in prices.
1.6
Solutions
41
4. a. Fill in the exchange table one column at a time. The entries in each column must sum to 1. Distribution of Output From:
output
Agric. .65 .10 .25 0
Energy .30 .10 .35 .25
Manuf . .30 .15 .15 .40
Transp. .20 .10 .30 .40
Purchased by : input Agric. Energy Manuf . Transp.
b. Denote the total annual output of the sectors by pA, pE, pM, and pT, respectively. From the first row of the table, the total input to Agriculture is .65pA + .30pE + .30pM + .20 pT. So the equilibrium prices must satisfy income expenses pA = .65 pA + .30 pE + .30 pM + .20 pT
From the second, third, and fourth rows of the table, the equilibrium equations are pE = .10 pA + .10 pE + .15 pM + .10 pT pM = .25 pA + .35 pE + .15 pM + .30 pT
pT
=
.25 pE + .40 pM + .40 pT
Move all variables to the left side and combine like terms: .35 pA - .30 pE - .30 pM - .20 pT = 0 -.10 pA + .90 pE - .15 pM - .10 pT = 0 -.25 pA - .35 pE + .85 pM - .30 pT = 0 -.25 pE - .40 pM + .60 pT = 0 Use gauss, bgauss, and scale operations to reduce the augmented matrix to reduced echelon form
.35 0 0 0
-.3 .81 0 0
-.3 -.24 1.0 0
-.2 -.16 -1.17 0
0 .35 0 0 ~ 0 0 0 0
-.3 .81 0 0
0 0 1 0
-.55 -.43 -1.17 0
0 .35 0 0 ~ 0 0 0 0
0 1 0 0
0 0 1 0
-.71 -.53 -1.17 0
0 0 0 0
Scale the first row and solve for the basic variables in terms of the free variable pT, and obtain pA = 2.03pT, pE = .53pT, and pM = 1.17pT. The data probably justifies at most two significant figures, so take pT = 100 and round off the other prices to pA = 200, pE = 53, and pM = 120.
5. The following vectors list the numbers of atoms of boron (B), sulfur (S), hydrogen (H), and oxygen (O): 2 3 B2S3 : , H 2 O: 0 0 0 0 , H 3 BO3 : 2 1 1 0 , H 2S: 3 3 0 1 2 0
boron sulfur hydrogen oxygen
The coefficients in the equation x1B2S3 + x2H20 x3H3BO3 + x4H2S satisfy
42
CHAPTER 1
Linear Equations in Linear Algebra
2 0 1 0 3 0 0 1 x1 + x2 = x3 + x4 0 2 3 2 0 1 3 0
Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and row reduce the augmented matrix of the homogeneous system:
2 3 0 0 2 0 ~ 0 0
0 0 2 1 0 1 0 0
-1 0 -3 -3 -1 -3 1 3
0 -1 -2 0
0 2 0 0 ~ 0 0 0 0 0
0 0 2 1
-1 3/ 2 -3 -3 0 1 0 0 0 0 1 0
0 -1 -2 0
0 2 0 0 ~ 0 0 0 0
0 1 0 2
-1 -3 3/ 2 -3 0 1 0 0 0 0 1 0
0 0 -1 -2
0 2 0 0 ~ 0 0 0 0 0 0 0 0
0 1 0 0
-1 -3 3/ 2 3
0 0 -1 -2
0 0 0 0
0 -2 / 3 -2
0 2 0 0 ~ 0 0 0 0
-2 / 3 -2 -2 / 3 0
0 1 0 0 ~ 0 0 0 0
-1/ 3 -2 -2 / 3 0
The general solution is x1 = (1/3) x4, x2 = 2x4, x3 = (2/3) x4, with x4 free. Take x4 = 3. Then x1 = 1, x2 = 6, and x3 = 2. The balanced equation is B2S3 + 6H20 2H3BO3 + 3H2S
6. The following vectors list the numbers of atoms of sodium (Na), phosphorus (P), oxygen (O), barium (Ba), and nitrogen(N): 3 1 Na 3 PO 4 : 4 , Ba(NO3 ) 2 : 0 0 0 0 6 , Ba 3 (PO 4 ) 2 : 1 2 0 2 8 , NaNO3 : 3 0 1 0 3 0 1
sodium phosphorus oxygen barium nitrogen
The coefficients in the equation x1Na3PO4 + x2Ba(NO3)2 x3Ba3(PO4)2 + x4NaNO3 satisfy
3 0 0 1 1 0 2 0 x1 4 + x2 6 = x3 8 + x4 3 0 1 3 0 0 2 0 1
Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and row reduce the augmented matrix of the homogeneous system:
3 1 4 0 0
0 0 6 1 2
0 -2 -8 -3 0
-1 0 -3 0 -1
0 1 0 3 0 ~ 4 0 0 0 0
0 0 6 1 2
-2 0 -8 -3 0
0 -1 -3 0 -1
0 1 0 0 0 ~ 0 0 0 0 0
0 0 6 1 2
-2 6 0 -3 0
0 -1 -3 0 -1
0 1 0 0 0 ~ 0 0 0 0 0
0 1 6 0 2
-2 -3 0 6 0
0 0 -3 -1 -1
0 0 0 0 0
1.6
Solutions
43
1 0 ~ 0 0 0
0 1 0 0 0
-2 -3 18 6 6
0 0 -3 -1 -1
0 1 0 0 0 ~ 0 0 0 0 0
0 1 0 0 0
-2 -3 1 0 0
0 0 -1/ 6 0 0
0 1 0 0 0 ~ 0 0 0 0 0
0 1 0 0 0
0 0 1 0 0
-1/ 3 -1/ 2 -1/ 6 0 0
0 0 0 0 0
The general solution is x1 = (1/3)x4, x2 = (1/2)x4, x3 = (1/6)x4, with x4 free. Take x4 = 6. Then x1 = 2, x2 = 3, and x3 = 1. The balanced equation is 2Na3PO4 + 3Ba(NO3)2 Ba3(PO4)2 + 6NaNO3
7. The following vectors list the numbers of atoms of sodium (Na), hydrogen (H), carbon (C), and oxygen (O): 1 1 NaHCO3 : , H 3C6 H 5O7 : 1 3 0 8 , Na 3C6 H 5O7 : 6 7 3 0 0 5 2 0 , H 2 O : , CO 2 : 6 0 1 7 1 2
sodium hydrogen carbon oxygen
The order of the various atoms is not important. The list here was selected by writing the elements in the order in which they first appear in the chemical equation, reading left to right: x1 NaHCO3 + x2 H3C6H5O7 x3 Na3C6H5O7 + x4 H2O + x5 CO2. The coefficients x1, ..., x5 satisfy the vector equation
1 0 3 0 0 1 8 5 2 0 x1 + x2 = x3 + x4 + x5 1 6 6 0 1 3 7 7 1 2
Move all the terms to the left side (changing the sign of each entry in the third, fourth, and fifth vectors) and reduce the augmented matrix:
1 1 1 3
0 8 6 7
-3 -5 -6 -7
0 -2 0 -1
0 0 -1 -2
0 1 0 0 ~ ~ 0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
-1 -1/ 3 -1/ 3 -1
0 0 0 0
The general solution is x1 = x5, x2 = (1/3)x5, x3 = (1/3)x5, x4 = x5, and x5 is free. Take x5 = 3. Then x1 = x4 = 3, and x2 = x3 = 1. The balanced equation is 3NaHCO3 + H3C6H5O7 Na3C6H5O7 + 3H2O + 3CO2
8. The following vectors list the numbers of atoms of potassium (K), manganese (Mn), oxygen (O), sulfur (S), and hydrogen (H): 1 0 0 1 1 0 KMnO 4 : 4 , MnSO 4 : 4 , H 2 O: 1 , MnO 2 : 0 1 0 0 0 2 The coefficients in the chemical equation 0 1 2 , K 2SO 4 : 0 0 2 0 4 , H 2SO 4 : 1 0 0 0 4 1 2
potassium manganese oxygen sulfur hydrogen
44
CHAPTER 1
Linear Equations in Linear Algebra
x1KMnO4 + x2MnSO4 + x3H2O x4MnO2 + x5K2SO4 + x6H2SO4 satisfy the vector equation
1 1 x1 4 + x2 0 0 0 0 0 2 1 0 1 0 4 + x3 1 = x4 2 + x5 4 + x6 1 0 0 1 0 2 0 0 0 0 4 1 2
Move the terms to the left side (changing the sign of each entry in the last three vectors) and reduce the augmented matrix:
1 1 4 0 0
0 1 4 1 0
0 0 1 0 2
0 -1 -2 0 0
-2 0 -4 -1 0
0 0 -4 -1 -2
0 1 0 0 0 ~ 0 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
-1.0 -1.5 -1.0 -2.5 -.5
0 0 0 0 0
The general solution is x1 = x6, x2 = (1.5)x6, x3 = x6, x4 = (2.5)x6, x5 = .5x6, and x6 is free. Take x6 = 2. Then x1 = x3 = 2, and x2 = 3, x4 = 5, and x5 = 1. The balanced equation is 2KMnO4 + 3MnSO4 + 2H2O 5MnO2 + K2SO4 + 2H2SO4
9. [M] Set up vectors that list the atoms per molecule. Using the order lead (Pb), nitrogen (N), chromium (Cr), manganese (Mn), and oxygen (O), the vector equation to be solved is 1 0 3 0 0 0 6 0 0 0 0 1 x1 0 + x2 1 = x3 0 + x4 2 + x5 0 + x6 0 0 2 0 0 1 0 0 8 4 3 2 1
lead nitrogen chromium manganese oxygen
The general solution is x1 = (1/6)x6, x2 = (22/45)x6, x3 = (1/18)x6, x4 = (11/45)x6, x5 = (44/45)x6, and x6 is free. Take x6 = 90. Then x1 = 15, x2 = 44, x3 = 5, x4 = 22, and x5 = 88. The balanced equation is 15PbN6 + 44CrMn2O8 5Pb3O4 + 22Cr2O3 + 88MnO2 + 90NO
10. [M] Set up vectors that list the atoms per molecule. Using the order manganese (Mn), sulfur (S), arsenic (As), chromium (Cr), oxygen (O), and hydrogen (H), the vector equation to be solved is 1 0 0 1 0 0 0 1 0 1 0 0 3 0 0 2 0 0 1 0 0 x1 + x2 + x3 = x4 + x5 + x6 + x7 0 10 0 0 0 1 0 0 35 4 4 0 12 1 0 0 2 1 3 0 2
manganese sulfur arsenic chromium oxygen hydrogen
In rational format, the general solution is x1 = (16/327)x7, x2 = (13/327)x7, x3 = (374/327)x7, x4 = (16/327)x7, x5 = (26/327)x7, x6 = (130/327)x7, and x7 is free. Take x7 = 327 to make the other variables whole numbers. The balanced equation is 16MnS + 13As2Cr10O35 + 374H2SO4 16HMnO4 + 26AsH3 + 130CrS3O12 + 327H2O
1.6
Solutions
45
Note that some students may use decimal calculation and simply "round off" the fractions that relate x1, ..., x6 to x7. The equations they construct may balance most of the elements but miss an atom or two. Here is a solution submitted by two of my students: 5MnS + 4As2Cr10O35 + 115H2SO4 5HMnO4 + 8AsH3 + 40CrS3O12 + 100H2O Everything balances except the hydrogen. The right side is short 8 hydrogen atoms. Perhaps the students thought that the 4H2 (hydrogen gas) escaped!
11. Write the equations for each node:
Node A B C Total flow:
Flow in
Flow out = = = = 20 x3 + x4
20 x1 80
A x3 B x2 C x4
x1 + x3 x2 80 80
+ -
x1 + x2 x 4 + 20
= = = = 0 1 0 0 20 0 80 60 1 -1 0 0 0 0 1 0
Rearrange the equations:
x1 x1
+
x2 x2
x3 x3
-
x4 x4
Reduce the augmented matrix:
1 0 1 0
0 1 1 0
1 -1 0 0
0 -1 0 1
20 1 0 ~ ~ 0 0 80 60 0
20 60 60 0
For this type of problem, the best description of the general solution uses the style of Section 1.2 rather than parametric vector form:
x1 = 20 - x3 x = 60 + x 2 3 . Since x1 cannot be negative, the largest value of x3 is 20. x3 is free x 4 = 60 12. Write the equations for each intersection:
200 B x1 40 A x4 D 60 x3 x5 x2 C 100
Intersection A B C D Total flow:
Flow in
x1 200 x2 + x3 x4 + x5 200
Flow out = = = = =
x3 + x4 + 40 x1 + x2 x5 + 100 60 200
46
CHAPTER 1
Linear Equations in Linear Algebra
Rearrange the equations:
x1 x1
- +
x2 x2
x3 x3
-
x4
= -
x5 x5
40 200 100 60 1 100 -1 100 1 60 0 0
+
= = = 0 0 1 0
x4
+ 0 1 0 0
Reduce the augmented matrix:
1 1 0 0
0 1 1 0
-1 0 1 0
-1 0 0 1
0 0 -1 1
40 1 200 0 ~ 100 0 60 0
-1 1 0 0
The general solution (written in the style of Section 1.2) is
x1 = 100 + x3 - x5 x = 100 - x + x 3 5 2 x3 is free x = 60 - x 5 4 x5 is free x1 = 40 + x3 x = 160 - x 3 2 b. When x4 = 0, x5 must be 60, and x3 is free x = 0 4 x5 = 60
c. The minimum value of x1 is 40 cars/minute, because x3 cannot be negative. 13. Write the equations for each intersection: Intersection Flow in Flow out A x2 + 30 = x1 + 80 B x3 + x5 x2 + x4 =
30
40
80 60
A x1 E
x2 B x3
x5 x4
C x6 D
100 90
C D E Total flow:
x1
x6 + 100 x4 + 40 x1 + 60 230
= = = =
x5 + 40 x6 + 90 x3 + 20 230
20
40
Rearrange the equations: -
x2 x2
= -
x3
-50 0 60 50 -40 0 -1 0 0 0 0 1 1 0 0 0 -1 0 1 0 0 0 -1 -1 0 -50 0 50 60 0
+
x4 x4
-
x5 x5
- -
x6 x6
= = =
x1
- -1 1 0 0 0 0 -1 0 0 -1 0 1 0 1 0
x3
Reduce the augmented matrix:
1 0 0 0 1
0 -1 1 0 0
0 0 -1 -1 0
-50 1 0 0 60 ~ ~ 0 50 0 0 -40
-1 1 0 0 0
1.6
Solutions
47
1 0 ~ ~ 0 0 0
0 1 0 0 0
-1 -1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 -1 -1 0
-40 10 50 60 0
x1 = x3 - 40 x = x + 10 3 2 x3 is free a. The general solution is x4 = x6 + 50 x5 = x6 + 60 x6 is free b. To find minimum flows, note that since x1 cannot be negative, x3 > 40. This implies that x2 > 50. Also, since x6 cannot be negative, x4 > 50 and x5 > 60. The minimum flows are x2 = 50, x3 = 40, x4 = 50, x5 = 60 (when x1 = 0 and x6 = 0). 14. Write the equations for each intersection. Intersection Flow in Flow out A x1 = x2 + 100
120 150 C x3 B x2 A x1 x4 D x 5 E x6 F 80 100
B C D E F
x2 + 50 x3 x4 + 150 x5 x6 + 100
= = = = =
x3 x4 + 120 x5 x6 + 80 x1
50 100
Rearrange the equations: x1 - x2
x2
=
x3 x3
100 - 50 120 -150 80 -100 0 -1 1 0 0 0 0 0 -1 1 0 0 0 0 0 -1 1 0 0 0 0 0 -1 0 100 -50 120 -150 80 0
-
-
x4 x4
= = -
x5 x5
- +
x6 x6
= = = -1 1 0 0 0 0
- x1 Reduce the augmented matrix:
1 0 0 0 0 -1
-1 1 0 0 0 0
0 -1 1 0 0 0
0 0 -1 1 0 0
0 0 0 -1 1 0
0 0 0 0 -1 1
100 1 0 -50 0 120 ~ ~ -150 0 0 80 -100 0
48
CHAPTER 1
Linear Equations in Linear Algebra
1 0 0 ~ ~ 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
-1 -1 -1 -1 -1 0
100 0 50 . The general solution is -70 80 0
x1 = 100 + x6 x = x 6 2 x3 = 50 + x6 . x4 = -70 + x6 x5 = 80 + x6 x6 is free
Since x4 cannot be negative, the minimum value of x6 is 70.
Note: The MATLAB box in the Study Guide discusses rational calculations, needed for balancing the chemical equations in Exercises 9 and 10. As usual, the appendices cover this material for Maple, Mathematica, and the TI and HP graphic calculators.
1.7
SOLUTIONS
Note: Key exercises are 920 and 2330. Exercise 30 states a result that could be a theorem in the text. There is a danger, however, that students will memorize the result without understanding the proof, and then later mix up the words row and column. Exercises 37 and 38 anticipate the discussion in Section 1.9 of one-to-one transformations. Exercise 44 is fairly difficult for my students.
1. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are the 7 9 0 5 7 9 0 5 0 three given vectors. Since 2 4 0 ~ 0 2 4 0 , there are no free variables. So the 0 -6 -8 0 0 0 4 0
homogeneous equation (*) has only the trivial solution. The vectors are linearly independent.
2. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are the 0 -3 0 2 -8 1 0 0 0 ~ 0 three given vectors. Since 5 4 0 5 4 0 , there are no free variables. So the 2 -8 1 0 0 0 -3 0
homogeneous equation (*) has only the trivial solution. The vectors are linearly independent.
3. Use the method of Example 3 (or the box following the example). By comparing entries of the vectors, one sees that the second vector is 3 times the first vector. Thus, the two vectors are linearly dependent. -1 -2 4. From the first entries in the vectors, it seems that the second vector of the pair , may be 2 4 -8 times the first vector. But there is a sign problem with the second entries. So neither of the vectors is a multiple of the other. The vectors are linearly independent. 5. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0: 0 3 -1 1
-8 -7 5 -3
5 4 -4 2
0 1 0 3 ~ 0 -1 0 0
-3 -7 5 -8
2 4 -4 5
0 1 0 0 ~ 0 0 0 0
-3 2 2 -8
2 -2 -2 5
0 1 0 0 ~ 0 0 0 0
-3 2 0 0
2 -2 0 -3
0 1 0 0 ~ 0 0 0 0
-3 2 0 0
2 -2 -3 0
0 0 0 0
1.7
Solutions
49
There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A are linearly independent.
6. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0: -4 0 1 5
-3 -1 0 4
0 4 3 6
0 1 0 0 ~ 0 -4 0 5
0 -1 -3 4
3 4 0 6
0 1 0 0 ~ 0 0 0 0
0 -1 -3 4
3 4 12 -9
0 1 0 0 ~ 0 0 0 0
0 -1 0 0
3 4 0 7
0 1 0 0 ~ 0 0 0 0
0 -1 0 0
3 4 7 0
0 0 0 0
There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A are linearly independent.
7. Study the equation Ax = 0. Some people may start with the method of Example 2: 1 -2 -4
4 -7 -5
-3 5 7
0 1 5
0 1 4 0 ~ 0 1 0 11 0
-3 -1 -5
0 1 5
0 1 0 ~ 0 0 0
4 1 0
-3 -1 6
0 1 -6
0 0 0
But this is a waste of time. There are only 3 rows, so there are at most three pivot positions. Hence, at least one of the four variables must be free. So the equation Ax = 0 has a nontrivial solution and the columns of A are linearly dependent.
8. Same situation as with Exercise 7. The (unnecessary) row operations are 1 -3 0
-3 7 1
3 -1 -4
-2 2 3
0 1 0 ~ 0 0 0
-3 -2 1
3 8 -4
-2 -4 3
0 1 0 ~ 0 0 0
-3 -2 0
3 8 0
-2 -4 1
0 0 0
Again, because there are at most three pivot positions yet there are four variables, the equation Ax = 0 has a nontrivial solution and the columns of A are linearly dependent.
9. a. The vector v3 is in Span{v1, v2} if and only if the equation x1v1 + x2v2 = v3 has a solution. To find out, row reduce [v1 v2 v3], considered as an augmented matrix: 1 -3 2
-3 9 -6
5 1 -7 ~ 0 h 0
-3 0 0
8 h - 10
5
At this point, the equation 0 = 8 shows that the original vector equation has no solution. So v3 is in Span{v1, v2} for no value of h. b. For {v1, v2, v3} to be linearly independent, the equation x1v1 + x2v2 + x3v3 = 0 must have only the trivial solution. Row reduce the augmented matrix [v1 v2 v3 0]
1 -3 2
-3 9 -6
5 -7 h
0 1 0 ~ 0 0 0
-3 0 0
5 8 h - 10
0 1 0 ~ 0 0 0
-3 0 0
5 8 0
0 0 0
For every value of h, x2 is a free variable, and so the homogeneous equation has a nontrivial solution. Thus {v1, v2, v3} is a linearly dependent set for all h.
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10. a. The vector v3 is in Span{v1, v2} if and only if the equation x1v1 + x2v2 = v3 has a solution. To find out, row reduce [v1 v2 v3], considered as an augmented matrix: 1 -5 -3
-2 10 6
2 1 -9 ~ 0 h 0
-2 0 0
2 1 h + 6
At this point, the equation 0 = 1 shows that the original vector equation has no solution. So v3 is in Span{v1, v2} for no value of h. b. For {v1, v2, v3} to be linearly independent, the equation x1v1 + x2v2 + x3v3 = 0 must have only the trivial solution. Row reduce the augmented matrix [v1 v2 v3 0]
1 -5 -3
-2 10 6
2 -9 h
0 1 0 ~ 0 0 0
-2 0 0
2 1 h+6
0 1 0 ~ 0 0 0
-2 0 0
2 1 0
0 0 0
For every value of h, x2 is a free variable, and so the homogeneous equation has a nontrivial solution. Thus {v1, v2, v3} is a linearly dependent set for all h.
11. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 -1 4
3 -5 7
-1 5 h
0 1 0 ~ 0 0 0
3 -2 -5
-1 4 h+4
0 1 0 ~ 0 0 0
3 -2 0
-1 4 h-6
0 0 0
The equation x1v1 + x2v2 + x3v3 = 0 has a nontrivial solution if and only if h 6 = 0 (which corresponds to x3 being a free variable). Thus, the vectors are linearly dependent if and only if h = 6.
12. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 2 -4 1
-6 7 -3
8
h 4
0 2 0 ~ 0 0 0
-6 -5 0
8
h + 16 0
0 0 0
The equation x1v1 + x2v2 + x3v3 = 0 has a free variable and hence a nontrivial solution no matter what the value of h. So the vectors are linearly dependent for all values of h.
13. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 5 -3
-2 -9 6
3
h -9
0 1 0 ~ 0 0 0
-2 1 0
3
h - 15 0
0 0 0
The equation x1v1 + x2v2 + x3v3 = 0 has a free variable and hence a nontrivial solution no matter what the value of h. So the vectors are linearly dependent for all values of h.
1.7
Solutions
51
14. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 -1 -3
-5 7 8
1 1 h
0 1 0 ~ 0 0 0
-5 2 -7
1 2 h+3
0 1 0 ~ 0 0 0
-5 2 0
1 2 h + 10
0 0 0
The equation x1v1 + x2v2 + x3v3 = 0 has a nontrivial solution if and only if h + 10 = 0 (which corresponds to x3 being a free variable). Thus, the vectors are linearly dependent if and only if h = 10.
15. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two entries in each vector. 16. The set is linearly dependent because the second vector is 3/2 times the first vector. 17. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector. 18. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two entries in each vector. 19. The set is linearly independent because neither vector is a multiple of the other vector. [Two of the entries in the first vector are 4 times the corresponding entry in the second vector. But this multiple does not work for the third entries.] 20. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector. 21. a. b. c. d.
False. A homogeneous system always has the trivial solution. See the box before Example 2. False. See the warning after Theorem 7. True. See Fig. 3, after Theorem 8. True. See the remark following Example 4.
22. a. True. See Fig. 1. 1 2 -2 and 4 is linearly dependent. See the warning after b. False. For instance, the set consisting of 3 6 Theorem 8. c. True. See the remark following Example 4. d. False. See Example 3(a). * * 23. 0 * 0 0 * 0 0 and 0 0 25. 0 0 0 0 0 0 0 0
* 0 0 24. , , 0 0 0 0 0
0 0
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* * 0 * . The columns must linearly independent, by Theorem 7, because the first column is not 26. 0 0 0 0 0 zero, the second column is not a multiple of the first, and the third column is not a linear combination of the preceding two columns (because a3 is not in Span{a1, a2}). 27. All five columns of the 75 matrix A must be pivot columns. Otherwise, the equation Ax = 0 would have a free variable, in which case the columns of A would be linearly dependent. 28. If the columns of a 57 matrix A span R5, then A has a pivot in each row, by Theorem 4. Since each pivot position is in a different column, A has five pivot columns. 29. A: any 32 matrix with two nonzero columns such that neither column is a multiple of the other. In this case the columns are linearly independent and so the equation Ax = 0 has only the trivial solution. B: any 32 matrix with one column a multiple of the other. 30. a. n b. The columns of A are linearly independent if and only if the equation Ax = 0 has only the trivial solution. This happens if and only if Ax = 0 has no free variables, which in turn happens if and only if every variable is a basic variable, that is, if and only if every column of A is a pivot column. 31. Think of A = [a1 a2 a3]. The text points out that a3 = a1 + a2. Rewrite this as a1 + a2 a3 = 0. As a matrix equation, Ax = 0 for x = (1, 1, 1). 32. Think of A = [a1 a2 a3]. The text points out that a1 + 2a2 = a3. Rewrite this as a1 + 2a2 a3 = 0. As a matrix equation, Ax = 0 for x = (1, 2, 1). 33. True, by Theorem 7. (The Study Guide adds another justification.) 34. True, by Theorem 9. 35. False. The vector v1 could be the zero vector. 36. False. Counterexample: Take v1, v2, and v4 all to be multiples of one vector. Take v3 to be not a multiple of that vector. For example, 1 2 1 4 1 2 0 4 v1 = , v 2 = , v 3 = , v 4 = 1 2 0 4 1 2 0 4 37. True. A linear dependence relation among v1, v2, v3 may be extended to a linear dependence relation among v1, v2, v3, v4 by placing a zero weight on v4. 38. True. If the equation x1v1 + x2v2 + x3v3 = 0 had a nontrivial solution (with at least one of x1, x2, x3 nonzero), then so would the equation x1v1 + x2v2 + x3v3 + 0v4 = 0. But that cannot happen because {v1, v2, v3, v4} is linearly independent. So {v1, v2, v3} must be linearly independent. This problem can also be solved using Exercise 37, if you know that the statement there is true.
1.7
Solutions
53
39. If for all b the equation Ax = b has at most one solution, then take b = 0, and conclude that the equation Ax = 0 has at most one solution. Then the trivial solution is the only solution, and so the columns of A are linearly independent. 40. An mn matrix with n pivot columns has a pivot in each column. So the equation Ax = b has no free variables. If there is a solution, it must be unique. 8 -9 41. [M] A = 6 5 8 0 ~ 0 0
-3 4 -2 -1 0 5 0 0
0 5 2 7 -7
-7 11 -4 0 2
2 8 -7 0 ~ 4 0 10 0
-3 5/8 1/ 4 7/8 -3 5/8 0 0
0 5 2 7 0 5 0 0
-7 25 / 8 5/ 4 35 / 8 -7 25 / 8 0 0
2 -19 / 4 5/ 2 35 / 4
-19 / 4 22 / 5 0
-3 5/8 0 0
25 / 8 0 0
8 -19 / 4 0 ~ 22 / 5 0 77 / 5 0
2
8 -9 The pivot columns of A are 1, 2, and 5. Use them to form B = 6 5 8 -9 Other likely choices use columns 3 or 4 of A instead of 2: 6 5
-3 4 -2 -1
2 -7 . 4 10 -7 11 -4 0 2 -7 . 4 10
0 5 2 7
2 8 -7 -9 , 4 6 10 5
Actually, any set of three columns of A that includes column 5 will work for B, but the concepts needed to prove that are not available now. (Column 5 is not in the two-dimensional subspace spanned by the first four columns.)
42. [M] 12 -7 9 -4 8
10 -6 9 -3 7
-6 4 -9 1 -5
-3 7 -5 6 -9
7 -9 5 -8 11
10 12 0 5 -1 ~ ~ 0 9 0 0 -8
10 -1/ 6 0 0 0
-6 1/ 2 0 0 0
-3 21/ 4 89 / 2 0 0
7 -59 /12 -89 / 2 0 0 10 -6 9 -3 7
10 65 / 6 89 3 0 -3 7 -5 6 -9 10 5 -1 . 9 -8
12 -7 The pivot columns of A are 1, 2, 4, and 6. Use them to form B = 9 -4 8
Other likely choices might use column 3 of A instead of 2, and/or use column 5 instead of 4.
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43. [M] Make v any one of the columns of A that is not in B and row reduce the augmented matrix [B v]. The calculations will show that the equation Bx = v is consistent, which means that v is a linear combination of the columns of B. Thus, each column of A that is not a column of B is in the set spanned by the columns of B. 44. [M] Calculations made as for Exercise 43 will show that each column of A that is not a column of B is in the set spanned by the columns of B. Reason: The original matrix A has only four pivot columns. If one or more columns of A are removed, the resulting matrix will have at most four pivot columns. (Use exactly the same row operations on the new matrix that were used to reduce A to echelon form.) If v is a column of A that is not in B, then row reduction of the augmented matrix [B v] will display at most four pivot columns. Since B itself was constructed to have four pivot columns, adjoining v cannot produce a fifth pivot column. Thus the first four columns of [B v] are the pivot columns. This implies that the equation Bx = v has a solution.
Note: At the end of Section 1.7, the Study Guide has another note to students about "Mastering Linear Algebra Concepts." The note describes how to organize a review sheet that will help students form a mental image of linear independence. The note also lists typical misuses of terminology, in which an adjective is applied to an inappropriate noun. (This is a major problem for my students.) I require my students to prepare a review sheet as described in the Study Guide, and I try to make helpful comments on their sheets. I am convinced, through personal observation and student surveys, that the students who prepare many of these review sheets consistently perform better than other students. Hopefully, these students will remember important concepts for some time beyond the final exam.
1.8
SOLUTIONS
Notes: The key exercises are 1720, 25 and 31. Exercise 20 is worth assigning even if you normally assign
only odd exercises. Exercise 25 (and 27) can be used to make a few comments about computer graphics, even if you do not plan to cover Section 2.6. For Exercise 31, the Study Guide encourages students not to look at the proof before trying hard to construct it. Then the Guide explains how to create the proof. Exercises 19 and 20 provide a natural segue into Section 1.9. I arrange to discuss the homework on these exercises when I am ready to begin Section 1.9. The definition of the standard matrix in Section 1.9 follows naturally from the homework, and so I've covered the first page of Section 1.9 before students realize we are working on new material. The text does not provide much practice determining whether a transformation is linear, because the time needed to develop this skill would have to be taken away from some other topic. If you want your students to be able to do this, you may need to supplement Exercises 29, 30, 32 and 33. If you skip the concepts of one-to-one and "onto" in Section 1.9, you can use the result of Exercise 31 to show that the coordinate mapping from a vector space onto Rn (in Section 4.4) preserves linear independence and dependence of sets of vectors. (See Example 6 in Section 4.4.)
2 1. T(u) = Au = 0 .5 2. T(u) = Au = 0 0
0 1 2 = , T(v) = 2 -3 -6 0 .5 0
2 0
0 a 2a = 2 b 2b
.5 0 0
0 1 .5 0 0 = 0 , T(v) = .5 -4 -2
0 .5 0
0 a .5a 0 b = .5b .5 c .5c
1.8
Solutions
55
1 3. [ A b ] = -2 3 1 ~ 0 0
0 1 -2
-2 6 -5
-1 1 7 ~ 0 -3 0 0 1 0 0 0 1
0 1 -2 3 1 2 -3 1 4 -5 -3 1 -5 1
-2 2 1
-1 1 5 ~ 0 0 0
0 1 0
-2 2 5
-1 5 10
0 1 0
-2 2 1 -3
-1 1 5 ~ 0 2 0 2 -4 -9
3 x = 1 , unique solution 2
1 4. [ A b ] = 0 3 1 ~ 0 0
1 -5 0 0 1
6 1 -7 ~ 0 -9 0 0 1 0 0 0 1
2 -4 -15
6 1 -7 ~ 0 -27 0
-3 1 0
2 -4 1
6 -7 1
-3 1 0
4 1 -3 ~ 0 1 0 -5 7 -7 5
-5 x = -3 , unique solution 1
1 5. [ A b ] = -3
-2 1 ~ -2 0
-7 2
-2 1 ~ 1 0
0 1
3 2
3 1
3 Note that a solution is not . To avoid this common error, write the equations: 1
x1 x2
+ 3 x3 + 2 x3
x1 = 3 - 3x3 and solve for the basic variables: x2 = 1 - 2 x3 = 1 x is free 3
= 3
x1 3 - 3 x3 3 -3 x = 1 - 2 x = 1 + x -2 . For a particular solution, one might choose General solution x = 2 3 3 x3 x3 0 1 3 x3 = 0 and x = 1 . 0 1 3 6. [ A b ] = 0 -3
x1 x2
-2 -4 1 5
1 5 1 -4 .
1 1 9 0 ~ 3 0 -6 0
-2 2 1 -1
1 2 1 -1
1 1 6 0 ~ 3 0 -3 0
-2 1 0 0
1 1 0 0
1 1 3 0 ~ 0 0 0 0
0 1 0 0
3 1 0 0
7 3 0 0
+ 3x3 +
x3
= 7 = 3
x1 = 7 - 3x3 x2 = 3 - x3 x is free 3 7 3 . 0
x1 7 - 3x3 7 -3 General solution: x = x2 = 3 - x3 = 3 + x3 -1 , one choice: x3 x3 0 1
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7. a = 5; the domain of T is R5, because a 65 matrix has 5 columns and for Ax to be defined, x must be in R5. b = 6; the codomain of T is R6, because Ax is a linear combination of the columns of A, and each column of A is in R6. 8. A must have 5 rows and 4 columns. For the domain of T to be R4, A must have four columns so that Ax is defined for x in R4. For the codomain of T to be R5, the columns of A must have five entries (in which case A must have five rows), because Ax is a linear combination of the columns of A. 1 9. Solve Ax = 0. 0 2 1 ~ 0 0
-4 1 -6 7 3 0
7 -4 6 0 0 0
x1
-5 3 -4
0 1 0 ~ 0 0 0 - 9 x3 - 4 x3
-4 1 2
7 -4 -8
-5 3 6
0 1 0 ~ 0 0 0
-4 1 0
7 -4 0
-5 3 0
0 0 0
0 1 0
-9 -4 0
x2
+ 7 x4 + 3 x4
x = 9 x3 - 7 x4 = 0 1 x = 4 x3 - 3x4 = 0, 2 x is free 0 = 0 3 x4 is free
x1 9 x3 - 7 x4 9 -7 x 4 x - 3x 4 -3 4 x = 2 = 3 = x3 + x4 x3 1 0 x3 0 1 x4 x4 1 1 10. Solve Ax = 0. 0 -2 1 0 ~ 0 0
x1 x2
3 0 1 3 2 3 3
9 3 2 0 0 1 0 0 ~ 0 0 0 0 = 0 = 0
2 -4 3 5 3 1 0 0
0 1 0 0 ~ 0 0 0 0 9 2 0 0 0 0 1 0
3 -3 1 9
9 -6 2 18
2 -6 3 9 0 1 0 0
0 1 0 0 ~ 0 0 0 0 3 2 0 0 0 0 1 0
3 1 -3 9 0 0 0 0
9 2 -6 18
2 3 -6 9
0 0 0 0
3 1 0 0
9 2 0 0 + 3 x3 + 2 x3
-18
0 1 0 0 ~ 0 0 0 0
x4
= 0
x1 = -3x3 x = -2 x 2 3 x3 is free x4 = 0
-3x3 -3 -2 x -2 3 x= = x3 x3 1 0 0
11. Is the system represented by [A b] consistent? Yes, as the following calculation shows. 1 0 2
-4 1 -6
7 -4 6
-5 3 -4
-1 1 1 ~ 0 0 0
-4 1 2
7 -4 -8
-5 3 6
-1 1 1 ~ 0 2 0
-4 1 0
7 -4 0
-5 3 0
-1 1 0
The system is consistent, so b is in the range of the transformation x 6 Ax .
1.8
Solutions
57
12. Is the system represented by [A b] consistent? 1 1 0 -2 1 0 ~ 0 0
3 0 1 3 3 1 0 0
9 3 2 0 9 2 0 0
2 -4 3 5 2 3 3 -18
-1 1 3 0 ~ -1 0 4 0 -1 1 -1 0 ~ 1 0 11 0
3 -3 1 9 3 1 0 0
9 -6 2 18 9 2 0 0
2 -6 3 9 2 3 3 0
-1 1 4 0 ~ -1 0 2 0 -1 -1 1 17
3 1 -3 9
9 2 -6 18
2 3 -6 9
-1 -1 4 2
The system is inconsistent, so b is not in the range of the transformation x 6 Ax .
13.
x2 v u x1 T(u) T(v)
14.
x2 v T(v) T(u) u x1
A reflection through the origin.
A contraction by the factor .5.
The transformation in Exercise 13 may also be described as a rotation of radians about the origin or a rotation of radians about the origin.
15.
x2 v T(u) T(v) u x1 T(v) v u x1
16.
x2 T(u)
A projection onto the x2-axis
A reflection through the line x2 = x1.
2 6 -1 -2 17. T(3u) = 3T(u) = 3 = , T(2v) = 2T(v) = 2 = , and 1 3 3 6 6 -2 T(3u + 2v) = 3T(u) = 2T(v) = + = 3 6
4 . 9
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18. Draw a line through w parallel to v, and draw a line through w parallel to u. See the left part of the figure below. From this, estimate that w = u + 2v. Since T is linear, T(w) = T(u) + 2T(v). Locate T(u) and 2T(v) as in the right part of the figure and form the associated parallelogram to locate T(w).
x2 w u x1 T(u) x2 T(v) 2T(v) T(w)
2v v
x1
19. All we know are the images of e1 and e2 and the fact that T is linear. The key idea is to write 5 1 0 x = = 5 - 3 = 5 e1 - 3 e 2 . Then, from the linearity of T, write -3 0 1 2 -1 13 T(x) = T(5e1 3e2) = 5T(e1) 3T(e2) = 5y1 3y2 = 5 - 3 = . 5 6 7 x x 1 0 To find the image of 1 , observe that x = 1 = x1 + x2 = x1e1 + x2e 2 . Then 0 1 x2 x2 2 -1 2 x - x T(x) = T(x1e1 + x2e2) = x1T(e1) + x2T(e2) = x1 + x2 = 1 2 5 6 5 x1 + 6 x2 20. Use the basic definition of Ax to construct A. Write x -2 T (x) = x1 v1 + x2 v 2 = [ v1 v 2 ] 1 = x2 5
7 -2 x, A = 5 -3
7 -3
21. a. True. Functions from Rn to Rm are defined before Fig. 2. A linear transformation is a function with certain properties. b. False. The domain is R5. See the paragraph before Example 1. c. False. The range is the set of all linear combinations of the columns of A. See the paragraph before Example 1. d. False. See the paragraph after the definition of a linear transformation. e. True. See the paragraph following the box that contains equation (4). 22. a. True. See the paragraph following the definition of a linear transformation. b. False. If A is an mn matrix, the codomain is Rm. See the paragraph before Example 1. c. False. The question is an existence question. See the remark about Example 1(d), following the solution of Example 1. d. True. See the discussion following the definition of a linear transformation. e. True. See the paragraph following equation (5).
1.8
Solutions
59
23.
x2 u v x1 T(v) T(u) T(u + v) T(u) T (cu) u x1 u+v x2 cu
24. Given any x in Rn, there are constants c1, ..., cp such that x = c1v1 + cpvp, because v1, ..., vp span Rn. Then, from property (5) of a linear transformation, T(x) = c1T(v1) + + cpT(vp) = c10 + + cp0 = 0 25. Any point x on the line through p in the direction of v satisfies the parametric equation x = p + tv for some value of t. By linearity, the image T(x) satisfies the parametric equation T(x) = T(p + tv) = T(p) + tT(v) (*) If T(v) = 0, then T(x) = T(p) for all values of t, and the image of the original line is just a single point. Otherwise, (*) is the parametric equation of a line through T(p) in the direction of T(v). 26. Any point x on the plane P satisfies the parametric equation x = su + tv for some values of s and t. By linearity, the image T(x) satisfies the parametric equation T(x) = sT(u) + tT(v) (s, t in R) (*) The set of images is just Span{T(u), T(v)}. If T(u) and T(v) are linearly independent, Span{T(u), T(v)} is a plane through T(u), T(v), and 0. If T(u) and T(v) are linearly dependent and not both zero, then Span{T(u), T(v)} is a line through 0. If T(u) = T(v) = 0, then Span{T(u), T(v)} is {0}. 27. a. From Fig. 7 in the exercises for Section 1.5, the line through T(p) and T(q) is in the direction of q p, and so the equation of the line is x = p + t(q p) = p + tq tp = (1 t)p + tq. b. Consider x = (1 t)p + tq for t such that 0 < t < 1. Then, by linearity of T, T(x) = T((1 t)p + tq) = (1 t)T(p) + tT(q) 0<t<1 (*) If T(p) and T(q) are distinct, then (*) is the equation for the line segment between T(p) and T(q), as shown in part (a) Otherwise, the set of images is just the single point T(p), because (1 t)T(p) + tT(q) =(1 t)T(p) + tT(p) = T(p) 28. Consider a point x in the parallelogram determined by u and v, say x = au + bv for 0 < a < 1, 0 < b < 1. By linearity of T, the image of x is T(x) = T(au + bv) = aT(u) + bT(v), for 0 < a < 1, 0 < b < 1 (*) This image point lies in the parallelogram determined by T(u) and T(v). Special "degenerate" cases arise when T(u) and T(v) are linearly dependent. If one of the images is not zero, then the "parallelogram" is actually the line segment from 0 to T(u) + T(v). If both T(u) and T(v) are zero, then the parallelogram is just {0}. Another possibility is that even u and v are linearly dependent, in which case the original parallelogram is degenerate (either a line segment or the zero vector). In this case, the set of images must be degenerate, too. 29. a. When b = 0, f (x) = mx. In this case, for all x,y in R and all scalars c and d, f (cx + dy) = m(cx + dy) = mcx + mdy = c(mx) + d(my) = cf (x) + df (y) This shows that f is linear.
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b. When f (x) = mx + b, with b nonzero, f(0) = m(0) = b = b 0. This shows that f is not linear, because every linear transformation maps the zero vector in its domain into the zero vector in the codomain. (In this case, both zero vectors are just the number 0.) Another argument, for instance, would be to calculate f (2x) = m(2x) + b and 2f (x) = 2mx + 2b. If b is nonzero, then f (2x) is not equal to 2f (x) and so f is not a linear transformation. c. In calculus, f is called a "linear function" because the graph of f is a line. 30. Let T(x) = Ax + b for x in Rn. If b is not zero, T(0) = A0 + b = b 0. Actually, T fails both properties of a linear transformation. For instance, T(2x) = A(2x) + b = 2Ax + b, which is not the same as 2T(x) = 2(Ax + b) = 2Ax + 2b. Also, T(x + y) = A(x + y) + b = Ax + Ay + b which is not the same as T(x) + T(y) = Ax + b + Ay + b 31. (The Study Guide has a more detailed discussion of the proof.) Suppose that {v1, v2, v3} is linearly dependent. Then there exist scalars c1, c2, c3, not all zero, such that c1v1 + c2v2 + c3v3 = 0 Then T(c1v1 + c2v2 + c3v3) = T(0) = 0. Since T is linear, c1T(v1) + c2T(v2) + c3T(v3) = 0 Since not all the weights are zero, {T(v1), T(v2), T(v3)} is a linearly dependent set. 32. Take any vector (x1, x2) with x2 0, and use a negative scalar. For instance, T(0, 1) = (2, 3), but T(1(0, 1)) = T(0, 1) = (2, 3) (1)T(0, 1). 33. One possibility is to show that T does not map the zero vector into the zero vector, something that every linear transformation does do. T(0, 0) = (0, 4, 0). 34. Suppose that {u, v} is a linearly independent set in Rn and yet T(u) and T(v) are linearly dependent. Then there exist weights c1, c2, not both zero, such that c1T(u) + c2T(v) = 0 Because T is linear, T(c1u + c2v) = 0. That is, the vector x = c1u + c2v satisfies T(x) = 0. Furthermore, x cannot be the zero vector, since that would mean that a nontrivial linear combination of u and v is zero, which is impossible because u and v are linearly independent. Thus, the equation T(x) = 0 has a nontrivial solution. 35. Take u and v in R3 and let c and d be scalars. Then cu + dv = (cu1 + dv1, cu2 + dv2, cu3 + dv3). The transformation T is linear because T(cu + dv) = (cu1 + dv1, cu2 + dv2, (cu3 + dv3)) = (cu1 + dv1, cu2 + dv2, cu3 dv3) = (cu1, cu2, cu3) + (dv1, dv2, dv3) = c(u1, u2, u3) + d(v1, v2, v3) = cT(u) + dT(v) 36. Take u and v in R3 and let c and d be scalars. Then cu + dv = (cu1 + dv1, cu2 + dv2, cu3 + dv3). The transformation T is linear because T(cu + dv) = (cu1 + dv1, 0, cu3 + dv3) = (cu1, 0, cu3) + (dv1, 0, dv3) = c(u1, 0, u3) + d(v1, 0, v3) = cT(u) + dT(v)
1.8
Solutions
61
4 -9 37. [M] -6 5 -9 5 38. [M] 7 9
-2 7 4 -3 -4 -8 11 -7
5
-8 5 8 -9 -7 16 -4
-5 0 3 -4
4 6 -9 5
0 1 0 0 ~ 0 0 0 0 0 1 0 0 ~ 0 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 0 0 1 0
-7 / 2 -9 / 2 0 0
3/ 4 5/ 4 -7 / 4 0
0 0 , 0 0 0 0 , 0 0
x1 = (7 / 2) x4 x = (9 / 2) x 2 4 x3 = 0 x4 is free x1 = -(3/ 4) x4 x = -(5 / 4) x 2 4 x3 = (7 / 4) x4 x4 is free
7 / 2 9 / 2 x = x4 0 1 -3/ 4 -5 / 4 x = x4 7/4 1
5 -5 7 1 0 0 -7 / 2 4 4 -2 -9 7 -8 0 5 0 1 0 -9 / 2 7 ~ , yes, b is in the range of the transformation, 39. [M] -6 4 5 3 9 0 0 1 0 1 8 -4 7 0 0 0 0 0 5 -3 because the augmented matrix shows a consistent system. In fact,
x1 = 4 + (7 / 2) x4 4 x = 7 + (9 / 2) x 7 2 4 ; when x4 = 0 a solution is x = . the general solution is 1 x3 = 1 x4 is free 0
4 - 7 1 0 0 3/ 4 -5 / 4 -9 -4 -9 5 -8 -7 0 1 0 -7 6 5 / 4 -11/ 4 , yes, b is in the range of the 40. [M] ~ 7 11 16 -9 13 0 0 1 -7 / 4 13/ 4 -5 0 0 0 5 0 0 9 -7 -4 transformation, because the augmented matrix shows a consistent system. In fact,
x1 = -5 / 4 - (3/ 4) x4 -2 x = -11/ 4 - (5 / 4) x -4 4 ; when x4 = 1 a solution is x = . the general solution is 2 5 x3 = 13/ 4 + (7 / 4) x4 x4 is free 1
Notes: At the end of Section 1.8, the Study Guide provides a list of equations, figures, examples, and connections with concepts that will strengthen a student's understanding of linear transformations. I encourage my students to continue the construction of review sheets similar to those for "span" and "linear independence," but I refrain from collecting these sheets. At some point the students have to assume the responsibility for mastering this material. If your students are using MATLAB or another matrix program, you might insert the definition of matrix multiplication after this section, and then assign a project that uses random matrices to explore properties of matrix multiplication. See Exercises 3436 in Section 2.1. Meanwhile, in class you can continue with your plans for finishing Chapter 1. When you get to Section 2.1, you won't have much to do. The Study Guide's MATLAB note for Section 2.1 contains the matrix notation students will need for a project on matrix multiplication. The appendices in the Study Guide have the corresponding material for Mathematica, Maple, and the T-83+/86/89 and HP-48G graphic calculators.
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1.9
SOLUTIONS
Notes: This section is optional if you plan to treat linear transformations only lightly, but many instructors
will want to cover at least Theorem 10 and a few geometric examples. Exercises 15 and 16 illustrate a fast way to solve Exercises 1722 without explicitly computing the images of the standard basis. The purpose of introducing one-to-one and onto is to prepare for the term isomorphism (in Section 4.4) and to acquaint math majors with these terms. Mastery of these concepts would require a substantial digression, and some instructors prefer to omit these topics (and Exercises 2540). In this case, you can use the result of Exercise 31 in Section 1.8 to show that the coordinate mapping from a vector space onto Rn (in Section 4.4) preserves linear independence and dependence of sets of vectors. (See Example 6 in Section 4.4.) The notions of one-to-one and onto appear in the Invertible Matrix Theorem (Section 2.3), but can be omitted there if desired Exercises 2528 and 3136 offer fairly easy writing practice. Exercises 31, 32, and 35 provide important links to earlier material.
3 1 1. A = [T(e1) T(e2)] = 3 1
-5 2 0 0
4
1 2. A = [T(e1) T(e2) T(e3)] = 3 3. T(e1) = e2, T(e2) = e1. A = [ -e 2 1/ 2 4. T(e1) = , T(e2) = -1/ 2
-7
-5 4
1 0 1/ 2 1/ 2
0 e1 ] = -1
1/ 2 ,A= 1/ 2
1/ 2 -1/ 2
1 1 5. T(e1) = e1 2e2 = , T(e2) = e2, A = -2 -2 3 6. T(e1) = e1, T(e2) = e2 + 3e1 = , A = 1 1 0
0 1 3 1
7. Follow what happens to e1 and e2. Since e1 is on the unit circle in the plane, it rotates through 3 /4 radians into a point on the unit circle that lies in the third quadrant and on the line x2 = x1 (that is, y = x in more familiar notation). The point (1,1) is on the ine x2 = x1 , but its distance
from the origin is 2. So the rotational image of e1 is (1/ 2, 1/ 2) . Then this image reflects in the horizontal axis to (1/ 2,1/ 2) . Similarly, e2 rotates into a point on the unit circle that lies in the second quadrant and on the line x2 = x1 , namely,
1.9
Solutions
63
(1/ 2, 1/ 2) . Then this image reflects in the horizontal axis to (1/ 2,1/ 2) . When the two calculations described above are written in vertical vector notation, the transformation's standard matrix [T(e1) T(e2)] is easily seen:
-1/ 2 -1/ 2 1/ 2 1/ 2 -1/ 2 e1 , e2 , A= -1/ 2 1/ 2 -1/ 2 1/ 2 1/ 2 8. e1 e1 e 2 and e 2 -e 2 -e1 , so A = [e2 0 -e1 ] = 1
1/ 2 1/ 2
-1 0
9. The horizontal shear maps e1 into e1, and then the reflection in the line x2 = x1 maps e1 into e2. (See Table 1.) The horizontal shear maps e2 into e2 into e2 2e1. To find the image of e2 2e1 when it is reflected in the line x2 = x1, use the fact that such a reflection is a linear transformation. So, the image of e2 2e1 is the same linear combination of the images of e2 and e1, namely, e1 2(e2) = e1 + 2e2. To summarize, 0 e1 e1 -e 2 and e 2 e2 - 2e1 -e1 + 2e2 , so A = -1
-1 2
To find the image of e2 2e1 when it is reflected through the vertical axis use the fact that such a reflection is a linear transformation. So, the image of e2 2e1 is the same linear combination of the images of e2 and e1, namely, e2 + 2e1.
0 10. e1 -e1 -e 2 and e 2 e 2 -e1 , so A = -1
-1 0
11. The transformation T described maps e1 e1 -e1 and maps e 2 -e 2 -e 2 . A rotation through radians also maps e1 into e1 and maps e2 into e2. Since a linear transformation is completely determined by what it does to the columns of the identity matrix, the rotation transformation has the same effect as T on every vector in 2 .
12. The transformation T in Exercise 8 maps e1 e1 e2 and maps e 2 -e2 -e1 . A rotation about the origin through / 2 radians also maps e1 into e2 and maps e2 into e1. Since a linear transformation is completely determined by what it does to the columns of the identity matrix, the rotation transformation has the same effect as T on every vector in 2 . 13. Since (2, 1) = 2e1 + e2, the image of (2, 1) under T is 2T(e1) + T(e2), by linearity of T. On the figure in the exercise, locate 2T(e1) and use it with T(e2) to form the parallelogram shown below.
x2 T(2, 1) 2T(e 1 ) T(e 1 ) T(e 2 ) x1
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14. Since T(x) = Ax = [a1 a2]x = x1a1 + x2a2 = a1 + 3a2, when x = (1, 3), the image of x is located by forming the parallelogram shown below.
x2 T(1, 3)
a1
a2 a1
x1
3 15. By inspection, 4 1 1 16. By inspection, -2 1
0 0 -1
-2 x1 3x1 - 2 x3 0 x2 = 4 x1 1 x3 x1 - x2 + x3
-1 x1 - x2 x1 = -2 x + x 1 1 2 x x1 0 2
17. To express T(x) as Ax , write T(x) and x as column vectors, and then fill in the entries in A by inspection, as done in Exercises 15 and 16. Note that since T(x) and x have four entries, A must be a 44 matrix.
0 x + x T(x) = 1 2 = x2 + x3 x3 + x4 x1 0 x 1 2 = x3 0 x4 0
0 1 1 0
0 0 1 1
A
0 x1 0 x2 0 x3 1 x4
18. As in Exercise 17, write T(x) and x as column vectors. Since x has 2 entries, A has 2 columns. Since T(x) has 4 entries, A has 4 rows.
2 x2 - 3x1 -3 x - 4x x 1 2 1 = A 1 = x2 0 0 0 x2
2 -4 x1 0 x2 1
19. Since T(x) has 2 entries, A has 2 rows. Since x has 3 entries, A has 3 columns.
x1 - 5 x2 + 4 x3 = x2 - 6 x3
A
x1 1 x2 = x 0 3
-5 1
x1 4 x2 -6 x3
20. Since T(x) has 1 entry, A has 1 row. Since x has 4 entries, A has 4 columns.
x1 x ] 2 = [2 x3 x4 x1 x -4] 2 x3 x4
[2 x1 + 3x3 - 4 x4 ] = [
A
0
3
1.9
Solutions
65
x +x 21. T(x) = 1 2 = A 4 x1 + 5 x2 1 1 3 1 1 4 5 8 ~ 0 1
3 x1 1 1 x1 x = 4 5 x . To solve T(x) = 8 , row reduce the augmented matrix: 2 2 3 1 0 7 7 ~ 0 1 -4 , x = -4 . -4
x1 - 2 x2 1 - x + 3 x = A x1 = -1 22. T(x) = 1 2 x 3x1 - 2 x2 2 3 matrix: 1 -1 3
-2 x 3 1 . To solve T(x) = x -2 2 -1 1 3 ~ 0 0 0
0 1 0
-1 4 , row reduce the augmented 9
-2 3 -2
-1 1 4 ~ 0 9 0
-2 1 4
-1 1 3 ~ 0 12 0
-2 1 0
5 5 3 , x = . 3 0
23. a. b. c. d. e. 24. a. b. c. d.
True. See Theorem 10. True. See Example 3. False. See the paragraph before Table 1. False. See the definition of onto. Any function from Rn to Rm maps each vector onto another vector. False. See Example 5.
False. See the paragraph preceding Example 2. True. See Theorem 10. True. See Table 1. False. See the definition of one-to-one. Any function from Rn to Rm maps a vector onto a single (unique) vector. e. True. See the solution of Example 5.
25. Three row interchanges on the standard matrix A of the transformation T in Exercise 17 produce 1 1 0 0 0 1 1 0 . This matrix shows that A has only three pivot positions, so the equation Ax = 0 has a 0 0 1 1 0 0 0 0
nontrivial solution. By Theorem 11, the transformation T is not one-to-one. Also, since A does not have a pivot in each row, the columns of A do not span R4. By Theorem 12, T does not map R4 onto R4.
26. The standard matrix A of the transformation T in Exercise 2 is 23. Its columns are linearly dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A is row 4 -5 1 2 3 equivalent to , which shows that the rows of A span R . By Theorem 12, T maps R 0 -19 19 onto R2. 4 1 -5 27. The standard matrix A of the transformation T in Exercise 19 is . The columns of A are 1 -6 0 linearly dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A has a pivot in each row, so the rows of A span R2. By Theorem 12, T maps R3 onto R2.
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28. The standard matrix A of the transformation T in Exercise 14 has linearly independent columns, because the figure in that exercise shows that a1 and a2 are not multiples. So T is one-to-one, by Theorem 12. Also, A must have a pivot in each column because the equation Ax = 0 has no free variables. Thus, the * 2 2 echelon form of A is . Since A has a pivot in each row, the columns of A span R . So T maps R 0 onto R2. An alternate argument for the second part is to observe directly from the figure in Exercise 14 that a1 and a2 span R2. This is more or less evident, based on experience with grids such as those in Figure 8 and Exercise 7 of Section 1.3. 29. By Theorem 12, the columns of the standard matrix A must be linearly independent and hence the * 0 equation Ax = 0 has no free variables. So each column of A must be a pivot column: A ~ 0 0 0 0
Note that T cannot be onto because of the shape of A. * * . 0
30. By Theorem 12, the columns of the standard matrix A must span 3. By Theorem 4, the matrix must
have a pivot in each row. There are four possibilities for the echelon form: * * * * * * * * * 0 * * 0 * * , 0 * * , 0 0 * , 0 0 * 0 0 * 0 0 0 0 0 0 0 0 0 Note that T cannot be one-to-one because of the shape of A.
31. "T is one-to-one if and only if A has n pivot columns." By Theorem 12(b), T is one-to-one if and only if the columns of A are linearly independent. And from the statement in Exercise 30 in Section 1.7, the columns of A are linearly independent if and only if A has n pivot columns. 32. The transformation T maps Rn onto Rm if and only if the columns of A span Rm, by Theorem 12. This happens if and only if A has a pivot position in each row, by Theorem 4 in Section 1.4. Since A has m rows, this happens if and only if A has m pivot columns. Thus, "T maps Rn onto Rm if and only A has m pivot columns." 33. Define T : n m by T(x) = Bx for some mn matrix B, and let A be the standard matrix for T. By definition, A = [T(e1) T(en)], where ej is the jth column of In. However, by matrix-vector multiplication, T(ej) = Bej = bj, the jth column of B. So A = [b1 bn] = B. 34. The transformation T maps Rn onto Rm if and only if for each y in Rm there exists an x in Rn such that y = T(x). 35. If T : n m maps n onto m , then its standard matrix A has a pivot in each row, by Theorem 12 and by Theorem 4 in Section 1.4. So A must have at least as many columns as rows. That is, m < n. When T is one-to-one, A must have a pivot in each column, by Theorem 12, so m > n. 36. Take u and v in Rp and let c and d be scalars. Then T(S(cu + dv)) = T(cS(u) + dS(v)) because S is linear = cT(S(u)) + dT(S(v)) because T is linear This calculation shows that the mapping x T(S(x)) is linear. See equation (4) in Section 1.8.
1.10 Solutions
67
4 -5 10 -5 1 0 0 44 / 35 1 0 0 1.2571 8 0 1 0 79 / 35 0 1 0 2.2571 3 -4 7 ~ . There is no pivot in the 37. [M] ~ ~ 4 -9 0 0 1 86 / 35 0 0 1 2.4571 5 -3 5 4 0 0 0 0 0 -3 -2 0 0 0 fourth column of the standard matrix A, so the equation Ax = 0 has a nontrivial solution. By Theorem 11, the transformation T is not one-to-one. (For a shorter argument, use the result of Exercise 31.) 5 4 -9 0 7 0 7 1 10 0 6 16 -4 1 -9 0 . No. There is no pivot in the third column of the 38. [M] ~ ~ 12 0 8 12 7 0 0 1 5 0 0 0 -8 -6 -2 0 standard matrix A, so the equation Ax = 0 has a nontrivial solution. By Theorem 11, the transformation T is not one-to-one. (For a shorter argument, use the result of Exercise 31.) 3 7 5 0 0 5 0 4 -7 1 6 -8 0 5 12 -8 1 0 1 0 39. [M] -7 10 -8 -9 14 ~ ~ 0 0 1 -2 0 . There is not a pivot in every row, so 4 2 -6 0 0 0 1 3 -5 0 -5 0 6 -6 -7 3 0 0 0 0 5 the columns of the standard matrix do not span R . By Theorem 12, the transformation T does not map R5 onto R5. 5 6 -1 0 0 0 5 9 13 1 14 15 -7 -6 0 4 1 0 0 -4 40. [M] -8 -9 12 -5 -9 ~ ~ 0 0 1 0 0 . There is not a pivot in every row, so 9 8 0 0 1 1 -5 -6 -8 0 13 14 15 0 2 11 0 0 0 0 the columns of the standard matrix do not span R5. By Theorem 12, the transformation T does not map R5 onto R5.
1.10
SOLUTIONS
1. a. If x1 is the number of servings of Cheerios and x2 is the number of servings of 100% Natural Cereal, then x1 and x2 should satisfy
nutrients nutrients quantities x1 per serving + x2 per serving of = of nutrients of Cheerios 100% Natural required That is, 110 130 295 4 3 9 x1 + x2 = 20 18 48 2 5 8
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110 4 b. The equivalent matrix equation is 20 2 matrix for this equation. 110 4 20 2 1 0 ~ 0 0 130 3 18 5 2.5 -7 -16 -145 295 2 9 4 ~ 48 20 8 110 4 1 -7 0 ~ -16 0 -145 0 5 3 18 130 2.5 1 0 0
130 295 x 3 1 9 . To solve this, row reduce the augmented = 18 x2 48 5 8 4 9 24 295
8 1 2.5 9 4 3 ~ 48 10 9 295 110 130 4 1 1 0 ~ 0 0 0 0 0 1 0 0 1.5 1 0 0
The desired nutrients are provided by 1.5 servings of Cheerios together with 1 serving of 100% Natural Cereal.
2. Set up nutrient vectors for one serving of Kellogg's Cracklin' Oat Bran (COB) and Kellogg's Crispix (Crp):
Nutrients: calories COB Crp
110 3 protein carbohydrate 21 fat 3
110 2 . 25 .4
110 110 3 2 , u = 3 . a. Let B = [ COB Crp ] = 2 21 25 .4 3 Then Bu lists the amounts of calories, protein, carbohydrate, and fat in a mixture of three servings of Cracklin' Oat Bran and two servings of Crispix. b. Let u1 and u2 be the number of servings of Cracklin' Oat Bran and Crispix, respectively. Can these 110 2.25 u ? To find out, row reduce the augmented matrix numbers satisfy the equation B 1 = u2 24 1 110 110 3 2 21 25 .4 3 110 1 2.25 3 ~ 24 21 1 3 1 2 25 .4 1 1 2.25 0 ~ 24 0 1 0 1 -1 4 -2.6 1 1 -.75 0 ~ 3 0 -2 0 1 -1 0 0 1 -.75 0 -.05
1.10 Solutions
69
The last row identifies an inconsistent system, because 0 = .05 is impossible. So, technically, there is no mixture of the two cereals that will supply exactly the desired list of nutrients. However, one could tentatively ignore the final equation and see what the other equations prescribe. They reduce to u1 = .25 and u2 = .75. What does the corresponding mixture provide? 110 110 110 3 2 2.25 .25 COB + .75 Crp = .25 + .75 = 21 25 24 3 .4 1.05 The error of 5% for fat might be acceptable for practical purposes. Actually, the data in COB and Crp are certainly not precise and may have some errors even greater than 5%.
3. Here are the data, assembled from Table 1 and Exercise 3:
Mg of Nutrients/Unit soy soy milk flour whey prot. 36 52 0 1.26 51 34 7 .19 13 74 1.1 .8 80 0 3.4 .18 Nutrients Required (milligrams) 33 45 3 .8
Nutrient protein carboh. fat calcium
a. Let x1, x2, x3, x4 represent the number of units of nonfat milk, soy flour, whey, and isolated soy protein, respectively. These amounts must satisfy the following matrix equation
36 52 0 1.26 51 34 7 .19 13 74 1.1 .8 51 34 7 .19 80 x1 33 0 x2 45 = 3.4 x3 3 .18 x4 .8 13 74 1.1 .8 80 0 3.4 .18 1 33 45 0 ~ ~ 0 3 0 .8 0 1 0 0 0 0 1 0 0 0 0 1 .64 .54 -.09 -.21
36 52 b. [M] 0 1.26
The "solution" is x1 = .64, x2 = .54, x3 = .09, x4 = .21. This solution is not feasible, because the mixture cannot include negative amounts of whey and isolated soy protein. Although the coefficients of these two ingredients are fairly small, they cannot be ignored. The mixture of .64 units of nonfat milk and .54 units of soy flour provide 50.6 g of protein, 51.6 g of carbohydrate, 3.8 g of fat, and .9 g of calcium. Some of these nutrients are nowhere close to the desired amounts. 4. Let x1, x2, and x3 be the number of units of foods 1, 2, and 3, respectively, needed for a meal. The values of x1, x2, and x3 should satisfy nutrients nutrients nutrients milligrams (in mg) + x (in mg) + x (in mg) = of nutrients x1 per unit 2 per unit 3 per unit of Food 1 of Food 2 of Food 3 required
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From the given data, 10 20 20 100 50 + x 40 + x 10 = 300 x1 2 3 30 10 40 200 To solve, row reduce the corresponding augmented matrix: 10 50 30 1 ~ 0 0 20 40 10 2 1 0 20 10 40 2 3/ 2 1 100 10 300 ~ 0 200 0 20 -60 -50 2 1 0 20 -90 -20 0 0 1 100 1 -200 ~ 0 -100 0 250 / 33 1 50 / 33 ~ 0 40 / 33 0 2 1 5 0 1 0 2 3/ 2 2 0 0 1 10 10 / 3 10 50 /11 50 / 33 40 / 33
10 1 10 / 3 ~ 0 40 / 33 0
50 /11 4.55 units of Food 1 x = 50 / 33 = 1.52 = units of Food 2 40 / 33 1.21 units of Food 3
5. Loop 1: The resistance vector is
5 -2 r1 = 0 0 -2 11 r2 = -3 0 Total of four RI voltage drops for current I 1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Current I4 does not flow in loop 1
Loop 2: The resistance vector is Voltage drop for I1 is negative; I1 flows in opposite direction Total of four RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2 0 0 , and R = [r1 r2 r3 r4] = -4 25 5 -2 0 0 -2 11 -3 0 0 -3 17 -4 0 0 . -4 25
0 -3 Also, r3 = , r4 = 17 -4
Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the currents in other loops adjacent to loop j to flow in the direction opposite to current Ij.) 40 -30 Next, set v = . The voltages in loops 2 and 4 are negative because the battery orientation in each 20 -10 loop is opposite to the direction chosen for positive current flow. Thus, the equation Ri = v becomes
1.10 Solutions
71
5 -2 0 0
-2 11 -3 0
0 -3 17 -4
0 I1 40 0 I 2 -30 . = -4 I 3 20 25 I 4 -10
I1 7.56 I -1.10 [M]: The solution is i = 2 = . I 3 .93 I 4 -.25
6. Loop 1: The resistance vector is
4 -1 r1 = 0 0 Total of four RI voltage drops for current I1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Current I 4 does not flow in loop 1
Loop 2: The resistance vector is -1 6 r2 = -2 0 Voltage drop for I 1 is negative; I 1 flows in opposite direction Total of four RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2 40 30 . Then Ri = v becomes 20 10
0 0 -2 , r4 = 0 , and R = [r1 r2 r3 r4]. Set v = Also, r3 = 10 -3 12 -3 4 -1 0 0 -1 6 -2 0 0 -2 10 -3
0 I1 40 I1 12.11 I 8.44 I 30 0 2 . . [M]: The solution is i = 2 = = I 3 4.26 -3 I 3 20 12 I 4 10 I 4 1.90
7. Loop 1: The resistance vector is 12 -7 r1 = 0 -4 Total of three RI voltage drops for current I 1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Voltage drop for I 4 is negative; I 4 flows in opposite direction
Loop 2: The resistance vector is -7 15 r2 = -6 0 Voltage drop for I1 is negative; I1 flows in opposite direction Total of three RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2
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0 -6 Also, r3 = , r4 = 14 -5
-4 0 , and R = [r1 r2 r3 r4] = -5 13
12 -7 0 -4
-7 15 -6 0
0 -6 14 -5
-4 0 . -5 13
Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the currents in other loops adjacent to loop j to flow in the direction opposite to current Ij.) 40 30 . Note the negative voltage in loop 4. The current direction chosen in loop 4 is Next, set v = 20 -10 opposed by the orientation of the voltage source in that loop. Thus Ri = v becomes 12 -7 0 -4 -7 15 -6 0 0 -6 14 -5 -4 I1 40 I1 11.43 I 30 0 2 . [M]: The solution is i = I 2 = 10.55 . = I 3 8.04 -5 I 3 20 13 I 4 -10 I 4 5.84
8. Loop 1: The resistance vector is 15 -5 r1 = 0 -5 -1 -5 15 r2 = -5 0 -2 Total of four RI voltage drops for current I1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Voltage drop for I 4 is negative; I 4 flows in opposite direction Voltage drop for I 5 is negative; I 5 flows in opposite direction
Loop 2: The resistance vector is Voltage drop for I1 is negative; I1 flows in opposite direction Total of four RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I4 does not flow in loop 2 Voltage drop for I5 is negative; I5 flows in opposite direction
0 -5 -1 0 -5 -1 15 -5 40 -5 0 -2 -5 15 -5 -30 0 -2 Also, r3 = 15 , r4 = -5 , r5 = -3 , and R = 0 -5 15 -5 -3 . Set v = 20 . Note the 0 -5 15 -4 -5 15 -4 -5 -10 -3 -4 10 -1 -2 -3 -4 10 0 negative voltages for loops where the chosen current direction is opposed by the orientation of the voltage source in that loop. Thus Ri = v becomes:
1.10 Solutions
73
15 -5 0 -5 -1
-5 15 -5 0 -2
0 -5 15 -5 -3
-5 0 -5 15 -4
-1 I1 40 -2 I 2 -30 -3 I 3 = 20 . [M] The solution is -4 I 4 -10 10 I 5 0
I1 3.37 I .11 2 I 3 = 2.27 . I 4 1.67 I 5 1.70
9. The population movement problems in this section assume that the total population is constant, with no migration or immigration. The statement that "about 5% of the city's population moves to the suburbs" means also that the rest of the city's population (95%) remain in the city. This determines the entries in the first column of the migration matrix (which concerns movement from the city). From: City Suburbs To: .95 .05 City Suburbs
Likewise, if 4% of the suburban population moves to the city, then the other 96% remain in the suburbs. .95 .04 This determines the second column of the migration matrix:, M = . The difference equation is .05 .96 600,000 xk+1 = Mxk for k = 0, 1, 2, .... Also, x0 = 400,000 .95 The population in 2001 (when k = 1) is x1 = Mx0 = .05 .95 The population in 2002 (when k = 2) is x2 = Mx1 = .05 .04 600,000 586,000 = .96 400,000 414,000 .04 586,000 573, 260 = .96 414,000 426,740
10. The data in the first sentence implies that the migration matrix has the form: From: City Suburbs To: .07 .03 City Suburbs
The remaining entries are determined by the fact that the numbers in each column must sum to 1. (For instance, if 7% of the city people move to the suburbs, then the rest, or 93%, remain in the city.) So the 800,000 .93 .03 migration matrix is M = . The initial population is x0 = 500,000 . .07 .97 .93 The population in 2001 (when k = 1) is x1 = Mx0 = .07 .93 The population in 2002 (when k = 2) is x2 = Mx1 = .07 .03 800,000 759,000 = .97 500,000 541,000 .03 759,000 722,100 = .97 541,000 577,900
11. The problem concerns two groups of peoplethose living in California and those living outside California (and in the United States). It is reasonable, but not essential, to consider the people living inside
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California first. That is, the first entry in a column or row of a vector will concern the people living in California. With this choice, the migration matrix has the form: From: Calif. Outside To: Calif. Outside
a. For the first column of the migration matrix M, compute persons {Calif.moved } who = 509,500 = .017146 29,726,000
{Total Calif. pop.}
The other entry in the first column is 1 .017146 = .982854. The exercise requests that 5 decimal places be used. So this number should be rounded to .98285. Whatever number of decimal places is used, it is important that the two entries sum to 1. So, for the first fraction, use .01715. For the second column of M, compute persons {outsidemoved } who = 564,100 = .00258 . The other entry 218,994,000
{Total outside pop.}
is 1 .00258 = .99742. Thus, the migration matrix is From: Calif. Outside To: .98285 .01715 .00258 .99742 Calif. Outside
b. [M] The initial vector is x0 = (29.716, 218.994), with data in millions of persons. Since x0 describes the population in 1990, and x1 describes the population in 1991, the vector x10 describes the projected population for the year 2000, assuming that the migration rates remain constant and there are no deaths, births, or migration. Here are some of the vectors in the calculation, with only the first 4 or 5 figures displayed. Numbers are in millions of persons: 29.7 29.8 29.8 30.1 30.18 30.223 219.0 , 218.9 , 218.9 , , 218.6 , 218.53 , 218.487 = x10. .97 12. Set M = .00 .03 .10 305 .97 .05 .10 305 308 and x = 48 . Then x = .00 .90 .05 48 48 , and .90 .05 1 0 98 .03 .05 .85 98 95 .05 .85 .97 .05 .10 308 311 x2 = .00 .90 .05 48 48 . The entries in x2 give the approximate distribution of cars on .03 .05 .85 95 92 Wednesday, two days after Monday. .05
13. [M] The order of entries in a column of a migration matrix must match the order of the columns. For instance, if the first column concerns the population in the city, then the first entry in each column must be the fraction of the population that moves to (or remains in) the city. In this case, the data in the 600,000 .95 .03 exercise leads to M = and x0 = 400,000 .05 .97
1.10 Solutions
75
a. Some of the population vectors are 523, 293 472,737 439, 417 417, 456 x5 = , x10 = 527, 263 , x15 = 560,583 , x 20 = 582,544 476,707 The data here shows that the city population is declining and the suburban population is increasing, but the changes in population each year seem to grow smaller. 350,000 b. When x0 = , the situation is different. Now 650,000 358,523 364,140 367,843 370, 283 x5 = , x10 = , x15 = , x 20 = 641, 477 635,860 632,157 629,717 The city population is increasing slowly and the suburban population is decreasing. No other conclusions are expected. (This example will be analyzed in greater detail later in the text.) 14. Here are Figs. (a) and (b) for Exercise 13, followed by the figure for Exercise 34 in Section 1.1:
20 0 0 1 4 20 2 3 0 0 10 10 0 1 4 0 2 3 40 40 10 10 20 1 4 20 2 3 40 40
20 (a)
20
10 (b)
10
30
30
Section 1.1
For Fig. (a), the equations are: 4T1 = 0 + 20 + T2 + T4 4T2 = T1 + 20 + 0 + T3 4T3 = T4 + T2 + 0 + 20 4T4 = 0 + T1 + T3 + 20 To solve the system, rearrange the equations and row reduce the augmented matrix. Interchanging rows 1 and 4 speeds up the calculations. The first five steps are shown in detail.
4 -1 0 -1 1 0 ~ 0 0
-1 4 -1 0 0 -1 4 -1 -1 0 -1 4 20
1 -1 20 ~ 20 0 20 4
-20
0 4 -1 -1
1 -1 4 0
-4 0 -1 -1
-20
1 0 20 ~ 20 0 20 0
-20
0 4 -1 -1
1 0 4 -4
-4 -4 -1 15
-20
1 0 0 ~ 20 0 100 0
10 10 10 10
0 1 -1 -1
1 0 4 -4
-4 -1 -1 15
-20 0 20 100
0 1 0 0
1 0 4 -4
-4 -1 -2 14
1 0 0 ~ 20 0 100 0
0 1 0 0
1 0 4 0
-4 -1 -2 12
1 0 0 ~ ~ 20 0 0 120
0 1 0 0
0 0 1 0
0 0 0 1
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For Fig (b), the equations are 4T1 = 10 + 0 + T2 + T4 4T2 = T1 + 0 + 40 + T3 4T3 = T4 + T2 + 40 + 10 4T4 = 10 + T1 + T3 + 10 Rearrange the equations and row reduce the augmented matrix:
4 -1 0 -1
-1 4 -1 0 0 -1 4 -1 -1 0 -1 4 10
1 0 40 ~ ~ 50 0 20 0
0 1 0 0
0 0 1 0
0 0 0 1
17.5 20 12.5
10
a. Here are the solution temperatures for the three problems studied: Fig. (a) in Exercise 14 of Section 1.10: (10, 10, 10, 10) Fig. (b) in Exercise 14 of Section 1.10: (10, 17.5, 20, 12.5) Figure for Exercises 34 in Section 1.1 (20, 27.5, 30, 22.5) When the solutions are arranged this way, it is evident that the third solution is the sum of the first two solutions. What might not be so evident is that list of boundary temperatures of the third problem is the sum of the lists of boundary temperatures of the first two problems. (The temperatures are listed clockwise, starting at the left of T1.) Fig. (a): ( 0, 20, 20, 0, 0, 20, 20, 0) Fig. (b): (10, 0, 0, 40, 40, 10, 10, 10) Fig. from Section 1.1: (10, 20, 20, 40, 40, 30, 30, 10) b. When the boundary temperatures in Fig. (a) are multiplied by 3, the new interior temperatures are also multiplied by 3. c. The correspondence from the list of eight boundary temperatures to the list of four interior temperatures is a linear transformation. A verification of this statement is not expected. However, it can be shown that the solutions of the steady-state temperature problem here satisfy a superposition principle. The system of equations that approximate the interior temperatures can be written in the form Ax = b, where A is determined by the arrangement of the four interior points on the plate and b is a vector in R4 determined by the boundary temperatures.
Note: The MATLAB box in the Study Guide for Section 1.10 discusses scientific notation and shows how
to generate a matrix whose columns list the vectors x0, x1, x2, ..., determined by an equation xk+1 = Mxk for k = 0 , 1, ....
Chapter 1
SUPPLEMENTARY EXERCISES
1. a. False. (The word "reduced" is missing.) Counterexample:
1 A= 3
2 1 , B= 4 0
2 1 , C= -2 0
2 1
The matrix A is row equivalent to matrices B and C, both in echelon form.
Chapter 1
Supplementary Exercises
77
b. False. Counterexample: Let A be any nn matrix with fewer than n pivot columns. Then the equation Ax = 0 has infinitely many solutions. (Theorem 2 in Section 1.2 says that a system has either zero, one, or infinitely many solutions, but it does not say that a system with infinitely many solutions exists. Some counterexample is needed.) c. True. If a linear system has more than one solution, it is a consistent system and has a free variable. By the Existence and Uniqueness Theorem in Section 1.2, the system has infinitely many solutions. d. False. Counterexample: The following system has no free variables and no solution: x1 + x2 = 1 x2 = 5 x1 + x2 = 2 e. True. See the box after the definition of elementary row operations, in Section 1.1. If [A b] is transformed into [C d] by elementary row operations, then the two augmented matrices are row equivalent. f. True. Theorem 6 in Section 1.5 essentially says that when Ax = b is consistent, the solution sets of the nonhomogeneous equation and the homogeneous equation are translates of each other. In this case, the two equations have the same number of solutions. g. False. For the columns of A to span Rm, the equation Ax = b must be consistent for all b in Rm, not for just one vector b in Rm. h. False. Any matrix can be transformed by elementary row operations into reduced echelon form, but not every matrix equation Ax = b is consistent. i. True. If A is row equivalent to B, then A can be transformed by elementary row operations first into B and then further transformed into the reduced echelon form U of B. Since the reduced echelon form of A is unique, it must be U. j. False. Every equation Ax = 0 has the trivial solution whether or not some variables are free. k. True, by Theorem 4 in Section 1.4. If the equation Ax = b is consistent for every b in Rm, then A must have a position in every one of its m rows. If A has m pivot positions, then A has m pivot columns, each containing one pivot position. l. False. The word "unique" should be deleted. Let A be any matrix with m pivot columns but more than m columns altogether. Then the equation Ax = b is consistent and has m basic variables and at least one free variable. Thus the equation does not does not have a unique solution. m. True. If A has n pivot positions, it has a pivot in each of its n columns and in each of its n rows. The reduced echelon form has a 1 in each pivot position, so the reduced echelon form is the nn identity matrix. n. True. Both matrices A and B can be row reduced to the 33 identity matrix, as discussed in the previous question. Since the row operations that transform B into I3 are reversible, A can be transformed first into I3 and then into B. o. True. The reason is essentially the same as that given for question f. p. True. If the columns of A span Rm, then the reduced echelon form of A is a matrix U with a pivot in each row, by Theorem 4 in Section 1.4. Since B is row equivalent to A, B can be transformed by row operations first into A and then further transformed into U. Since U has a pivot in each row, so does B. By Theorem 4, the columns of B span Rm. q. False. See Example 5 in Section 1.6. r. True. Any set of three vectors in R2 would have to be linearly dependent, by Theorem 8 in Section 1.6.
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s. False. If a set {v1, v2, v3, v4} were to span R5, then the matrix A = [v1 v2 v3 v4] would have a pivot position in each of its five rows, which is impossible since A has only four columns. t. True. The vector u is a linear combination of u and v, namely, u = (1)u + 0v. u. False. If u and v are multiples, then Span{u, v} is a line, and w need not be on that line. v. False. Let u and v be any linearly independent pair of vectors and let w = 2v. Then w = 0u + 2v, so w is a linear combination of u and v. However, u cannot be a linear combination of v and w because if it were, u would be a multiple of v. That is not possible since {u, v} is linearly independent. w. False. The statement would be true if the condition v1 is not zero were present. See Theorem 7 in Section 1.7. However, if v1 = 0, then {v1, v2, v3} is linearly dependent, no matter what else might be true about v2 and v3. x. True. "Function" is another word used for "transformation" (as mentioned in the definition of "transformation" in Section 1.8), and a linear transformation is a special type of transformation. y. True. For the transformation x 6 Ax to map R5 onto R6, the matrix A would have to have a pivot in every row and hence have six pivot columns. This is impossible because A has only five columns. z. False. For the transformation x 6 Ax to be one-to-one, A must have a pivot in each column. Since A has n columns and m pivots, m might be less than n. 2. If a 0, then x = b/a; the solution is unique. If a = 0, and b 0, the solution set is empty, because 0x = 0 b. If a = 0 and b = 0, the equation 0x = 0 has infinitely many solutions. 3. a. Any consistent linear system whose echelon form is
* * 0 * 0 0 0
* * * * 0 * * or 0 0 * or 0 0 * * 0 0 0 0 0 0 0 0 0
b. Any consistent linear system whose coefficient matrix has reduced echelon form I3. c. Any inconsistent linear system of three equations in three variables. 4. Since there are three pivots (one in each row), the augmented matrix must reduce to the form
* * * 0 * * . A solution of Ax = b exists for all b because there is a pivot in each row of A. Each 0 0 * solution is unique because there are no free variables.
3 k 1 3 k 1 5. a. ~ 0 h - 12 8 - 4k . If h = 12 and k v 2, the second row of the augmented matrix 4 h 8 indicates an inconsistent system of the form 0x2 = b, with b nonzero. If h = 12, and k = 2, there is only one nonzero equation, and the system has infinitely many solutions. Finally, if h v 12, the coefficient matrix has two pivots and the system has a unique solution. 1 h -2 h 1 -2 b. ~ 0 k + 3h 1 . If k + 3h = 0, the system is inconsistent. Otherwise, the 6 k -2 coefficient matrix has two pivots and the system has a unique solution.
Chapter 1
Supplementary Exercises
79
4 -2 7 -5 6. a. Set v1 = , v 2 = , v 3 = , and b = . "Determine if b is a linear combination of v1, v2, 8 -3 10 -3 v3." Or, "Determine if b is in Span{v1, v2, v3}." To do this, compute 7 -5 4 -2 7 -5 4 -2 . The system is consistent, so b is in Span{v1, v2, v3}. 8 -3 10 -3 ~ 0 1 -4 7 4 b. Set A = 8
-2
7 -5 , b = . "Determine if b is a linear combination of the columns of A." -3 10 -3
c. Define T(x) = Ax. "Determine if b is in the range of T."
2 -4 -2 b1 -5 , v = 1 , v = 1 and b = b . "Determine if v , v , v span R3." To do this, row 7. a. Set v1 = 2 3 1 2 3 2 7 -5 -3 b3 reduce [v1 v2 v3]: 2 -4 -2 2 -4 -2 2 -4 -2 -5 1 1 ~ 0 -9 -4 ~ 0 -9 -4 . The matrix does not have a pivot in each row, so 7 -5 -3 0 9 4 0 0 0 its columns do not span R3, by Theorem 4 in Section 1.4. 2 b. Set A = -5 7
-4 1 -5
-2 1 . "Determine if the columns of A span R3." -3
* * b. 0 * 0 0
c. Define T(x) = Ax. "Determine if T maps R3 onto R3."
* * * * 0 * 8. a. , , 0 * 0 0 0 0
1 2 9. The first line is the line spanned by . The second line is spanned by . So the problem is to write 2 1 5 1 2 6 as the sum of a multiple of 2 and a multiple of 1 . That is, find x1 and x2 such that 2 1 5 x1 + x2 = . Reduce the augmented matrix for this equation: 1 2 6 2 1
1 2
5 1 ~ 6 2
2 1
6 1 ~ 5 0
2 -3
6 1 ~ -7 0
2 1
6 1 ~ 7 / 3 0
0 1
4 / 3 7 / 3
5 4 2 7 1 5 8 / 3 7 / 3 Thus, = + or = + . 6 3 1 3 2 6 4 / 3 14 / 3
10. The line through a1 and the origin and the line through a2 and the origin determine a "grid" on the x1x2-plane as shown below. Every point in R2 can be described uniquely in terms of this grid. Thus, b can
80
CHAPTER 1
Linear Equations in Linear Algebra
be reached from the origin by traveling a certain number of units in the a1-direction and a certain number of units in the a2-direction.
x2
b a1 x1 a2
11. A solution set is a line when the system has one free variable. If the coefficient matrix is 23, then two of 1 2 * the columns should be pivot columns. For instance, take . Put anything in column 3. The 0 3 * resulting matrix will be in echelon form. Make one row replacement operation on the second row to 1 2 1 1 2 1 create a matrix not in echelon form, such as ~ 0 3 1 1 5 2 12. A solution set is a plane where there are two free variables. If the coefficient matrix is 23, then only one column can be a pivot column. The echelon form will have all zeros in the second row. Use a row 1 2 3 replacement to create a matrix not in echelon form. For instance, let A = . 1 2 3
1 0 * 13. The reduced echelon form of A looks like E = 0 1 * . Since E is row equivalent to A, the equation 0 0 0 1 0 * 3 0 Ex = 0 has the same solutions as Ax = 0. Thus 0 1 * -2 = 0 . 0 0 0 1 0 1 By inspection, E = 0 0
0 1 0
-3 2 . 0
1 a 0 14. Row reduce the augmented matrix for x1 + x2 = (*). a a + 2 0 1 a
a a+2
0 1 ~ 0 0
a a+2-a
2
0 1 = 0 0
a (2 - a)(1 + a)
0 0
The equation (*) has a nontrivial solution only when (2 a)(1 + a) = 0. So the vectors are linearly independent for all a except a = 2 and a = 1. 15. a. If the three vectors are linearly independent, then a, c, and f must all be nonzero. (The converse is true, too.) Let A be the matrix whose columns are the three linearly independent vectors. Then
Chapter 1
Supplementary Exercises
81
A must have three pivot columns. (See Exercise 30 in Section 1.7, or realize that the equation Ax = 0 has only the trivial solution and so there can be no free variables in the system of equations.) Since A is 33, the pivot positions are exactly where a, c, and f are located. b. The numbers a, ..., f can have any values. Here's why. Denote the columns by v1, v2, and v3. Observe that v1 is not the zero vector. Next, v2 is not a multiple of v1 because the third entry of v2 is nonzero. Finally, v3 is not a linear combination of v1 and v2 because the fourth entry of v3 is nonzero. By Theorem 7 in Section 1.7, {v1, v2, v3} is linearly independent. 16. Denote the columns from right to left by v1, ..., v4. The "first" vector v1 is nonzero, v2 is not a multiple of v1 (because the third entry of v2 is nonzero), and v3 is not a linear combination of v1 and v2 (because the second entry of v3 is nonzero). Finally, by looking at first entries in the vectors, v4 cannot be a linear combination of v1, v2, and v3. By Theorem 7 in Section 1.7, the columns are linearly independent. 17. Here are two arguments. The first is a "direct" proof. The second is called a "proof by contradiction." i. Since {v1, v2, v3} is a linearly independent set, v1 0. Also, Theorem 7 shows that v2 cannot be a multiple of v1, and v3 cannot be a linear combination of v1 and v2. By hypothesis, v4 is not a linear combination of v1, v2, and v3. Thus, by Theorem 7, {v1, v2, v3, v4} cannot be a linearly dependent set and so must be linearly independent. ii. Suppose that {v1, v2, v3, v4} is linearly dependent. Then by Theorem 7, one of the vectors in the set is a linear combination of the preceding vectors. This vector cannot be v4 because v4 is not in Span{v1, v2, v3}. Also, none of the vectors in {v1, v2, v3} is a linear combinations of the preceding vectors, by Theorem 7. So the linear dependence of {v1, v2, v3, v4} is impossible. Thus {v1, v2, v3, v4} is linearly independent. 18. Suppose that c1 and c2 are constants such that (*) c1v1 + c2(v1 + v2) = 0 Then (c1 + c2)v1 + c2v2 = 0. Since v1 and v2 are linearly independent, both c1 + c2 = 0 and c2 = 0. It follows that both c1 and c2 in (*) must be zero, which shows that {v1, v1 + v2} is linearly independent. 19. Let M be the line through the origin that is parallel to the line through v1, v2, and v3. Then v2 v1 and v3 v1 are both on M. So one of these two vectors is a multiple of the other, say v2 v1 = k(v3 v1). This equation produces a linear dependence relation (k 1)v1 + v2 kv3 = 0. A second solution: A parametric equation of the line is x = v1 + t(v2 v1). Since v3 is on the line, there is some t0 such that v3 = v1 + t0(v2 v1) = (1 t0)v1 + t0v2. So v3 is a linear combination of v1 and v2, and {v1, v2, v3} is linearly dependent. 20. If T(u) = v, then since T is linear, T(u) = T((1)u) = (1)T(u) = v. 21. Either compute T(e1), T(e2), and T(e3) to make the columns of A, or write the vectors vertically in the definition of T and fill in the entries of A by inspection:
? Ax = ? ?
? A ?
? x1 x1 ? x2 = - x2 , ? x3 x3
1 A = 0 0
0 -1 0
0 0 1
22. By Theorem 12 in Section 1.9, the columns of A span R3. By Theorem 4 in Section 1.4, A has a pivot in each of its three rows. Since A has three columns, each column must be a pivot column. So the equation
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Ax = 0 has no free variables, and the columns of A are linearly independent. By Theorem 12 in Section 1.9, the transformation x 6 Ax is one-to-one.
a 23. b 4 3
4a - b 4 5 implies that = a 3 0 3a -3 4 5 4 ~ 0 0 -3 25 / 4 5
- +
3b 4b -3 1
= =
5 0
. Solve: 0 1 16 / 5 1 ~ -3/ 5 0 0 1 4 / 5 -3/ 5
4 ~ -15 / 4 0
5 4 ~ -3/ 5 0
Thus a = 4/5 and b = 3/5. 24. The matrix equation displayed gives the information 2a - 4b = 2 5 and 4a + 2b = 0. Solve for a and b: 2 -4 2 5 2 -4 2 5 1 -2 5 1 0 1/ 5 ~ ~ ~ 2 0 0 10 -4 5 0 1 -2 / 5 0 1 -2 / 5 4
So a = 1/ 5, b = -2 / 5.
25. a. The vector lists the number of three-, two-, and one-bedroom apartments provided when x1 floors of plan A are constructed.
3 4 5 7 + x 4 + x 3 b. x1 2 3 8 8 9 3 4 5 66 7 + x 4 + x 3 = 74 c. [M] Solve x1 2 3 8 8 9 136 3 7 8
4 4 8
5 3 9
66 1 ~ 0 74 0 136
0 1 0
-1/ 2 13/ 8 0
2 15 0
x1 x2
- +
(1/ 2) x3 (13/ 8) x3 0
= = =
2 15 0
The general solution is
x1 2 + (1/ 2) x3 2 1/ 2 x = 15 - (13/ 8) x = 15 + x -13/ 8 x = 2 3 3 x3 0 1 x3
However, the only feasible solutions must have whole numbers of floors for each plan. Thus, x3 must be a multiple of 8, to avoid fractions. One solution, for x3 = 0, is to use 2 floors of plan A and 15 floors of plan B. Another solution, for x3 = 8, is to use 6 floors of plan A , 2 floors of plan B, and 8 floors of plan C. These are the only feasible solutions. A larger positive multiple of 8 for x3 makes x2 negative. A negative value for x3, of course, is not feasible either.
2.1
SOLUTIONS
Notes: The definition here of a matrix product AB gives the proper view of AB for nearly all matrix calculations. (The dual fact about the rows of A and the rows of AB is seldom needed, mainly because vectors here are usually written as columns.) I assign Exercise 13 and most of Exercises 1722 to reinforce the definition of AB. Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem, in Section 2.3. Exercises 2325 are mentioned in a footnote in Section 2.2. A class discussion of the solutions of Exercises 2325 can provide a transition to Section 2.2. Or, these exercises could be assigned after starting Section 2.2. Exercises 27 and 28 are optional, but they are mentioned in Example 4 of Section 2.4. Outer products also appear in Exercises 3134 of Section 4.6 and in the spectral decomposition of a symmetric matrix, in Section 7.1. Exercises 2933 provide good training for mathematics majors.
2 1. -2 A = (-2) 4 0 -5 -1 -4 = 2 -8 -5 -4 0 10 2 . Next, use B 2A = B + (2A): -4 0 10 2 3 = -4 -7 -5 6 3 -7
7 B - 2A = 1
1 -4 + -3 -8
The product AC is not defined because the number of columns of A does not match the number of rows 1 5 + 2 4 1 13 1 2 3 5 1 3 + 2(-1) of C. CD = -1 4 = -2 3 + 1(-1) -2 5 + 1 4 = -7 -6 . For mental computation, the -2 1 row-column rule is probably easier to use than the definition. 2 2. A + 2 B = 4 0 -5 -1 7 + 21 2 -5 -4 1 2 + 14 = -3 4 + 2 0 - 10 -5 - 8 -1 + 2 16 = 2 - 6 6 -10 -13 1 -4
The expression 3C E is not defined because 3C has 2 columns and E has only 1 column. 1 1 7 + 2 1 1(-5) + 2(-4) 1 1 + 2(-3) 9 -13 -5 1 2 7 -5 CB = 1 -4 -3 = -2 7 + 1 1 -2(-5) + 1(-4) -2 1 + 1(-3) = -13 6 -5 -2 1 The product EB is not defined because the number of columns of E does not match the number of rows of R.
83
84 CHAPTER 2 Matrix Algebra
3 3. 3 I 2 - A = 0
0 4 - 3 5
-1 3 - 4 = -2 0 - 5
0 - (-1) -1 = 3 - (-2) -5 -3 , or -6
1 5
4 (3I 2 ) A = 3( I 2 A) = 3 5 3 (3I 2 ) A = 0
9 4. A - 5I 3 = -8 -4 -1 7 1
-1 12 = -2 15 -1 3 4 + 0 = -2 0 + 3 5
0 5 0
0 4 3 5
3(-1) + 0 12 = 0 + 3(-2) 15
-1 2 1 3 -6 3 -5 35 5
-3 -6
3 5 -6 - 0 8 0
0 4 0 = -8 5 -4 -1 7 1 3 -6 8
9 (5I 3 ) A = 5( I 3 A) = 5 A = 5 -8 -4 5 (5I 3 ) A = 0 0 0 5 0 0 9 0 -8 5 -4 -1 7 1
3 45 -6 = -40 8 -20
15 -30 , or 40
59 + 0 + 0 = 0 + 5(-8) + 0 0 + 0 + 5(-4) -1 5 5. a. Ab 1 = 2 AB = [ Ab1 -1 b. 5 2
5(-1) + 0 + 0 0 + 57 + 0 0 + 0 + 5 1 -1 5 Ab 2 = 2
5 3 + 0 + 0 45 0 + 5(-6) + 0 = -45 0 + 0 + 5 8 -20 2 4 -2 = -6 4 1 -3 -7
-5 35 5
15 -30 40
2 -7 3 = 7 , 4 -2 -3 12 -7 Ab 2 ] = 7 12 4 -6 -7
2 3 4 -2 -3
-1 3 + 2(-2) -2 = 5 3 + 4(-2) 1 2 3 - 3(-2)
-1( -2) + 2 1 -7 5(-2) + 4 1 = 7 2(-2) - 3 1 12 -2 14 3 = -9 0 -1 5 4
4 -6 -7
4 6. a. Ab1 = -3 3 AB = [ Ab1 4 b. -3 3
-2 0 1 = -3 , 0 2 5 13 0 Ab 2 ] = -3 13 14 -9 4
4 Ab 2 = -3 3
-2 1 0 2 5
4 1 - 2 2 3 = -3 1 + 0 2 -1 3 1 + 5 2
4 3 - 2(-1) 0 -3 3 + 0(-1) = -3 3 3 + 5(-1) 13
14 -9 4
2.1
Solutions
85
7. Since A has 3 columns, B must match with 3 rows. Otherwise, AB is undefined. Since AB has 7 columns, so does B. Thus, B is 37. 8. The number of rows of B matches the number of rows of BC, so B has 3 rows. 2 9. AB = -3 2 10. AB = -4 1 11. AD = 1 1 2 DA = 0 0 5 4 1 3 -3 8 6 5 1 2 4 0 3 0 1 2 3 0 5 0 0 1 0 1 5 1 -5 23 = k -9 4 1 = 5 -2 0 3 0 1 2 4 -10 + 5k 4 , while BA = 3 15 + k -7 2 , AC = 14 -4 3 6 12 2 6 20 5 15 25 2 9 25 -3 5 6 3 -5 2 k -3 -2 1 = 1 -2 5 23 = 1 6 - 3k -7 14 15 . 15 + k
Then AB = BA if and only if 10 + 5k = 15 and 9 = 6 3k, which happens if and only if k = 5.
0 2 0 = 2 5 2 1 2 3 = 3 5 5
Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D. To make AB = BA, one can take B to be a multiple of I3. For instance, if B = 4I3, then AB and BA are both the same as 4A. 12. Consider B = [b1 b2]. To make AB = 0, one needs Ab1 = 0 and Ab2 = 0. By inspection of A, a suitable 2 6 2 2 b1 is , or any multiple of . Example: B = . 1 3 1 1 13. Use the definition of AB written in reverse order: [Ab1 Abp] = A[b1 bp]. Thus [Qr1 Qrp] = QR, when R = [r1 rp]. 14. By definition, UQ = U[q1 q4] = [Uq1 Uq4]. From Example 6 of Section 1.8, the vector Uq1 lists the total costs (material, labor, and overhead) corresponding to the amounts of products B and C specified in the vector q1. That is, the first column of UQ lists the total costs for materials, labor, and overhead used to manufacture products B and C during the first quarter of the year. Columns 2, 3, and 4 of UQ list the total amounts spent to manufacture B and C during the 2nd, 3rd, and 4th quarters, respectively. 15. a. False. See the definition of AB. b. False. The roles of A and B should be reversed in the second half of the statement. See the box after Example 3. c. True. See Theorem 2(b), read right to left. d. True. See Theorem 3(b), read right to left. e. False. The phrase "in the same order" should be "in the reverse order." See the box after Theorem 3. 16. a. False. AB must be a 33 matrix, but the formula for AB implies that it is 31. The plus signs should be just spaces (between columns). This is a common mistake. b. True. See the box after Example 6. c. False. The left-to-right order of B and C cannot be changed, in general.
86 CHAPTER 2 Matrix Algebra
d. False. See Theorem 3(d). e. True. This general statement follows from Theorem 3(b). 2 -1 -1 = AB = [ Ab1 Ab 2 Ab3 ] , the first column of B satisfies the equation 17. Since 3 6 -9 7 -1 1 -2 -1 1 0 7 Ax = . Row reduction: [ A Ab1 ] ~ ~ 0 1 4 . So b1 = 4 . Similarly, 5 6 6 -2 2 1 0 - 8 1 -2 -8 [ A Ab 2 ] ~ ~ 0 1 -5 and b2 = -5 . 5 -9 -2
Note: An alternative solution of Exercise 17 is to row reduce [A Ab1 Ab2] with one sequence of row operations. This observation can prepare the way for the inversion algorithm in Section 2.2.
18. The first two columns of AB are Ab1 and Ab2. They are equal since b1 and b2 are equal. 19. (A solution is in the text). Write B = [b1 b2 b3]. By definition, the third column of AB is Ab3. By hypothesis, b3 = b1 + b2. So Ab3 = A(b1 + b2) = Ab1 + Ab2, by a property of matrix-vector multiplication. Thus, the third column of AB is the sum of the first two columns of AB. 20. The second column of AB is also all zeros because Ab2 = A0 = 0. 21. Let bp be the last column of B. By hypothesis, the last column of AB is zero. Thus, Abp = 0. However, bp is not the zero vector, because B has no column of zeros. Thus, the equation Abp = 0 is a linear dependence relation among the columns of A, and so the columns of A are linearly dependent.
Note: The text answer for Exercise 21 is, "The columns of A are linearly dependent. Why?" The Study Guide
supplies the argument above, in case a student needs help.
22. If the columns of B are linearly dependent, then there exists a nonzero vector x such that Bx = 0. From this, A(Bx) = A0 and (AB)x = 0 (by associativity). Since x is nonzero, the columns of AB must be linearly dependent. 23. If x satisfies Ax = 0, then CAx = C0 = 0 and so Inx = 0 and x = 0. This shows that the equation Ax = 0 has no free variables. So every variable is a basic variable and every column of A is a pivot column. (A variation of this argument could be made using linear independence and Exercise 30 in Section 1.7.) Since each pivot is in a different row, A must have at least as many rows as columns. 24. Take any b in Rm. By hypothesis, ADb = Imb = b. Rewrite this equation as A(Db) = b. Thus, the vector x = Db satisfies Ax = b. This proves that the equation Ax = b has a solution for each b in Rm. By Theorem 4 in Section 1.4, A has a pivot position in each row. Since each pivot is in a different column, A must have at least as many columns as rows. 25. By Exercise 23, the equation CA = In implies that (number of rows in A) > (number of columns), that is, m > n. By Exercise 24, the equation AD = Im implies that (number of rows in A) < (number of columns), that is, m < n. Thus m = n. To prove the second statement, observe that DAC = (DA)C = InC = C, and also DAC = D(AC) = DIm = D. Thus C = D. A shorter calculation is C = InC = (DA)C = D(AC) = DIn = D 26. Write I3 =[e1 e2 e3] and D = [d1 d2 d3]. By definition of AD, the equation AD = I3 is equivalent |to the three equations Ad1 = e1, Ad2 = e2, and Ad3 = e3. Each of these equations has at least one solution because the columns of A span R3. (See Theorem 4 in Section 1.4.) Select one solution of each equation and use them for the columns of D. Then AD = I3.
2.1
Solutions
87
27. The product uTv is a 11 matrix, which usually is identified with a real number and is written without the matrix brackets. uT v = [ -2
3 a -4] b = -2a + 3b - 4c , vT u = [ a c b -2a c ] = 3a -4a -2a -4] = -2b -2c
-2b
b
-2 c ] 3 = -2a + 3b - 4c -4
-2 uv = 3 [ a -4
T
3b -4b 3a 3b 3c
-2c 3c -4c -4a -4b -4c
a vu = b [ -2 c
T
3
28. Since the inner product uTv is a real number, it equals its transpose. That is, uTv = (uTv)T = vT (uT)T = vTu, by Theorem 3(d) regarding the transpose of a product of matrices and by Theorem 3(a). The outer product uvT is an nn matrix. By Theorem 3, (uvT)T = (vT)TuT = vuT. 29. The (i, j)-entry of A(B + C) equals the (i, j)-entry of AB + AC, because
aik (bkj + ckj ) = aik bkj + aik ckj
k =1 k =1 k =1
n
n
n
The (i, j)-entry of (B + C)A equals the (i, j)-entry of BA + CA, because
(bik + cik )akj = bik akj + cik akj
k =1 k =1 k =1
n
n
n
30. The (i, j))-entries of r(AB), (rA)B, and A(rB) are all equal, because
r aik bkj = (raik )bkj = aik (rbkj )
k =1 k =1 k =1 n n n
31. Use the definition of the product ImA and the fact that Imx = x for x in Rm. ImA = Im[a1 an] = [Ima1 Iman] = [a1 an] = A 32. Let ej and aj denote the jth columns of In and A, respectively. By definition, the jth column of AIn is Aej, which is simply aj because ej has 1 in the jth position and zeros elsewhere. Thus corresponding columns of AIn and A are equal. Hence AIn = A. 33. The (i, j)-entry of (AB)T is the ( j, i)-entry of AB, which is a j1b1i + + a jn bni
The entries in row i of BT are b1i, ... , bni, because they come from column i of B. Likewise, the entries in column j of AT are aj1, ..., ajn, because they come from row j of A. Thus the (i, j)-entry in BTAT is a j1b1i + " + a jnbni , as above.
34. Use Theorem 3(d), treating x as an n1 matrix: (ABx)T = xT(AB)T = xTBTAT. 35. [M ] The answer here depends on the choice of matrix program. For MATLAB, use the help command to read about zeros, ones, eye, and diag. For other programs see the appendices in the Study Guide. (The TI calculators have fewer single commands that produce special matrices.)
88 CHAPTER 2 Matrix Algebra
36. [M] The answer depends on the choice of matrix program. In MATLAB, the command rand(6,4) creates a 64 matrix with random entries uniformly distributed between 0 and 1. The command round(19*(rand(6,4).5))
creates a random 64 matrix with integer entries between 9 and 9. The same result is produced by the command randomint in the Laydata Toolbox on text website. For other matrix programs see the appendices in the Study Guide.
37. [M] (A + I)(A I) (A2 I) = 0 for all 44 matrices. However, (A + B)(A B) A2 B2 is the zero matrix only in the special cases when AB = BA. In general, (A + B)(A B) = A(A B) + B(A B) = AA AB + BA BB. 38. [M] The equality (AB)T = ATBT is very likely to be false for 44 matrices selected at random. 39. [M] The matrix S "shifts" the entries in a vector (a, b, c, d, e) to yield (b, c, d, e, 0). The entries in S2 result from applying S to the columns of S, and similarly for S 3 , and so on. This explains the patterns of entries in the powers of S:
0 0 S 2 = 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 , S 3 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 , S 4 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
S 5 is the 55 zero matrix. S 6 is also the 55 zero matrix. .3318 40. [M] A5 = .3346 .3336 .3346 .3323 .3331 .3336 .333337 .3331 , A10 = .333330 .333333 .3333 .333330 .333336 .333334 .333333 .333334 .333333
The entries in A20 all agree with .3333333333 to 9 or 10 decimal places. The entries in A30 all agree with .33333333333333 to at least 14 decimal places. The matrices appear to approach the matrix 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 . Further exploration of this behavior appears in Sections 4.9 and 5.2. 1/ 3 1/ 3 1/ 3
Note: The MATLAB box in the Study Guide introduces basic matrix notation and operations, including the commands that create special matrices needed in Exercises 35, 36 and elsewhere. The Study Guide appendices treat the corresponding information for the other matrix programs.
2.2
SOLUTIONS
Notes: The text includes the matrix inversion algorithm at the end of the section because this topic is popular. Students like it because it is a simple mechanical procedure. However, I no longer cover it in my classes because technology is readily available to invert a matrix whenever needed, and class time is better spent on more useful topics such as partitioned matrices. The final subsection is independent of the inversion algorithm and is needed for Exercises 35 and 36. Key Exercises: 8, 1124, 35. (Actually, Exercise 8 is only helpful for some exercises in this section. Section 2.3 has a stronger result.) Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem (IMT) in Section 2.3, along with Exercises 23 and 24 in Section 2.1. I recommend letting students work on two or more of these four exercises before proceeding to Section 2.3. In this way students participate in the
2.2
Solutions
89
proof of the IMT rather than simply watch an instructor carry out the proof. Also, this activity will help students understand why the theorem is true.
8 1. 5 3 2. 7 8 3. -7 3 4. 7
6 4
-1
=
-1
1 4 32 - 30 -5 1 4 12 - 14 -7 =
-6 2 = 8 -5 / 2 -2 -2 = 3 7 / 2
-3 4 1 -3/ 2 -5 1 or -1.4 8 1 -1.6
2 4
=
-1
5 -5 -4 -8
-5 1 -40 - (-35) 7
-5 1 -5 = -5 7 8 4 1 -8 = 3 4 -7
-1
=
-8 1 -24 - ( -28) -7
4 -2 or -7 / 4 3
1 3/ 4
8 5. The system is equivalent to Ax = b, where A = 5 2 x = A1b = -5 / 2
6 2 and b = -1 , and the solution is 4
-3 2 7 = . Thus x1 = 7 and x2 = 9. 4 -1 -9
5 8 9 1 6. The system is equivalent to Ax = b, where A = and b = 11 , and the solution is x = A b. To -7 -5 compute this by hand, the arithmetic is simplified by keeping the fraction 1/det(A) in front of the matrix for A1. (The Study Guide comments on this in its discussion of Exercise 7.) From Exercise 3, 1 -5 x = A1b = - 5 7 1 2 7. a. 5 12
-1
-5 -9 1 -10 2 =- = . Thus x1 = 2 and x2 = 5. 8 11 5 25 -5 -2 1 12 = 1 2 -5 -2 6 or -2.5 1 -1 .5
=
12 1 1 12 - 2 5 -5
x = A1b1 =
1 12 2 -5
-2 -1 1 -18 -9 = = . Similar calculations give 1 3 2 8 4
11 6 13 A-1b 2 = , A-1b3 = , A-1b 4 = . -5 -2 -5 1 2 b. [A b1 b2 b3 b4] = 5 12 1 ~ 0 1 ~ 0 2 2 0 1 -1 8 -9 4 1 -10 11 -5 2 -4 6 -2 -1 3 1 -5 2 6 2 1 3 5 -1 4 1 -5 2 -2 3 -5
3 1 ~ -10 0 13 -5
-9 11 6 13 The solutions are , , , and , the same as in part (a). 4 -5 -2 -5
90 CHAPTER 2 Matrix Algebra
Note: The Study Guide also discusses the number of arithmetic calculations for this Exercise 7, stating that
when A is large, the method used in (b) is much faster than using A1.
8. Left-multiply each side of the equation AD = I by A1 to obtain A1AD = A1I, ID = A1, and D = A1. Parentheses are routinely suppressed because of the associative property of matrix multiplication. 9. a. True, by definition of invertible. b. False. See Theorem 6(b).
1 1 c. False. If A = , then ab cd = 1 0 0, but Theorem 4 shows that this matrix is not invertible, 0 0 because ad bc = 0. d. True. This follows from Theorem 5, which also says that the solution of Ax = b is unique, for each b. e. True, by the box just before Example 6.
10. a. False. The product matrix is invertible, but the product of inverses should be in the reverse order. See Theorem 6(b). b. True, by Theorem 6(a). c. True, by Theorem 4. d. True, by Theorem 7. e. False. The last part of Theorem 7 is misstated here. 11. (The proof can be modeled after the proof of Theorem 5.) The np matrix B is given (but is arbitrary). Since A is invertible, the matrix A1B satisfies AX = B, because A(A1B) = A A1B = IB = B. To show this solution is unique, let X be any solution of AX = B. Then, left-multiplication of each side by A1 shows that X must be A1B: A1 (AX) = A1B, IX = A1B, and X = A1B. 12. If you assign this exercise, consider giving the following Hint: Use elementary matrices and imitate the proof of Theorem 7. The solution in the Instructor's Edition follows this hint. Here is another solution, based on the idea at the end of Section 2.2. Write B = [b1 bp] and X = [u1 up]. By definition of matrix multiplication, AX = [Au1 Aup]. Thus, the equation AX = B is equivalent to the p systems: Au1 = b1, ... Aup = bp Since A is the coefficient matrix in each system, these systems may be solved simultaneously, placing the augmented columns of these systems next to A to form [A b1 bp] = [A B]. Since A is invertible, the solutions u1, ..., up are uniquely determined, and [A b1 bp] must row reduce to [I u1 up] = [I X]. By Exercise 11, X is the unique solution A1B of AX = B. 13. Left-multiply each side of the equation AB = AC by A1 to obtain A1AB = A1AC, IB = IC, and B = C. This conclusion does not always follow when A is singular. Exercise 10 of Section 2.1 provides a counterexample. 14. Right-multiply each side of the equation (B C ) D = 0 by D1 to obtain (B C)DD1 = 0D1, (B C)I = 0, B C = 0, and B = C. 15. The box following Theorem 6 suggests what the inverse of ABC should be, namely, C1B1A1. To verify that this is correct, compute: (ABC) C1B1A1 = ABCC1B1A1 = ABIB1A1 = ABB1A1 = AIA1 = AA1 = I and C1B1A1 (ABC) = C1B1A1ABC = C1B1IBC = C1B1BC = C1IC = C1C = I
2.2
Solutions
91
16. Let C = AB. Then CB1 = ABB1, so CB1 = AI = A. This shows that A is the product of invertible matrices and hence is invertible, by Theorem 6.
Note: The Study Guide warns against using the formula (AB) 1 = B1A1 here, because this formula can be
used only when both A and B are already known to be invertible.
17. Right-multiply each side of AB = BC by B1: ABB1 = BCB1, AI = BCB1, A = BCB1. 18. Left-multiply each side of A = PBP1 by P1: P1A = P1PBP1, P1A = IBP1, P1A = BP1 Then right-multiply each side of the result by P: P1AP = BP1P, P1AP = BI, P1AP = B 19. Unlike Exercise 17, this exercise asks two things, "Does a solution exist and what is it?" First, find what the solution must be, if it exists. That is, suppose X satisfies the equation C1(A + X)B1 = I. Left-multiply each side by C, and then right-multiply each side by B: CC1(A + X)B1 = CI, I(A + X)B1 = C, (A + X)B1B = CB, (A + X)I = CB Expand the left side and then subtract A from both sides: AI + XI = CB, A + X = CB, X = CB A If a solution exists, it must be CB A. To show that CB A really is a solution, substitute it for X: C1[A + (CB A)]B1 = C1[CB]B1 = C1CBB1 = II = I.
proofs. After some practice with algebra, an expression such as CC1(A + X)B1 could be simplified directly to (A + X)B1 without first replacing CC1 by I. However, you may wish this detail to be included in the homework for this section.
Note: The Study Guide suggests that students ask their instructor about how many details to include in their
20. a. Left-multiply both sides of (A AX)1 = X1B by X to see that B is invertible because it is the product of invertible matrices. b. Invert both sides of the original equation and use Theorem 6 about the inverse of a product (which applies because X1 and B are invertible): A AX = (X1B)1 = B1(X1)1 = B1X Then A = AX + B1X = (A + B1)X. The product (A + B1)X is invertible because A is invertible. Since X is known to be invertible, so is the other factor, A + B1, by Exercise 16 or by an argument similar to part (a). Finally, (A + B1)1A = (A + B1)1(A + B1)X = X
Note: This exercise is difficult. The algebra is not trivial, and at this point in the course, most students will
not recognize the need to verify that a matrix is invertible.
21. Suppose A is invertible. By Theorem 5, the equation Ax = 0 has only one solution, namely, the zero solution. This means that the columns of A are linearly independent, by a remark in Section 1.7. 22. Suppose A is invertible. By Theorem 5, the equation Ax = b has a solution (in fact, a unique solution) for each b. By Theorem 4 in Section 1.4, the columns of A span Rn. 23. Suppose A is nn and the equation Ax = 0 has only the trivial solution. Then there are no free variables in this equation, and so A has n pivot columns. Since A is square and the n pivot positions must be in different rows, the pivots in an echelon form of A must be on the main diagonal. Hence A is row equivalent to the nn identity matrix.
92 CHAPTER 2 Matrix Algebra
24. If the equation Ax = b has a solution for each b in Rn, then A has a pivot position in each row, by Theorem 4 in Section 1.4. Since A is square, the pivots must be on the diagonal of A. It follows that A is row equivalent to In. By Theorem 7, A is invertible.
0 0 x1 0 a b 25. Suppose A = and ad bc = 0. If a = b = 0, then examine c d x = 0 This has the c d 2 d solution x1 = . This solution is nonzero, except when a = b = c = d. In that case, however, A is the -c -b zero matrix, and Ax = 0 for every vector x. Finally, if a and b are not both zero, set x2 = . Then a a b -b - ab + ba 0 Ax 2 = = = , because cb + da = 0. Thus, x2 is a nontrivial solution of Ax = 0. c d a - cb + da 0 So, in all cases, the equation Ax = 0 has more than one solution. This is impossible when A is invertible (by Theorem 5), so A is not invertible. d 26. -c -b a a c a c
b da - bc = d 0
0 . Divide both sides by ad bc to get CA = I. -cb + ad 0 . -cb + da
b 1 d d ad - bc -c
b d d -c
-b ad - bc = a 0
b d d -c
Divide both sides by ad bc. The right side is I. The left side is AC, because 1 a ad - bc c -b a = a c -b = AC a
27. a. Interchange A and B in equation (1) after Example 6 in Section 2.1: rowi (BA) = rowi (B)A. Then replace B by the identity matrix: rowi (A) = rowi (IA) = rowi (I)A. b. Using part (a), when rows 1 and 2 of A are interchanged, write the result as
row 2 ( A) row 2 ( I ) A row 2 ( I ) row ( A) = row ( I ) A = row ( I ) A = EA 1 1 1 row 3 ( A) row 3 ( I ) A row 3 ( I ) (*)
Here, E is obtained by interchanging rows 1 and 2 of I. The second equality in (*) is a consequence of the fact that rowi (EA) = rowi (E)A.
c. Using part (a), when row 3 of A is multiplied by 5, write the result as
row1 ( A) row1 ( I ) A row1 ( I ) row ( A) = row ( I ) A = row ( I ) A = EA 2 2 2 5 row 3 ( A) 5 row 3 ( I ) A 5 row 3 ( I ) Here, E is obtained by multiplying row 3 of I by 5.
28. When row 3 of A is replaced by row3(A) 4row1(A), write the result as
row1 ( A) row1 ( I ) A = row 2 ( A) row 2 ( I ) A row 3 ( A) - 4 row1 ( A) row 3 ( I ) A - 4 row1 ( I ) A
2.2
Solutions
93
row1 ( I ) A row1 ( I ) = A = EA = row 2 ( I ) A row 2 ( I ) [row 3 ( I ) - 4 row1 ( I )] A row 3 ( I ) - 4 row1 ( I ) Here, E is obtained by replacing row3(I) by row3(I) 4row1(I ). 1 29. [ A I ] = 4 2 7 1 0 2 -1 1 0 0 1 ~ 1 4 0 1 ~ -1 0
-2 4 4 1 3 7 8 10 7/2 1 3 -4 1 -1 0 0 1 0 0 1 -4 10 1 0 0 0 1 3 3 4 3/ 2 1 0 0 0 1 -2 0 1 0 0 1 0
0 1 ~ 1 0
2 -1
1 -4
0 1 ~ 1 0
2 1
1 4
0 1 ~ -1 0
0 1
-7 4
2 -1
-7 A1 = 4
30. [ A
5 I] = 4 1 ~ 0 2 1
10 7 1/ 5 4/5
0 1 -3 -2 -2 2 0 0 1 -2 -7 6 1 0
2 7 0 1
1/ 5 0 -7 / 5 4/5
0 1 ~ 1 0
2 -1
1/ 5 -4 / 5
0 1 2 -1
0 0 1
2 -7 / 5 -1 . A = 4 / 5 -1
0 1 -3 -2 -2 8 8 10 7 1 3 -2 3 4 3 3 4 3/ 2 1 -1 -2 1 -4 2 0 1 0 1 1 1 1 1 1/ 2 0 1 0
31. [ A
1 I ] = -3 2 1 ~ 0 0 1 ~ 0 0 0 1 0 0 1 0
0 1 0 ~ 0 1 0 0 1 0
0 1 0 ~ 0 1 0
0 0 2
1 8 . A-1 = 10 1 7 / 2 1/ 2 0 1 0 ~ 0 1 0 -2 1 2
32. [ A
1 I] = 4 -2 1 ~ 0 0 -2
0 0 1
0 0 . The matrix A is not invertible. 1
1 -1 33. Let B = 0 # 0
0 0 , and for j = 1, ..., n, let aj, bj, and ej denote the jth columns of A, B, -1 1 % % # 0 " -1 1 and I, respectively. Note that for j = 1, ..., n 1, aj aj+1 = ej (because aj and aj+1 have the same entries except for the jth row), bj = ej ej+1 and an = bn = en. To show that AB = I, it suffices to show that Abj = ej for each j. For j = 1, ..., n 1, Abj = A(ej ej+1) = Aej Aej+1 = aj aj+1 = ej
"
94 CHAPTER 2 Matrix Algebra
and Abn = Aen = an = en. Next, observe that aj = ej + + en for each j. Thus, Baj = B(ej + + en) = bj + + bn = (ej ej+1) + (ej+1 ej+2) + + (en1 en) + en = ej This proves that BA = I. Combined with the first part, this proves that B = A1.
Note: Students who do this problem and then do the corresponding exercise in Section 2.4 will appreciate the Invertible Matrix Theorem, partitioned matrix notation, and the power of a proof by induction.
34. Let
1 1 A = 1 # 1 0 2 2 2 0 0 3 3
"
% "
0 1 -1/ 2 0 0 , and B = 0 # # 0 n
0 1/ 2 -1/ 3 0
0 0 1/ 3 %
"
%
-1/ n
0 # 1/ n
and for j = 1, ..., n, let aj, bj, and ej denote the jth columns of A, B, and I, respectively. Note that for 1 1 1 j = 1, ..., n1, aj = j(ej + + en), bj = e j - e j +1 , and b n = e n . j j +1 n
To show that AB = I, it suffices to show that Abj = ej for each j. For j = 1, ..., n1,
1 1 1 1 e j +1 = a j - a j +1 Abj = A e j - j j +1 j +1 j
= (ej + + en) (ej+1 + + en) = ej
1 1 Also, Abn = A e n = a n = e n . Finally, for j = 1, ..., n, the sum bj + + bn is a "telescoping sum" n n 1 whose value is e j . Thus, j 1 Baj = j(Bej + + Ben) = j(bj + + bn) = j e j = e j j which proves that BA = I. Combined with the first part, this proves that B = A1.
Note: If you assign Exercise 34, you may wish to supply a hint using the notation from Exercise 33: Express
each column of A in terms of the columns e1, ..., en of the identity matrix. Do the same for B. 35. Row reduce [A e3]: -2 2 1 1 ~ 0 0 -7 5 3 3 -1 0 -9 6 4 0 0 1 0 1 0 ~ 2 1 -2 -15 1 6 ~ 0 4 0 3 5 -7 3 1 0
1
4 6 -9 0 0 1
1 1 0 ~ 0 0 0 -15 1 -6 ~ 0 4 0
3 -1 -1 0 1 0
4 -2 -1 0 0 1
1 1 -2 ~ 0 2 0 3 -6 . 4
3 -1 0
4 -2 1
1 -2 4
3 Answer: The third column of A is -6 . 4
2.2
Solutions
95
36. [M] Write B = [A F], where F consists of the last two columns of I3, and row reduce: -25 B = 546 154 -9 180 50 -27 537 149 0 1 0
1
0 1 0 ~ 0 1 0
0 1 0
0 0 1
3/ 2 -433/ 6 68 / 3
-9 / 2 439 / 2 -69
1.5000 The last two columns of A are -72.1667 22.6667
-4.5000 219.5000 -69.0000
1 1 -1 37. There are many possibilities for C, but C = is the only one whose entries are 1, 1, and 0. 0 -1 1 With only three possibilities for each entry, the construction of C can be done by trial and error. This is probably faster than setting up a system of 4 equations in 6 unknowns. The fact that A cannot be invertible follows from Exercise 25 in Section 2.1, because A is not square.
0 0 . 38. Write AD = A[d1 d2] = [Ad1 0 1 [There are 25 possibilities for D if entries of D are allowed to be 1, 1, and 0.] There is no 42 matrix C such that CA = I4. If this were true, then CAx would equal x for all x in R4. This cannot happen because the columns of A are linearly dependent and so Ax = 0 for some nonzero vector x. For such an x, CAx = C(0) = 0. An alternate justification would be to cite Exercise 23 or 25 in Section 2.1. .005 .002 39. y = Df = .002 .004 .001 .002 and 3, respectively. .001 30 .27 .002 50 = .30 . The deflections are .27 in., .30 in., and .23 in. at points 1, 2, .005 20 .23
1 0 Ad2]. The structure of A shows that D = 0 0
40. [M] The stiffness matrix is D1. Use an "inverse" command to produce
0 2 -1 -1 D = 125 3 -1 0 -1 2 To find the forces (in pounds) required to produce a deflection of .04 cm at point 3, most students will use technology to solve Df = (0, 0, .04) and obtain (0, 5, 10). Here is another method, based on the idea suggested in Exercise 42. The first column of D1 lists the forces required to produce a deflection of 1 in. at point 1 (with zero deflection at the other points). Since the transformation y 6 D1y is linear, the forces required to produce a deflection of .04 cm at point 3 is
1
given by .04 times the third column of D1, namely (.04)(125) times (0, 1, 2), or (0, 5, 10) pounds.
41. To determine the forces that produce a deflections of .08, .12, .16, and .12 cm at the four points on the beam, use technology to solve Df = y, where y = (.08, .12, .16, .12). The forces at the four points are 12, 1.5, 21.5, and 12 newtons, respectively.
96 CHAPTER 2 Matrix Algebra
42. [M] To determine the forces that produce a deflection of .240 cm at the second point on the beam, use technology to solve Df = y, where y = (0, .24, 0, 0). The forces at the four points are 104, 167, 113, and 56.0 newtons, respectively (to three significant digits). These forces are .24 times the entries in the
second column of D1. Reason: The transformation y 6 D -1y is linear, so the forces required to produce a deflection of .24 cm at the second point are .24 times the forces required to produce a deflection of 1 cm at the second point. These forces are listed in the second column of D1. Another possible discussion: The solution of Dx = (0, 1, 0, 0) is the second column of D1. Multiply both sides of this equation by .24 to obtain D(.24x) = (0, .24, 0, 0). So .24x is the solution of Df = (0, .24, 0, 0). (The argument uses linearity, but students may not mention this.)
Note: The Study Guide suggests using gauss, swap, bgauss, and scale to reduce [A I], because
I prefer to postpone the use of ref (or rref) until later. If you wish to introduce ref now, see the Study Guide's technology notes for Sections 2.8 or 4.3. (Recall that Sections 2.8 and 2.9 are only covered when an instructor plans to skip Chapter 4 and get quickly to eigenvalues.)
2.3
SOLUTIONS
Notes: This section ties together most of the concepts studied thus far. With strong encouragement from an instructor, most students can use this opportunity to review and reflect upon what they have learned, and form a solid foundation for future work. Students who fail to do this now usually struggle throughout the rest of the course. Section 2.3 can be used in at least three different ways. (1) Stop after Example 1 and assign exercises only from among the Practice Problems and Exercises 1 to 28. I do this when teaching "Course 3" described in the text's "Notes to the Instructor. " If you did not cover Theorem 12 in Section 1.9, omit statements (f) and (i) from the Invertible Matrix Theorem. (2) Include the subsection "Invertible Linear Transformations" in Section 2.3, if you covered Section 1.9. I do this when teaching "Course 1" because our mathematics and computer science majors take this class. Exercises 2940 support this material. (3) Skip the linear transformation material here, but discusses the condition number and the Numerical Notes. Assign exercises from among 128 and 4145, and perhaps add a computer project on the condition number. (See the projects on our web site.) I do this when teaching "Course 2" for our engineers. The abbreviation IMT (here and in the Study Guide) denotes the Invertible Matrix Theorem (Theorem 8).
7 5 1. The columns of the matrix are not multiples, so they are linearly independent. By (e) in the -3 -6 IMT, the matrix is invertible. Also, the matrix is invertible by Theorem 4 in Section 2.2 because the determinant is nonzero. 6 -4 2. The fact that the columns of are multiples is not so obvious. The fastest check in this case 6 -9 may be the determinant, which is easily seen to be zero. By Theorem 4 in Section 2.2, the matrix is not invertible.
3. Row reduction to echelon form is trivial because there is really no need for arithmetic calculations: 0 0 5 0 0 5 0 0 5 -3 -7 ~ 0 -7 ~ 0 -7 0 0 0 The 33 matrix has 3 pivot positions and hence is 8 5 -1 0 5 -1 0 0 -1
invertible, by (c) of the IMT. [Another explanation could be given using the transposed matrix. But see the note below that follows the solution of Exercise 14.]
2.3
Solutions
97
-7 4. The matrix 3 2 0 5. 1 -4 3 0 -9
0 0 0
4 -1 obviously has linearly dependent columns (because one column is zero), and 9 0 3 -9 2 1 -5 ~ 0 7 0
-4 1 4 ~ 0 -12 0
so the matrix is not invertible (or singular) by (e) in the IMT.
-5 1 2 ~ 0 7 -4 -4 1 4 ~ 0 0 0
0 3 -9
2 1 -5 ~ 0 15 0
-4 4 0
0 3 0
2 -5 0
The matrix is not invertible because it is not row equivalent to the identity matrix. 1 6. 0 -3 -1 3 7. -2 0
-5 -5 -5
3 6
3 -9
3 0
The matrix is not invertible because it is not row equivalent to the identity matrix. -3 5 -6 -1 0 8 3 2 1 -1 -3 0 ~ 2 0 1 0 -3 -4 0 -1 0 8 3 2 1 -1 0 0 ~ 0 0 1 0 -3 -4 0 0 0 8 3 0 1 0 0 1
The 44 matrix has four pivot positions and so is invertible by (c) of the IMT. 1 0 8. The 44 matrix 0 0 4 -6 9. [M] 7 -1 -1 0 ~ 0 0 -1 0 ~ 0 0 0 1 -5 2 -7 11 10 3 3 5 0 0 7 9 2 0 4 6 is invertible because it has four pivot positions, by (c) of the IMT. 8 10 2 1 -5 0 3 11 10 -7 2 8 0 0 -1 -1 9 0 ~ 19 0 -7 0 3 5 25.375 -.1250 2 8 0 0 3 5 1 0 2 -11 9 8 3 -7 31 5 -1 15 12 -11 2 8 0 0 3 5 25.375 1 -1 -11 24.375 1
-7 -1 9 -6 ~ 19 7 -1 4 3 5 31 -7
2 8 9 -11 2 8 0 0
-1 -1 -11 0 ~ 12 0 15 0
-1 -1 -11 0 ~ 24.375 0 -.1250 0 -1 -11 1 -1
3 5 1 25.375
-1 -1 -11 0 ~ 1 0 24.375 0
The 44 matrix is invertible because it has four pivot positions, by (c) of the IMT.
98 CHAPTER 2 Matrix Algebra
5 6 10. [M] 7 9 8 5 0 ~ 0 0 0
3 4 5 6 5 3 .4 0 0 0
1 2 3 4 2 1 .8 0 1 0
7 8 10 -9 11
9 5 -8 0 9 ~ 0 -5 0 4 0
3 .4 .8 .6 .2
1 .8 1.6 2.2 .4 3 .4 0 0 0 1 .8 1 0 0
7 -.4 .2 -21.6 -.2 7 -.4 -21 1 0
9 -18.8 -3.6 -21.2 -10.4 9 -18.8 7 34 -1
7 -.4 1 -21 0
9 5 -18.8 0 34 ~ 0 7 0 -1 0
The 55 matrix is invertible because it has five pivot positions, by (c) of the IMT.
11. a. b. c. d.
True, by the IMT. If statement (d) of the IMT is true, then so is statement (b). True. If statement (h) of the IMT is true, then so is statement (e). False. Statement (g) of the IMT is true only for invertible matrices. True, by the IMT. If the equation Ax = 0 has a nontrivial solution, then statement (d) of the IMT is false. In this case, all the lettered statements in the IMT are false, including statement (c), which means that A must have fewer than n pivot positions. e. True, by the IMT. If AT is not invertible, then statement (1) of the IMT is false, and hence statement (a) must also be false. True. If statement (k) of the IMT is true, then so is statement ( j). True. If statement (e) of the IMT is true, then so is statement (h). True. See the remark immediately following the proof of the IMT. False. The first part of the statement is not part (i) of the IMT. In fact, if A is any nn matrix, the linear transformation x 6 Ax maps n into n, yet not every such matrix has n pivot positions. e. True, by the IMT. If there is a b in n such that the equation Ax b = is inconsistent, then statement (g) of the IMT is false, and hence statement (f) is also false. That is, the transformation x 6 Ax cannot be one-to-one.
12. a. b. c. d.
Note: The solutions below for Exercises 1330 refer mostly to the IMT. In many cases, however, part or all of an acceptable solution could also be based on various results that were used to establish the IMT.
13. If a square upper triangular nn matrix has nonzero diagonal entries, then because it is already in echelon form, the matrix is row equivalent to In and hence is invertible, by the IMT. Conversely, if the matrix is invertible, it has n pivots on the diagonal and hence the diagonal entries are nonzero. 14. If A is lower triangular with nonzero entries on the diagonal, then these n diagonal entries can be used as pivots to produce zeros below the diagonal. Thus A has n pivots and so is invertible, by the IMT. If one of the diagonal entries in A is zero, A will have fewer than n pivots and hence be singular.
Notes: For Exercise 14, another correct analysis of the case when A has nonzero diagonal entries is to apply
the IMT (or Exercise 13) to AT. Then use Theorem 6 in Section 2.2 to conclude that since AT is invertible so is its transpose, A. You might mention this idea in class, but I recommend that you not spend much time discussing AT and problems related to it, in order to keep from making this section too lengthy. (The transpose is treated infrequently in the text until Chapter 6.) If you do plan to ask a test question that involves AT and the IMT, then you should give the students some extra homework that develops skill using AT. For instance, in Exercise 14 replace "columns" by "rows."
2.3
Solutions
99
Also, you could ask students to explain why an nn matrix with linearly independent columns must also have linearly independent rows.
15. If A has two identical columns then its columns are linearly dependent. Part (e) of the IMT shows that A cannot be invertible. 16. Part (h) of the IMT shows that a 55 matrix cannot be invertible when its columns do not span R5. 17. If A is invertible, so is A1, by Theorem 6 in Section 2.2. By (e) of the IMT applied to A1, the columns of A1 are linearly independent. 18. By (g) of the IMT, C is invertible. Hence, each equation Cx = v has a unique solution, by Theorem 5 in Section 2.2. This fact was pointed out in the paragraph following the proof of the IMT. 19. By (e) of the IMT, D is invertible. Thus the equation Dx = b has a solution for each b in R7, by (g) of the IMT. Even better, the equation Dx = b has a unique solution for each b in R7, by Theorem 5 in Section 2.2. (See the paragraph following the proof of the IMT.) 20. By the box following the IMT, E and F are invertible and are inverses. So FE = I = EF, and so E and F commute. 21. The matrix G cannot be invertible, by Theorem 5 in Section 2.2 or by the box following the IMT. So (h) of the IMT is false and the columns of G do not span Rn. 22. Statement (g) of the IMT is false for H, so statement (d) is false, too. That is, the equation Hx = 0 has a nontrivial solution. 23. Statement (b) of the IMT is false for K, so statements (e) and (h) are also false. That is, the columns of K are linearly dependent and the columns do not span Rn. 24. No conclusion about the columns of L may be drawn, because no information about L has been given. The equation Lx = 0 always has the trivial solution. 25. Suppose that A is square and AB = I. Then A is invertible, by the (k) of the IMT. Left-multiplying each side of the equation AB = I by A1, one has A1AB = A1I, IB = A1, and B = A1. By Theorem 6 in Section 2.2, the matrix B (which is A1) is invertible, and its inverse is (A1)1, which is A. 26. If the columns of A are linearly independent, then since A is square, A is invertible, by the IMT. So A2, which is the product of invertible matrices, is invertible. By the IMT, the columns of A2 span Rn. 27. Let W be the inverse of AB. Then ABW = I and A(BW) = I. Since A is square, A is invertible, by (k) of the IMT.
Note: The Study Guide for Exercise 27 emphasizes here that the equation A(BW) = I, by itself, does not show
that A is invertible. Students are referred to Exercise 38 in Section 2.2 for a counterexample. Although there is an overall assumption that matrices in this section are square, I insist that my students mention this fact when using the IMT. Even so, at the end of the course, I still sometimes find a student who thinks that an equation AB = I implies that A is invertible.
28. Let W be the inverse of AB. Then WAB = I and (WA)B = I. By (j) of the IMT applied to B in place of A, the matrix B is invertible.
100 CHAPTER 2 Matrix Algebra
29. Since the transformation x 6 Ax is not one-to-one, statement (f) of the IMT is false. Then (i) is also false and the transformation x 6 Ax does not map Rn onto Rn. Also, A is not invertible, which implies that the transformation x 6 Ax is not invertible, by Theorem 9.
30. Since the transformation x 6 Ax is one-to-one, statement (f) of the IMT is true. Then (i) is also true and the transformation x 6 Ax maps Rn onto Rn. Also, A is invertible, which implies that the transformation x 6 Ax is invertible, by Theorem 9. 31. Since the equation Ax = b has a solution for each b, the matrix A has a pivot in each row (Theorem 4 in Section 1.4). Since A is square, A has a pivot in each column, and so there are no free variables in the equation Ax = b, which shows that the solution is unique.
Note: The preceding argument shows that the (square) shape of A plays a crucial role. A less revealing proof
is to use the "pivot in each row" and the IMT to conclude that A is invertible. Then Theorem 5 in Section 2.2 shows that the solution of Ax = b is unique.
32. If Ax = 0 has only the trivial solution, then A must have a pivot in each of its n columns. Since A is square (and this is the key point), there must be a pivot in each row of A. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for each b in Rn. Another argument: Statement (d) of the IMT is true, so A is invertible. By Theorem 5 in Section 2.2, the equation Ax = b has a (unique) solution for each b in Rn.
9 -5 33. (Solution in Study Guide) The standard matrix of T is A = , which is invertible because 4 -7 det A 0. By Theorem 9, the transformation T is invertible and the standard matrix of T1 is A1. From 7 9 the formula for a 22 inverse, A-1 = . So 4 5 7 T -1 ( x1 , x2 ) = 4 9 x1 = ( 7 x1 + 9 x2 , 4 x1 + 5 x2 ) 5 x2 -8 , which is invertible because det A = 2 0. By Theorem 9, 7
6 34. The standard matrix of T is A = -5
T is invertible, and T -1 (x) = Bx, where B = A-1 = T -1 ( x1 , x2 ) =
1 7 2 5
1 7 2 5
8 . Thus 6
8 x1 7 5 x = 2 x1 + 4 x2 , 2 x1 + 3x2 6 2
35. (Solution in Study Guide) To show that T is one-to-one, suppose that T(u) = T(v) for some vectors u and v in Rn. Then S(T(u)) = S(T(v)), where S is the inverse of T. By Equation (1), u = S(T(u)) and S(T(v)) = v, so u = v. Thus T is one-to-one. To show that T is onto, suppose y represents an arbitrary vector in Rn and define x = S(y). Then, using Equation (2), T(x) = T(S(y)) = y, which shows that T maps Rn onto Rn. Second proof: By Theorem 9, the standard matrix A of T is invertible. By the IMT, the columns of A are linearly independent and span Rn. By Theorem 12 in Section 1.9, T is one-to-one and maps Rn onto Rn. 36. If T maps Rn onto Rn, then the columns of its standard matrix A span Rn, by Theorem 12 in Section 1.9. By the IMT, A is invertible. Hence, by Theorem 9 in Section 2.3, T is invertible, and A1 is the standard matrix of T1. Since A1 is also invertible, by the IMT, its columns are linearly independent and span Rn. Applying Theorem 12 in Section 1.9 to the transformation T1, we conclude that T1 is a one-to-one mapping of Rn onto Rn.
2.3
Solutions
101
37. Let A and B be the standard matrices of T and U, respectively. Then AB is the standard matrix of the mapping x 6 T (U (x)) , because of the way matrix multiplication is defined (in Section 2.1). By hypothesis, this mapping is the identity mapping, so AB = I. Since A and B are square, they are invertible, by the IMT, and B = A1. Thus, BA = I. This means that the mapping x 6 U (T ( x)) is the identity mapping, i.e., U(T(x)) = x for all x in Rn. 38. Let A be the standard matrix of T. By hypothesis, T is not a one-to-one mapping. So, by Theorem 12 in Section 1.9, the standard matrix A of T has linearly dependent columns. Since A is square, the columns of A do not span Rn. By Theorem 12, again, T cannot map Rn onto Rn. 39. Given any v in Rn, we may write v = T(x) for some x, because T is an onto mapping. Then, the assumed properties of S and U show that S(v) = S(T(x)) = x and U(v) = U(T(x)) = x. So S(v) and U(v) are equal for each v. That is, S and U are the same function from Rn into Rn. 40. Given u, v in n, let x = S(u) and y = S(v). Then T(x)=T(S(u)) = u and T(y) = T(S(v)) = v, by equation (2). Hence S (u + v) = S (T (x) + T (y )) = S (T (x + y )) Because T is linear =x+y By equation (1) = S (u) + S ( v )
So, S preserves sums. For any scalar r, Because T is linear S (r u) = S (rT (x)) = S (T (r x)) = rx = rS (u) So S preserves scalar multiples. Thus S ia a linear transformation. 41. [M] a. The exact solution of (3) is x1 = 3.94 and x2 = .49. The exact solution of (4) is x1 = 2.90 and x2 = 2.00. b. When the solution of (4) is used as an approximation for the solution in (3) , the error in using the value of 2.90 for x1 is about 26%, and the error in using 2.0 for x2 is about 308%. c. The condition number of the coefficient matrix is 3363. The percentage change in the solution from (3) to (4) is about 7700 times the percentage change in the right side of the equation. This is the same order of magnitude as the condition number. The condition number gives a rough measure of how sensitive the solution of Ax = b can be to changes in b. Further information about the condition number is given at the end of Chapter 6 and in Chapter 7. By equation (1)
Note: See the Study Guide's MATLAB box, or a technology appendix, for information on condition number.
Only the TI-83+ and TI-89 lack a command for this. 42. [M] MATLAB gives cond(A) = 23683, which is approximately 104. If you make several trials with MATLAB, which records 16 digits accurately, you should find that x and x1 agree to at least 12 or 13 significant digits. So about 4 significant digits are lost. Here is the result of one experiment. The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places:
.9501 .21311 , b = Ax = x = rand(4,1) = .6068 .4860
-3.8493 5.5795 . The MATLAB solution is x1 = A\b = 20.7973 .8467
.9501 .2311 . .6068 .4860
102 CHAPTER 2 Matrix Algebra
.0171 .4858 1012. The computed solution x1 is accurate to about However, x x1 = -.2360 .2456 12 decimal places. 43. [M] MATLAB gives cond(A) = 68,622. Since this has magnitude between 104 and 105, the estimated accuracy of a solution of Ax = b should be to about four or five decimal places less than the 16 decimal places that MATLAB usually computes accurately. That is, one should expect the solution to be accurate to only about 11 or 12 decimal places. Here is the result of one experiment. The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places: .2190 .0470 x = rand(5,1) = .6789 , b = Ax = .6793 .9347 15.0821 .8165 19.0097 . The MATLAB solution is x1 = A\b = -5.8188 14.5557 .2190 .0470 .6789 . .6793 .9347
.3165 -.6743 However, x x1 = .3343 10-11 . The computed solution x1 is accurate to about 11 decimal places. .0158 -.0005 44. [M] Solve Ax = (0, 0, 0, 0, 1). MATLAB shows that cond( A) 4.8 105. Since MATLAB computes numbers accurately to 16 decimal places, the entries in the computed value of x should be accurate to at least 11 digits. The exact solution is (630, 12600, 56700, 88200, 44100). 45. [M] Some versions of MATLAB issue a warning when asked to invert a Hilbert matrix of order 12 or larger using floating-point arithmetic. The product AA1 should have several off-diagonal entries that are far from being zero. If not, try a larger matrix.
Note: All matrix programs supported by the Study Guide have data for Exercise 45, but only MATLAB and Maple have a single command to create a Hilbert matrix. The HP-48G data for Exercise 45 contain a program that can be edited to create other Hilbert matrices. Notes: The Study Guide for Section 2.3 organizes the statements of the Invertible Matrix Theorem in a table that imbeds these ideas in a broader discussion of rectangular matrices. The statements are arranged in three columns: statements that are logically equivalent for any mn matrix and are related to existence concepts, those that are equivalent only for any nn matrix, and those that are equivalent for any np matrix and are related to uniqueness concepts. Four statements are included that are not in the text's official list of statements, to give more symmetry to the three columns. You may or may not wish to comment on them. I believe that students cannot fully understand the concepts in the IMT if they do not know the correct wording of each statement. (Of course, this knowledge is not sufficient for understanding.) The Study Guide's Section 2.3 has an example of the type of question I often put on an exam at this point in the course. The section concludes with a discussion of reviewing and reflecting, as important steps to a mastery of linear algebra.
2.4
Solutions
103
2.4
SOLUTIONS
Notes: Partitioned matrices arise in theoretical discussions in essentially every field that makes use of matrices. The Study Guide mentions some examples (with references). Every student should be exposed to some of the ideas in this section. If time is short, you might omit Example 4 and Theorem 10, and replace Example 5 by a problem similar to one in Exercises 110. (A sample replacement is given at the end of these solutions.) Then select homework from Exercises 113, 15, and 21 24. The exercises just mentioned provide a good environment for practicing matrix manipulation. Also, students will be reminded that an equation of the form AB = I does not by itself make A or B invertible. (The matrices must be square and the IMT is required.)
1. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the entry of the left block-matrix on the left. I E
0 A I C B IA + 0C = D EA + IC IB + 0 D A = EB + ID EA + C
EB + D
B
2. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the entry of the left block-matrix on the left. E 0
0 A F C B EA + 0C = D 0 A + FC EB + 0 D EA = 0 B + FD FC EB FD
3. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the entry of the left block-matrix on the left. 0 I
I W 0 Y X 0W + IY = Z IW + 0Y 0 X + IZ Y = IX + 0 Z W Z X
4. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the entry of the left block-matrix on the left. I - X A C
0 A I C B I 0X B IA + 0C = D - XA + IC 0 AI + BX = Y CI + 0 X IB + 0 D A = - XA + C - XB + ID A0 + BY C 0 + 0Y BY = I 0=0
- XB + D
B
5. Compute the left side of the equation:
Set this equal to the right side of the equation:
A + BX C
BY 0 = 0 Z
I A + BX = 0 so that 0 C=Z
Since the (2, 1) blocks are equal, Z = C. Since the (1, 2) blocks are equal, BY = I. To proceed further, assume that B and Y are square. Then the equation BY =I implies that B is invertible, by the IMT, and Y = B1. (See the boxed remark that follows the IMT.) Finally, from the equality of the (1, 1) blocks, BX = A, B1BX = B1(A), and X = B1A. The order of the factors for X is crucial.
Note: For simplicity, statements (j) and (k) in the Invertible Matrix Theorem involve square matrices C and D. Actually, if A is nn and if C is any matrix such that AC is the nn identity matrix, then C must be nn, too. (For AC to be defined, C must have n rows, and the equation AC = I implies that C has n columns.) Similarly, DA = I implies that D is nn. Rather than discuss this in class, I expect that in Exercises 58, when
104 CHAPTER 2 Matrix Algebra
students see an equation such as BY = I, they will decide that both B and Y should be square in order to use the IMT.
6. Compute the left side of the equation: X Y
0A Z B 0 XA + 0 B = C YA + ZB X 0 + 0C XA = Y 0 + ZC YA + ZB 0 ZC
Set this equal to the right side of the equation:
XA YA + ZB
0 I = ZC 0
0 XA = I so that I YA + ZB = 0
0=0 ZC = I
To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A1. (See the boxed remark that follows the IMT.) Similarly, if C and Z are assumed to be square, then the equation ZC = I implies that C is invertible, by the IMT, and Z = C1. Finally, use the (2, 1) blocks and right-multiplication by A1: YA = ZB = C1B, YAA1 = (C1B)A1, and Y = C1BA1 The order of the factors for Y is crucial.
7. Compute the left side of the equation: X Y
0 0
A 0 0 I B
Z XA + 0 + 0 B 0 = YA + 0 + IB I
XZ + 0 + 0 I YZ + 0 + II
Set this equal to the right side of the equation:
XA YA + B
XZ I = YZ + I 0
0 XA = I so that YA + B = 0 I
XZ = 0 YZ + I = I
To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A1. (See the boxed remark that follows the IMT) Also, X is invertible. Since XZ = 0, X 1 XZ = X 1 0 = 0, so Z must be 0. Finally, from the equality of the (2, 1) blocks, YA = B. Right-multiplication by A1 shows that YAA1 = BA1 and Y = BA1. The order of the factors for Y is crucial.
8. Compute the left side of the equation: A B X Y Z AX + B 0 AY + B 0 0 I 0 0 I = 0 X + I 0 0Y + I 0 Set this equal to the right side of the equation: AX 0
AY 0 AZ + B I = I 0 0 0 0 I AZ + BI 0 Z + II
To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A1. (See the boxed remark that follows the IMT. Since AY = 0, from the equality of the (1, 2) blocks, left-multiplication by A1 gives A1AY = A10 = 0, so Y = 0. Finally, from the (1, 3) blocks, AZ = B. Left-multiplication by A1 gives A1AZ = A1(B), and Z = A1B. The order of the factors for Z is crucial.
Note: The Study Guide tells students, "Problems such as 510 make good exam questions. Remember to mention the IMT when appropriate, and remember that matrix multiplication is generally not commutative." When a problem statement includes a condition that a matrix is square, I expect my students to mention this fact when they apply the IMT.
2.4
Solutions
105
9. Compute the left side of the equation: I X Y
0 I 0 0 A11 0 A21 I A31 A12 IA11 + 0 A21 + 0 A31 A22 = XA11 + IA21 + 0 A31 A32 YA11 + 0 A21 + IA31 A12 B12 B22 B32 IA12 + 0 A22 + 0 A32 XA12 + IA22 + 0 A32 YA12 + 0 A22 + IA32
Set this equal to the right side of the equation: A11 XA + A 21 11 YA11 + A31
B11 XA12 + A22 = 0 YA12 + A32 0
A11 = B11 so that XA11 + A21 = 0 YA11 + A31 = 0
A12 = B12 XA12 + A22 = B22 YA12 + A32 = B32
Since the (2,1) blocks are equal, XA11 + A21 = 0 and XA11 = - A21. Since A11 is invertible, right
- - multiplication by A111 gives X = - A21 A111. Likewise since the (3,1) blocks are equal, - - YA11 + A31 = 0 and YA11 = - A31. Since A11 is invertible, right multiplication by A111 gives Y = - A31 A111. - - Finally, from the (2,2) entries, XA12 + A22 = B22 . Since X = - A21 A111 , B22 = - A21 A111 A12 + A22 .
10. Since the two matrices are inverses, I C A I C A
0 I B 0 I B
0 T 0 Z IX 0 I 0 Z IX
0 I Y 0 I Y 0 I B +Y
0 I 0 = 0 I 0
0 I 0
0 0 I I 0 + 0 I + 0Y C 0 + II + 0Y A0 + BI + IY I 0 + 00 + 0 I C 0 + I 0 + 0I A0 + B 0 + II
Compute the left side of the equation: 0 II + 0 Z + 0 X 0 = CI + IZ + 0 X I AI + BZ + I X 0 I 0 = 0 I 0 0 I 0 0 0 I 0=0 0=0 I =I
Set this equal to the right side of the equation: I C+Z A + BZ + X
I =I so that C+Z =0 A + BZ + X = 0
0=0 I =I B +Y = 0
Since the (2,1) blocks are equal, C + Z = 0 and Z = -C . Likewise since the (3, 2) blocks are equal, B + Y = 0 and Y = - B. Finally, from the (3,1) entries, A + BZ + X = 0 and X = - A - BZ . Since Z = -C , X = - A - B (-C ) = - A + BC .
11. a. True. See the subsection Addition and Scalar Multiplication. b. False. See the paragraph before Example 3. 12. a. True. See the paragraph before Example 4. b. False. See the paragraph before Example 3.
106 CHAPTER 2 Matrix Algebra
13. You are asked to establish an if and only if statement. First, supose that A is invertible, D and let A-1 = F B 0
E . Then G E BD = G CF BE I = CG 0 0 I
0 D C F
Since B is square, the equation BD = I implies that B is invertible, by the IMT. Similarly, CG = I implies that C is invertible. Also, the equation BE = 0 imples that E = B -1 0 = 0. Similarly F = 0. Thus
B A = 0
-1
0 C
-1
D = E
E B -1 = G 0
0 C -1
(*)
This proves that A is invertible only if B and C are invertible. For the "if " part of the statement, suppose that B and C are invertible. Then (*) provides a likely candidate for A-1 which can be used to show that A is invertible. Compute: B 0 0 B -1 C 0 0 BB -1 = C -1 0 0 I = CC -1 0 0 I
Since A is square, this calculation and the IMT imply that A is invertible. (Don't forget this final sentence. Without it, the argument is incomplete.) Instead of that sentence, you could add the equation: B -1 0 0 B C -1 0 0 B -1 B = C 0 0 I = C -1C 0 0 I
14. You are asked to establish an if and only if statement. First suppose that A is invertible. Example 5 shows that A11 and A22 are invertible. This proves that A is invertible only if A11 A22 are invertible. For the if part of this statement, suppose that A11 and A22 are invertible. Then the formula in Example 5 provides a likely candidate for A-1 which can be used to show that A is invertible . Compute:
A11 0
-1 A12 A 11 A22 0 -1 -1 -1 - A 11 A12 A 22 A11 A 11 + A12 0 = -1 -1 A 22 0 A11 + A22 0 -1 -1 -1 A11 (- A 11 ) A 12 A 22 + A12 A 22 -1 -1 -1 0(- A 11 ) A12 A 22 + A 22 A 22
I = 0 I = 0
-1 -1 -( A11 A 11 ) A12 A -1 + A12 A 22 22 I -1 -1 - A12 A 22 + A 12 A 22 I = I 0
0 I
Since A is square, this calculation and the IMT imply that A is invertible.
15. Compute the right side of the equation:
I X 0 A11 I 0 0 I S 0 Y A11 = I X A 11 0 I S 0 Y A11 = I X A11 A11Y X A11Y + S A 11Y = A 12 X A 11 Y + S = A 22
Set this equal to the left side of the equation: A 11 X A 11 A 11Y A 11 = X A 11Y + S A 21 A 11 = A 11 A 12 so that X A = A A 22 11 21
-1 Since the (1, 2) blocks are equal, A 11Y = A 12. Since A11 is invertible, left multiplication by A 11 gives -1 Y = A 11 A 12. Likewise since the (2,1) blocks are equal, X A11 = A21. Since A11 is invertible, right
2.4
Solutions
107
- - multiplication by A111 gives that X = A21 A111. One can check that the matrix S as given in the exercise satisfies the equation X A11Y + S = A22 with the calculated values of X and Y given above.
16. Suppose that A and A11 are invertible. First note that
I X and I 0 Y I I 0 -Y I = I 0 0 I 0 I I - X 0 I = I 0 0 I
I 0 I Y Since the matrices and 0 I X I are square, they are both invertible by the IMT. Equation (7) may be left multipled by -1 -1 I 0 I Y and right multipled by X I to find 0 I 0 I Y A 0 I I A11 0 Thus by Theorem 6, the matrix is invertible as the product of invertible matrices. Finally, S 0 Exercise 13 above may be used to show that S is invertible. A11 0 0 I = S X
-1 -1
17. The column-row expansions of Gk and Gk+1 are:
T Gk = X k X k T T = col1 ( X k ) row1 ( X k ) + ... + colk ( X k ) row k ( X k )
and
T Gk +1 = X k +1 X k +1 T T T = col1 ( X k +1 ) row1 ( X k +1 ) + ... + colk ( X k +1 ) row k ( X k +1 ) + colk +1 ( X k +1 ) row k +1 ( X k +1 ) T T T = col1 ( X k ) row1 ( X k ) + ... + colk ( X k ) row k ( X k ) + colk +1 ( X k +1 ) row k +1 ( X k ) T = Gk + colk +1 ( X k +1 ) row k +1 ( X k )
since the first k columns of Xk+1 are identical to the first k columns of Xk. Thus to update Gk to produce T Gk+1, the number colk+1 (Xk+1) rowk+1 ( X k ) should be added to Gk.
18. Since W = [ X x0 ] ,
X T X T X W T W = T [ X x0 ] = T x0 x0 X
S = xT x0 - xT X ( X T X ) -1 X T x0 0 0 = xT ( I m - X ( X T X ) -1 X T )x0 0 = xT Mx0 0
X T x0 xT x 0 0
By applying the formula for S from Exercise 15, S may be computed:
108 CHAPTER 2 Matrix Algebra
19. The matrix equation (8) in the text is equivalent to ( A - sI n )x + Bu = 0 and C x + u = y Rewrite the first equation as ( A - sI n )x = - Bu. When A - sI n is invertible,
x = ( A - sI n ) -1 (- Bu) = -( A - sI n ) -1 Bu
Substitute this formula for x into the second equation above: C (-( A - sI n ) -1 Bu ) + u = y, so that I m u - C ( A - sI n ) -1 Bu = y Thus y = ( I m - C ( A - sI n ) -1 B )u. If W ( s ) = I m - C ( A - sI n ) -1 B, then y = W ( s )u. The matrix W(s) is the Schur complement of the matrix A - sI n in the system matrix in equation (8)
20. The matrix in question is
A - BC - sI n B Im -C By applying the formula for S from Exercise 15, S may be computed:
S = I m - (-C )( A - BC - sI m ) -1 B = I m + C ( A - BC - sI m ) -1 B
1 21. a. A2 = 3 A b. M 2 = I
0 1 -1 3 0 A - A I
0 1 + 0 = -1 3 - 3
0 + 0 1 = 0 + (-1) 2 0
0 1 0 I
0 A2 + 0 = - A A - A
0 + 0 I = 0 + (- A) 2 0
I 22. Let C be any nonzero 23 matrix. Define A = 3 C I A2 = 3 C
0 I3 -I2 C 0 I3 + 0 = - I 2 CI 3 - I 2C
0 . Then -I2 0+0
2
I = 3 0 + (- I 2 ) 0
0 I2
23. The product of two 11 "lower triangular" matrices is "lower triangular." Suppose that for n = k, the product of two kk lower triangular matrices is lower triangular, and consider any (k+1) (k+1) matrices A1 and B1. Partition these matrices as
a A1 = v b 0T , B1 = A w
0T B
where A and B are kk matrices, v and w are in Rk, and a and b are scalars. Since A1 and B1 are lower triangular, so are A and B. Then
a A1B1 = v
0T b A w
T 0T ab + 0 w = B vb + Aw
a0T + 0T B ab = v0T + AB bv + Aw
0T AB
Since A and B are kk, AB is lower triangular. The form of A1B1 shows that it, too, is lower triangular. Thus the statement about lower triangular matrices is true for n = k +1 if it is true for n = k. By the principle of induction, the statement is true for all n > 1.
2.4
Solutions
109
Note: Exercise 23 is good for mathematics and computer science students. The solution of Exercise 23 in the Study Guide shows students how to use the principle of induction. The Study Guide also has an appendix on "The Principle of Induction," at the end of Section 2.4. The text presents more applications of induction in Section 3.2 and in the Supplementary Exercises for Chapter 3.
1 1 24. Let An = 1 # 1
0 1 1 1 0 0 1
"
% 1 "
0 1 -1 0 0 , Bn = 0 # 0 1
0 1 -1
0 0 1
"
% "
0T Bk
% -1
0 0 0 . 1
By direct computation A2B2 = I2. Assume that for n = k, the matrix AkBk is Ik, and write
1 Ak +1 = v
1 0T and Bk +1 = Ak w
where v and w are in Rk , vT = [1 1 1], and wT = [1 0 0]. Then
1 Ak +1 Bk +1 = v
0T 1 Ak w
T 0T 1 + 0 w = Bk v + Ak w
0T + 0T Bk 1 = v0T + Ak Bk 0
0T Ik
= I k +1
The (2,1)-entry is 0 because v equals the first column of Ak., and Akw is 1 times the first column of Ak. By the principle of induction, AnBn = In for all n > 2. Since An and Bn are square, the IMT shows that - these matrices are invertible, and Bn = An 1.
Note: An induction proof can also be given using partitions with the form shown below. The details are
slightly more complicated.
Ak Ak +1 = T v
0 Bk and Bk +1 = T 1 w 0 Bk 1 wT
0 1
0 Ak Bk + 0wT Ak 0 + 0 I k 0 = T = = I k +1 1 vT Bk + wT 1 vT 0 + 1 0 The (2,1)-entry is 0T because vT times a column of Bk equals the sum of the entries in the column, and all of such sums are zero except the last, which is 1. So vTBk is the negative of wT. By the principle of induction, AnBn = In for all n > 2. Since An and Bn are square, the IMT shows that these matrices are - invertible, and Bn = An 1.
Ak Ak +1 Bk +1 = T v
25. First, visualize a partition of A as a 22 blockdiagonal matrix, as below, and then visualize the (2,2)-block itself as a block-diagonal matrix. That is,
1 3 A = 0 0 0
2 5 0 0 0 0 0 2 0 0 0 0 0 7 5 0 0 A11 0 = 0 8 6
2 0 , where A22 = 0 A22 0
0 7 5
0 2 8 = 0 6
0 B
110 CHAPTER 2 Matrix Algebra
3 Observe that B is invertible and B1 = -2.5 invertible, and
-1 A22
-4 . By Exercise 13, the block diagonal matrix A22 is 3.5
0 -4 3.5 2 . By Exercise 13, A itself is invertible, -1
.5 = 0
0 3 -2.5
.5 -4 = 0 3.5 0
0 3 -2.5
-5 Next, observe that A11 is also invertible, with inverse 3 and its inverse is block diagonal:
A-1 A-1 = 11 0
-5 2 3 -1 0 = -1 A22 0
0 .5 0 0 0 3 -2.5
-5 3 0 = 0 -4 0 3.5 0
2 -1 0 0 0
0 0 .5 0 0
0 0 0 3 -2.5
0 0 0 -4 3.5
26. [M] This exercise and the next, which involve large matrices, are more appropriate for MATLAB, Maple, and Mathematica, than for the graphic calculators. a. Display the submatrix of A obtained from rows 15 to 20 and columns 5 to 10. MATLAB: A(15:20, 5:10)
Maple: Mathematica:
submatrix(A, 15..20, 5..10) Take[ A, {15,20}, {5,10} ]
b. Insert a 510 matrix B into rows 10 to 14 and columns 20 to 29 of matrix A: MATLAB: A(10:14, 20:29) = B ; The semicolon suppresses output display.
Maple: Mathematica:
copyinto(B, A, 10, 20): For [ i=10, i<=14, i++, For [ j=20, j<=29, j++, A [[ i,j ]] = B [[ i-9, j-19 ]] ] ]; Colon suppresses output.
The colon suppresses output display.
A 0 c. To create B = with MATLAB, build B out of four blocks: T 0 A B = [A zeros(30,20); zeros(20,30) A']; Another method: first enter B = A ; and then enlarge B with the command B(21:50, 31:50) = A'; This places AT in the (2, 2) block of the larger B and fills in the (1, 2) and (2, 1) blocks with zeros. For Maple: B := matrix(50,50,0): copyinto(A, B, 1, 1): copyinto( transpose(A), B, 21, 31): For Mathematica: B = BlockMatrix[ {{A, ZeroMatrix[30,20]}, ZeroMatrix[20,30], Transpose[A]}} ]
2.4
Solutions
111
27. a. [M] Construct A from four blocks, say C11, C12, C21, and C22, for example with C11 a 3030 matrix and C22 a 2020 matrix. MATLAB: C11 = A(1:30, 1:30) + B(1:30, 1:30) C12 = A(1:30, 31:50) + B(1:30, 31:50) C21 = A(31:50, 1:30)+ B(31:50, 1:30) C22 = A(31:50, 31:50) + B(31:50, 31:50) C = [C11 C12; C21 C22]
The commands in Maple and Mathematica are analogous, but with different syntax. The first commands are: C11 := submatrix(A, 1..30, 1..30) + submatrix(B, 1..30, 1..30) Maple: Mathematica: c11 := Take[ A, {1,30), {1,30} ] + Take[B, {1,30), {1,30} ]
b. The algebra needed comes from block matrix multiplication:
A12 B11 B12 A11 B11 + A12 B21 A11 B12 + A12 B22 A AB = 11 = A21 A22 B21 B22 A21 B11 + A22 B21 A21 B12 + A22 B22 Partition both A and B, for example with 3030 (1, 1) blocks and 2020 (2, 2) blocks. The four necessary submatrix computations use syntax analogous to that shown for (a). 0 x1 b1 A c. The algebra needed comes from the block matrix equation 11 = , where x1 and b1 A21 A22 x 2 b 2 are in R30 and x2 and b2 are in R20. Then A1 1x1 = b1, which can be solved to produce x1. Once x1 is found, rewrite the equation A21x1 + A22x2 = b2 as A22x2 = c, where c = b2 A21x1, and solve A22x2 = c for x2.
Notes: The following may be used in place of Example 5:
Example 5: Use equation (*) to find formulas for X, Y, and Z in terms of A, B, and C. Mention any assumptions you make in order to produce the formulas.
X Y 0 I ZA 0 I = B C 0 I (*)
Solution: This matrix equation provides four equations that can be used to find X, Y, and Z: X + 0 = I, 0=0 YI + ZA = C, Y0 + ZB = I (Note the order of the factors.) The first equation says that X = I. To solve the fourth equation, ZB = I, assume that B and Z are square. In this case, the equation ZB = I implies that B and Z are invertible, by the IMT. (Actually, it suffices to assume either that B is square or that Z is square.) Then, right-multiply each side of ZB = I to get ZBB1 = IB1 and Z = B1. Finally, the third equation is Y + ZA = C. So, Y + B1A = C, and Y = C B1A. The following counterexample shows that Z need not be square for the equation (*) above to be true.
1 0 1 3 0 1 2 4 0 0 1 1 0 0 3 0 1 0 0 0 1 1 1 -1 1 0 1 1 1 -1 0 0 2 -1 2 0 1 0 0 5 = 6 -3 3 4 0 1 5 6 0 0 1 0 0 0 0 1
112 CHAPTER 2 Matrix Algebra
Note that Z is not determined by A, B, and C, when B is not square. For instance, another Z that works in 5 0 3 this counterexample is Z = . -1 -2 0
2.5
SOLUTIONS
Notes: Modern algorithms in numerical linear algebra are often described using matrix factorizations. For
practical work, this section is more important than Sections 4.7 and 5.4, even though matrix factorizations are explained nicely in terms of change of bases. Computational exercises in this section emphasize the use of the LU factorization to solve linear systems. The LU factorization is performed using the algorithm explained in the paragraphs before Example 2, and performed in Example 2. The text discusses how to build L when no interchanges are needed to reduce the given matrix to U. An appendix in the Study Guide discusses how to build L in permuted unit lower triangular form when row interchanges are needed. Other factorizations are introduced in Exercises 2226. 1 1. L = -1 2 0 1 -5 0 3 0 , U = 0 0 1 0 1 -5 0 0 1 -7 -2 0 -2 -7 -1 , b = 5 . First, solve Ly = b. 2 -1 0 1 -5 0 0 1 -7 -2 The only arithmetic is in column 4 16
1 [ L b] = -1 2 1 0 0 0 1 0 0 0 1
-7 1 5 ~ 0 2 0
-7 -7 -2 , so y = -2 . 6 6 -7 -2 0 -7 1 0 -2 -1 -1 0 0 1 -7 3 -2 0 6 0 -19 3 4 0 -6 0 -7 -2 0 0 1 0 0 0 1 -2 -1 1 -7 3 -2 0 -6 0 0 1 0 -7 -2 0 0 0 1 0 0 1 3 4 -6 -19 -8 -6
Next, solve Ux = y, using back-substitution (with matrix notation). 3 [U y ] = 0 0 3 ~ 0 0
9 1 4 0 -6 0
So x = (3, 4, 6). To confirm this result, row reduce the matrix [A b]: 3 [ A b] = -3 6 -7 5 -4 -2 1 0 -7 3 5 0 2 0 -7 -2 10 -2 -1 4 -7 3 -2 0 16 0 -7 -2 0 -2 -1 -1 -7 -2 6
From this point the row reduction follows that of [U y] above, yielding the same result.
2.5
Solutions
113
1 2. L = -1 2
0 1 0
0 4 , U = 0 0 0 1 0 1 0 0 0 1
3 -2 0
-5 2 , b = -4 . First, solve Ly = b: 2 6 2 0 1 0 0 0 1 2 -2 , 2
1 [ L b] = -1 2 2 so y = -2 . 2
2 1 -4 0 6 0
Next solve Ux = y, using back-substitution (with matrix notation): 4 [U y ] = 0 0 4 0 0 4 [ A b] = -4 8 3 -2 0 3 1 0 0 0 1 -5 2 2 2 4 -2 0 2 0 7 4 2 0 1 0 -5 7 -8 0 1 0 3 -2 0 0 0 1 -5 2 1 2 4 -2 0 1 0 0 1 0 2 -2 2 3 -2 0 0 0 1 7 -4 1
1 1 2 0 1 0 3 -5 2 2
0 1/4 0 2 , 1 1
so x = (1/ 4, 2, 1). To confirm this result, row reduce the matrix [A b]: 3 -5 6 2 4 -4 0 6 0 -1 -3 0 -2 0
From this point the row reduction follows that of [U y] above, yielding the same result. 1 3. L = -3 4 0 1 -1 0 2 , U = 0 0 0 1 2 1 , b = 0 . First, solve Ly = b: 4 4 1
1 0 0 1 1 0 0 1 1 0 0 1 [ L b] = -3 1 0 0 0 1 0 3 0 1 0 3 , 4 -1 1 4 0 -1 1 0 0 0 1 0 1 so y = 3 . 3 Next solve Ux = y, using back-substitution (with matrix notation): 2 [U y ] = 0 0 2 0 0 so x = (1, 3, 3). -1 -3 0 0 1 0 0 0 1 2 4 1 1 2 3 0 3 0 -2 3 , 3 -1 -3 0 0 0 1 -5 2 -9 0 3 0 -1 1 0 0 0 1 -5 3 3
114 CHAPTER 2 Matrix Algebra
1 4. L = 1 / 2 3 / 2
0 1 -5
0 2 ,U = 0 0 0 1 0 1 -5 0 0 1
-2 -2 0
4 0 , b = -5 . First, solve Ly = b: -1 7 -6 0 1 -5 0 0 1 0 1 -5 0 7 0 0 1 0 0 0 1 0 -5 , -18
1 [ L b] = 1 / 2 3 / 2 0 so y = -5 . -18
0 1 -5 0 7 0
Next solve Ux = y, using back-substitution (with matrix notation): 4 0 2 -2 4 0 2 -2 2 -2 [U y ] = 0 -2 -1 -5 0 -2 -1 -5 0 -2 0 0 -6 -18 0 0 1 3 0 0 2 0 0 so x = (5, 1, 3). 1 2 5. L = -1 -4 0 1 0 3 0 1 0 0 0 ,U = 0 1 0 -5 1 0 0 0 1 0 3 0 1 0 0 0 0 1 -5 0 0 1 -5 0 0 0 1 0 0 0 1 -2 -3 0 0 -4 1 2 0 -2 1 0 0 0 1 -12 2 1 0 3 0 0 1 0 0 0 1 -10 1 1 0 3 0 0 1 0 0 0 1
0 0 1
-12 -2 3
-5 1 , 3
-3 1 7 0 , b = . First solve Ly = b: 0 1 1 3 0 1 0 3 0 1 0 0 0 0 1 0 0 0 1 -5 0 0 0 1 0 0 0 1 1 5 1 7 1 5 , 1 -3
1 2 [ L b] = -1 -4 1 0 0 0 1 5 so y = . 1 -3
1 1 7 0 0 0 3 0 1 1 5 0 1 0 -8 0
Next solve Ux = y, using back-substitution (with matrix notation): 1 0 [U y ] = 0 0 -2 -3 0 0 -4 1 2 0 -3 0 1 1 1 1 5 0 1 0 -3 0 -2 -3 0 0 -4 1 2 0 0 0 0 1 -8 5 4 -3
2.5
Solutions
115
1 0 0 0 1 0 0 0
-2 -3 0 0 -2 1 0 0
-4 1 1 0 0 0 1 0 0 0 0
0 0 0 1
-8 1 5 0 2 0 -3 0 0 1 -1 0 2 0 -3 0 0 1 0 0
-2 -3 0 0 0 0 1 0
0 0 1 0 0 0 0 1
0 0 0 1
0 3 2 -3
1
-2 -1 , 2 -3
so x = (2, 1, 2, 3). 1 -3 6. L = 3 -5 0 1 -2 4 0 0 1 -1 0 1 0 0 ,U = 0 0 1 0 0 1 -2 4 0 1 0 0 0 0 1 -1 0 0 1 -1 0 0 0 1 0 0 0 1 3 3 0 0 4 5 -2 0 0 1 -2 2 , b = . First, solve Ly = b: -1 0 1 2 0 1 -2 4 0 0 1 0 0 0 0 1 0 0 1 -1 0 0 0 1 1 1 , -2 1 1 1 -4 7
1 -3 [ L b] = 3 -5 1 0 0 0 1 1 so y = . -2 1
1 1 -2 0 -1 0 2 0 0 1 0 0
1 1 1 0 -2 0 3 0
Next solve Ux = y, using back-substitution (with matrix notation): 1 0 [U y ] = 0 0 1 0 0 0 1 0 0 0 3 3 0 0 3 3 0 0 3 1 0 0 4 5 1 0 0 0 1 0 4 5 -2 0 0 0 0 1 0 0 0 1 0 2 0 1 1 1 1 0 -2 0 1 0 1 1 -1 0 1 0 1 0 -3 1 -2 0 1 0 1 0 3 3 0 0 0 1 0 0 3 3 0 0 0 0 1 0 0 0 1 0 4 5 -2 0 0 0 0 1 0 0 0 1 0 0 0 1 -3 -6 1 1 3 -2 , 1 1 1 -1 -2 1
so x = (3, 2, 1, 1).
116 CHAPTER 2 Matrix Algebra
2 5 7. Place the first pivot column of into L, after dividing the column by 2 (the pivot), then add -3 -4 3/2 times row 1 to row 2, yielding U. 5 2 5 2 A= ~ 0 7/2 = U -3 - 4 2 -3 [7 / 2 ] 2 1 -3 /2 7 / 2 1 , L = -3 /2 1 0 1
8. Row reduce A to echelon form using only row replacement operations. Then follow the algorithm in Example 2 to find L.
6 A= 4 9 6 5 0 9 =U -1
6 4 [ -1] 6 - 1 1 2 / 3 3 9. A = -3 9 3 -3 9 3 1 -1 3 -1 -2 -5 0 1 , L= 1 2 / 3 1 2 3 10 0 6 0 2 3 -3 12 ~ 0 0 0 -2 -1 -1 -3 0 2 12 = U -8
-3 -2 [ -8] 3 8 1 , L = -1 3 1 0 1 2 /3 0 0 1
1 2 /3
2.5
Solutions
117
-5 10. A = 10 15 -5 10 15 -5 1 -2 -3 3 11. A = 6 -1
3 -8 1
4 -5 -9 ~ 0 2 0
3 -2 10
4 -5 -1 ~ 0 14 0
3 -2 0
4 -1 = U 9
-2 10 [9] -2 9 1 , L = -2 -3 1 3 3 2 0 0 0 -6 5 5 0 1 -5 0 0 1 -6 5 0 3 -4 = U 5
1 -5 -6 -7 7
3 3 -4 0 1 0
3 6 5 -1 5 [ 5 ] 3 5 5 1 2 -1 / 3 1 , L = 2 -1 / 3 1 0 1 1 0 0 1
1 1
12. Row reduce A to echelon form using only row replacement operations. Then follow the algorithm in Example 2 to find L. Use the last column of I3 to make L unit lower triangular.
2 A= 1 -6 2 1 -6 2 1 1/2 -3 -4 5 -2 2 2 -4 0 4 0 -4 7 -14 2 2 -5 0 10 0 -4 7 0 2 -5 = U 0
7 -14 7 1 , L = 1/2 -3 1 0 1 -2 0 0 1
1 -2
118 CHAPTER 2 Matrix Algebra
3 -5 -3 1 3 -5 -3 1 3 -5 -3 1 -1 -5 8 0 -2 3 4 1 0 -2 3 1 = U No more pivots! 13. 4 2 -5 -7 0 -10 15 5 0 0 0 0 2 -3 -1 0 0 0 0 -2 -4 7 5 0 1 -1 4 -2
-2 -10 2 Use the last two columns of I 4 to make L unit lower triangular.
1
2 0 0 1 0 0 0 0 1
1 1 0 -1 1 , L = -1 1 4 5 1 4 5 -2 -1 0 1 -2 -1
1 4 -1 5 1 4 -1 5 1 4 -1 5 3 7 -2 9 0 -5 1 -6 0 -5 1 -6 =U 14. A = -2 -3 1 -4 0 5 -1 6 0 0 0 0 -1 6 -1 7 0 10 -2 12 0 0 0 0
1 3 -2 -1 1 1 3 -2 -1
-5 5 10 Use the last two columns of I 4 to make L unit lower triangular. 5 1 , L = 3 -2 1 -1 0 1 -1 -2 0 0 1 0 0 0 0 1
1 -1 -2
1 0
2.5
Solutions
119
2 -4 4 -2 2 -4 4 -2 2 -4 4 -2 15. A = 6 -9 7 -3 0 3 -5 3 0 3 -5 3 = U -1 -4 8 0 0 -6 10 -1 0 0 0 5 2 6 -1 2
3 -6 [5] 3 5 1 , L = 3 -1 / 2 1 -6 -7 14 -14 21 0 1 -2 0 0 1 -6 -7 0 0 0 6 5 0 = U 0 0
1 3 -1 / 2 2 -4 16. A = 3 -6 8
2 -4 3 -6 8
1 -2
-6 5 5 4 -3
6 2 -7 0 -1 ~ 0 -8 0 9 0
6 2 5 0 -10 ~ 0 10 0 -15 0
-7 14 -14 21 Use the last three columns of I to make L unit lower triangular. 5
2 1 -2 3 / 2 -3 4 1 17. L = -1 2 reduce [L 0 1 0
7 1 -2 , L = 3 / 2 -3 4 1 3 -2 0 1 0 0 0 1 0 0 1 -2 2 -3 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
1 -2 2 -3 1 0 0 1 0
0 4 , U = 0 0 0 1 0 1 0 0 0 1
-5 2 To find L1, use the method of Section 2.2; that is, row 2 0 1 0 0 1 0 0 1 0 0 0 1 1 1 -2 0 1 0 0 0 = [ I L-1 ], 1
I ]:
1 [ L I ] = -1 2
120 CHAPTER 2 Matrix Algebra
1 so L = 1 -2
-1
0 1 0
0 0 . Likewise to find U 1, row reduce [U I ] : 1 3 -2 0 1 0 0 -5 2 2 3/ 2 1 0 1 0 0 0 1 0 0 4 0 0 1 0 0 1 0 0 0 1 3 -2 0 1/ 4 0 0 0 0 2 1 0 0 3/ 8 -1/ 2 0 0 1 0 5 / 2 -1 1
4 [U I ] = 0 0 4 0 0 so U
-1
0 -2 0
0 0 2
1 1 -1 0 1 0
1/ 4 1/ 2 = [ I U -1 ], 1/ 2
1/ 4 = 0 0
-1
3/ 8 -1/ 2 0
1/ 4 1/ 2 . Thus 1/ 2 3/ 8 -1/ 2 0 -1 -3 0 1 0 0 1 3 -1 0 1 1 0 1 0 1/ 4 1 1/ 2 1 1/ 2 -2 0 1 0 0 1/ 8 0 = -3/ 2 1 -1 3/ 8 -1/ 2 0 1/ 4 1/ 2 1/ 2
1/ 4 A =U L = 0 0
-1 -1
1 18. L = -3 4
0 1 -1
0 2 , U = 0 0 0 1 0 1 -1 0 1 0 0 0 1 0 0 1
2 4 To find L-1 , row reduce[ L I ]: 1 0 1 0 ~ 0 1 0 0 0 = I 1 0 1 -1 L-1 , 0 0 1 1 3 -4 0 1 0 0 0 1
[L
1 I ] = -3 4 1 ~ 0 0
1 so L-1 = 3 -1
0 1 1
0 0 . Likewise to find U -1 , row reduce [U 1 -1 -3 0 1 0 0 1/ 2 0 0 2 4 1 0 -1/ 3 0 -1/ 6 -1/ 3 0 1 0 0 0 1 0 0 2 0 ~ 0 1 0 0 1 0 -1 -3 0 0 0 1 1 0 0 0 0 1 1 0 0
I]: 0 1 0 -2 -4 1 -2 / 3 4 / 3 1
[U
2 ~ 0 0 1 ~ 0 0
2 I ] = 0 0 -1 1 0 0 1 0 0 0 1 0 0 1
-2 2 4 / 3 ~ 0 1 0
-1/ 3 -1/ 3 0
-1/ 3 4 / 3 = [ I U -1 ], 1
2.5
Solutions
121
so U
-1
1/ 2 = 0 0
-1
-1/ 6 -1/ 3 0
-1/ 3 4 / 3 . Thus 1 -1/ 6 -1/ 3 0 -1/ 3 1 4 / 3 3 1 -1 0 1 1 0 1/ 3 0 = -7 / 3 1 -1 -1/ 2 1 1 -1/ 3 4 / 3 1
1/ 2 A =U L = 0 0
-1 -1
19. Let A be a lower-triangular n n matrix with nonzero entries on the diagonal, and consider the augmented matrix [A I]. a. The (1, 1)-entry can be scaled to 1 and the entries below it can be changed to 0 by adding multiples of row 1 to the rows below. This affects only the first column of A and the first column of I. So the (2, 2)-entry in the new matrix is still nonzero and now is the only nonzero entry of row 2 in the first n columns (because A was lower triangular). The (2, 2)-entry can be scaled to 1, the entries below it can be changed to 0 by adding multiples of row 2 to the rows below. This affects only columns 2 and n + 2 of the augmented matrix. Now the (3, 3) entry in A is the only nonzero entry of the third row in the first n columns, so it can be scaled to 1 and then used as a pivot to zero out entries below it. Continuing in this way, A is eventually reduced to I, by scaling each row with a pivot and then using only row operations that add multiples of the pivot row to rows below. b. The row operations just described only add rows to rows below, so the I on the right in [A I] changes into a lower triangular matrix. By Theorem 7 in Section 2.2, that matrix is A1. 20. Let A = LU be an LU factorization for A. Since L is unit lower triangular, it is invertible by Exercise 19. Thus by the Invertible Matrix Theroem, L may be row reduced to I. But L is unit lower triangular, so it can be row reduced to I by adding suitable multiples of a row to the rows below it, beginning with the top row. Note that all of the described row operations done to L are row-replacement operations. If elementary matrices E1, E2, ... Ep implement these row-replacement operations, then E p ... E2 E1 A = ( E p ... E2 E1 ) LU = IU = U
This shows that A may be row reduced to U using only row-replacement operations.
21. (Solution in Study Guide.) Suppose A = BC, with B invertible. Then there exist elementary matrices E1, ..., Ep corresponding to row operations that reduce B to I, in the sense that Ep ... E1B = I. Applying the same sequence of row operations to A amounts to left-multiplying A by the product Ep ... E1. By associativity of matrix multiplication. E p ... E1 A = E p ... E1 BC = IC = C
so the same sequence of row operations reduces A to C.
22. First find an LU factorization for A. Row reduce A to echelon form using only row replacement operations:
2 6 A= 2 4 -6 -4 -9 -7 -2 3 -2 -5 -3 -2 3 3 2 8 0 9 ~ 0 -1 0 4 0 -4 3 -3 6 -9 -2 1 -1 2 -3 3 2 -1 0 6 ~ 0 -7 0 13 0 -4 3 0 0 0 -2 1 0 0 0 3 -1 5 -5 10
122 CHAPTER 2 Matrix Algebra
2 0 ~ 0 0 0
-4 3 0 0 0
-2 1 0 0 0
3 -1 5 = U 0 0
then follow the algorithm in Example 2 to find L. Use the last two columns of I5 to make L unit lower triangular.
2 6 2 4 -6 2
3 -3 6 -9 3
5 -5 10 5
0 0 0 0 1 1 3 3 1 1 0 0 0 1 -1 , L = 1 -1 1 1 0 0 2 -1 1 2 -1 1 0 2 2 -3 -3 -3 2 0 1 3 2 0 1 Now notice that the bottom two rows of U contain only zeros. If one uses the row-column method to find LU, the entries in the final two columns of L will not be used, since these entries will be multiplied zeros from the bottom two rows of U. So let B be the first three columns of L and let C be the top three rows of U. That is, 1 3 B= 1 2 -3 0 1 -1 2 3 0 0 2 1 , C = 0 0 -1 2 -4 3 0 -2 1 0 3 -1 5
Then B and C have the desired sizes and BC = LU = A. We can generalize this process to the case where A in m n, A = LU, and U has only three non-zero rows: let B be the first three columns of L and let C be the top three rows of U.
23. a. Express each row of D as the transpose of a column vector. Then use the multiplication rule for partitioned matrices to write
dT 1 T d 2 c4 ] T d3 T d 4 T T T T = c1d1 + c2d 2 + c3d3 + c 4d4
A = CD = [c1
c2
c3
which is the sum of four outer products. b. Since A has 400 100 = 40000 entries, C has 400 4 = 1600 entries and D has 4 100 = 400 entries, to store C and D together requires only 2000 entries, which is 5% of the amount of entries needed to store A directly.
2.5
Solutions
123
24. Since Q is square and QTQ = I, Q is invertible by the Invertible Matrix Theorem and Q1 = QT. Thus A is the product of invertible matrices and hence is invertible. Thus by Theorem 5, the equation Ax = b has a unique solution for all b. From Ax = b, we have QRx = b, QTQRx = QTb, Rx = QTb, and finally x = R1QTb. A good algorithm for finding x is to compute QTb and then row reduce the matrix [ R QTb ]. See Exercise 11 in Section 2.2 for details on why this process works. The reduction is fast in this case because R is a triangular matrix. 25. A = UDV T . Since U and V T are square, the equations U T U = I and V T V = I imply that U and V T are invertible, by the IMT, and hence U 1 = U T and (VT)1 = V. Since the diagonal entries 1 ,!, n in D are
- nonzero, D is invertible, with the inverse of D being the diagonal matrix with 1 1 ,!, -1 on the n
diagonal. Thus A is a product of invertible matrices. By Theorem 6, A is invertible and A1 = (UDV T ) 1 = (VT)1D1U1 = VD1UT.
26. If A = PDP1, where P is an invertible 3 3 matrix and D is the diagonal matrix
0 1 0 1/ 2 D= 0 0 then 0 0 1/ 3
A2 = ( PDP -1 )( PDP -1 ) = PD( P -1 P) DP -1 = PDIDP -1 = PD 2 P -1
and since
0 1 0 1/ 2 D = 0 0
2
0 1 0 0 1/ 2 0 1/ 3 0 0 0 0 P -1 1/ 9
0 0 1 = 0 1/ 22 0 1/ 3 0 0
0 1 0 0 = 0 1/ 4 0 1/ 32 0
0 0 1/ 9
0 1 0 1/ 4 A = P 0 0
2
Likewise, A3 = PD3P 1, so 0 1 A = P 0 1/ 23 0 0
3
0 0 1 0 P -1 = P 0 1/ 8 0 0 1/ 33
0 0 P -1 1/ 27
In general, Ak = PDkP1, so 0 0 1 k A = P 0 1/ 2 0 P -1 0 0 1/ 3k
k
27. First consider using a series circuit with resistance R1 followed by a shunt circuit with resistance R2 for the network. The transfer matrix for this network is
1 -1/R 2
0 1 1 0
- R1 1 = 1 -1/R2
- R1 ( R1 + R2 ) / R2
124 CHAPTER 2 Matrix Algebra
For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps, the transfer matrix must satisfy 1 -1/R 2 12 - 6 R1 - R1 12 9 6 = (-12 + 6 R + 6 R )/ R = 4 ( R1 + R2 )/ R2 1 2 2
Equate the top entries and obtain R1 = 1 ohm. Substitute this value in the bottom entry and solve to 2 obtain R2 = 9 ohms. The ladder network is 2
a.
v1 i1 1/ 2 ohm v2 i2 i2 9/ 2 ohms i3 v3
Next consider using a shunt circuit with resistance R1 followed by a series circuit with resistance R2 for the network. The transfer matrix for this network is 1 0 - R2 1 1 -1/R1 0 ( R1 + R2 )/ R1 = 1 -1/R1 - R2 1
For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps, the transfer matrix must satisfy ( R1 + R2 ) /R1 - R2 12 (12 R1 + 12 R2 ) /R1 - 6 R2 9 = -1/R = 4 1 6 -12 /R1 + 6 1 Equate the bottom entries and obtain R1 = 6 ohms. Substitute this value in the top entry and solve to obtain R2 = 3 ohms. The ladder network is 4
b.
v1 i1 6 ohms i2 v2 i2 3/4 ohm v3 i3
28. The three shunt circuits have transfer matrices
0 0 1 0 1 1 -1/R 1 , -1/R , and -1/R 1 1 1 2 3 respectively. To find the transfer matrix for the series of circuits, multiply these matrices 1 -1/R3 0 1 , 1 -1/R2 0 , and 1 1 -1/R1 1 0 = 1 -(1/R1 + 1/R2 + 1/R3 ) 0 1
Thus the resulting network is itself a shunt circuit with resistance 1/R1 + 1/R2 + 1/R3 . 0 1 29. a. The first circuit is a shunt circuit with resistance R1 ohms, so its transfer matrix is . -1/R1 1 1 - R2 . The second circuit is a series circuit with resistance R2 ohms, so its transfer matrix is 1 0
2.5
Solutions
125
1 The third circuit is a shunt circuit with resistance R3 ohms so its transfer matrix is -1/R3 The transfer matrix of the network is the product of these matrices, in right-to-left order:
0 . 1
- R2 0 1 - R2 1 0 ( R1 + R2 ) /R1 1 = -1/R 1 0 1 -1/R1 1 -( R1 + R2 + R3 ) /R3 ( R2 + R3 ) /R3 3 b. To find a ladder network with a structure like that in part (a) and with the given transfer matrix A, we must find resistances R1, R2, and R3 such that 4/3 A= -1/ 4 -12 ( R1 + R2 ) /R1 = 3 -( R1 + R2 + R3 ) /R3 - R2 ( R2 + R3 ) /R3
From the (1, 2) entries, R2 = 12 ohms. The (1, 1) entries now give ( R1 + 12) /R1 = 4 / 3, which may be solved to obtain R1 = 36 ohms. Likewise the (2, 2) entries give ( R3 + 12) /R3 = 3, which also may be solved to obtain R3 = 6 ohms. Thus the matrix A may be factored as 1 A= -1/R3 0 1 1 0 - R2 1 1 -1/R1 -12 1 1 -1/36 0 1 0 1
0 1 1 = -1/6 1 0 The ladder network is
i1 v1
i2 36 ohms v2
i2 12 ohms
i3 v3
i3 6 ohms
i4 v4
30. Answers may vary. The network below interchanges the series and shunt circuits.
i1 R1 v1 v2 R2 v3 i2 i2 i3 i3 R3 v4 i4
The transfer matrix of this network is the product of the individual transfer matrices, in right-to-left order. 1 0 - R3 1 1 -1/R2 0 1 1 0 - R1 = 1
( R2 + R3 ) /R2 - R3 - R1 ( R2 + R3 ) /R2 -1/R ( R1 + R2 ) /R2 2 By setting the matrix A from the previous exercise equal to this matrix, one may find that ( R2 + R3 ) /R2 - R3 - R1 ( R2 + R3 ) /R2 4 /3 -12 -1/R = -1/4 ( R1 + R2 ) /R2 3 2 Set the (2, 1) entries equal and obtain R2 = 4 ohms. Substitute this value for R2, equating the (2, 2) entries and solving gives R1 = 8 ohms. Likewise equating the (1, 1) entries gives R3 = 4/3 ohms.
126 CHAPTER 2 Matrix Algebra
The ladder network is
i1 8 ohms v1 v2 i2 i2 4 ohms i3 v3 i3 4/3 ohms v4 i4
Note: The Study Guide's MATLAB box for Section 2.5 suggests that for most LU factorizations in this section, students can use the gauss command repeatedly to produce U, and use paper and mental arithmetic to write down the columns of L as the row reduction to U proceeds. This is because for Exercises 716 the pivots are integers and other entries are simple fractions. However, for Exercises 31 and 32 this is not reasonable, and students are expected to solve an elementary programming problem. (The Study Guide provides no hints.)
31. [M] Store the matrix A in a temporary matrix B and create L initially as the 88 identity matrix. The following sequence of MATLAB commands fills in the entries of L below the diagonal, one column at a time, until the first seven columns are filled. (The eighth column is the final column of the identity matrix.) L(2:8, 1) = B(2:8, 1)/B(1, 1) B = gauss(B, 1) L(3:8, 2) = B(3:8, 2)/B(2, 2) B = gauss(B, 2) # L(8:8, 7) = B(8:8, 7)/B(7, 7) U = gauss(B,7)
Of course, some students may realize that a loop will speed up the process. The for..end syntax is illustrated in the MATLAB box for Section 5.6. Here is a MATLAB program that includes the initial setup of B and L: B = A L = eye(8) for j=1:7 L(j+1:8, j) = B(j+1:8, j)/B(j, j) B = gauss(B, j) end U = B
a. To four decimal places, the results of the LU decomposition are
1 -.25 -.25 0 L= 0 0 0 0 0 1 -.0667 -.2667 0 0 0 0 0 0 1 -.2857 -.2679 0 0 0 0 0 0 1 -.0833 -.2917 0 0 0 0 0 0 1 -.2921 -.2697 0 0 0 0 0 0 1 -.0861 -.2948 0 0 0 0 0 0 1 -.2931 0 0 0 0 0 0 0 1
2.5
Solutions
127
4 0 0 0 U = 0 0 0 0
-1 3.75 0 0 0 0 0 0
-1 -.25 3.7333 0 0 0 0 0
0 -1 -1.0667 3.4286 0 0 0 0
0 0 -1 -.2857 3.7083 0 0 0
0 0 0 -1 -1.0833 3.3919 0 0
0 0 0 0 -1 -.2921 3.7052 0
0 0 0 0 -1 -1.0861 3.3868 0
b. The result of solving Ly = b and then Ux = y is x = (3.9569, 6.5885, 4.2392, 7.3971, 5.6029, 8.7608, 9.4115, 12.0431)
.2953 .0866 .0945 .0509 c. A-1 = .0318 .0227 .0010 .0082 .0866 .2953 .0509 .0945 .0227 .0318 .0082 .0100 .0945 .0509 .3271 .1093 .1045 .0591 .0318 .0227 0 0 -1 3 -1 .0509 .0945 .1093 .3271 .0591 .1045 .0227 .0318 .0318 .0227 .1045 .0591 .3271 .1093 .0945 .0509 .0227 .0318 .0591 .1045 .1093 .3271 .0509 .0945 .0010 .0082 .0318 .0227 .0945 .0509 .2953 .0866 .0082 .0100 .0227 .0318 .0509 .0945 .0866 .2953
0 3 -1 -1 3 -1 32. [M] A = 0 -1 3 0 -1 0 0 0 0 matrices, produce
0 0 0 . The commands shown for Exercise 31, but modified for 55 -1 3
1 - 1 3 L= 0 0 0 3 0 U = 0 0 0
0 1 3 -8 0 0 -1
8 3
0 0 1 8 - 21 0 0 -1
21 8
0 0 0 1 21 - 55 0 0 -1
55 21
0 0 0 0 1
0 0 0
0 0
0
0 0 0 -1 144 55
128 CHAPTER 2 Matrix Algebra
b. Let sk+1 be the solution of Lsk+1 = tk for k = 0, 1, 2, .... Then tk+1 is the solution of Utk+1 = sk+1 for k = 0, 1, 2, .... The results are
10.0000 6.5556 6.5556 4.7407 15.3333 9.6667 11.8519 7.6667 s1 = 17.7500 , t1 = 10.4444 , s 2 = 14.8889 , t 2 = 8.5926 , 18.7619 9.6667 15.3386 7.6667 17.1636 6.5556 12.4121 4.7407 4.7407 3.5988 3.5988 2.7922 9.2469 6.0556 7.2551 4.7778 s3 = 12.0602 , t 3 = 6.9012 , s 4 = 9.6219 , t 4 = 5.4856 . 12.2610 6.0556 9.7210 4.7778 9.4222 3.5988 7.3104 2.7922
2.6
SOLUTIONS
Notes: This section is independent of Section 1.10. The material here makes a good backdrop for the series expansion of (IC)1 because this formula is actually used in some practical economic work. Exercise 8 gives an interpretation to entries of an inverse matrix that could be stated without the economic context.
1. The answer to this exercise will depend upon the order in which the student chooses to list the sectors. The important fact to remember is that each column is the unit consumption vector for the appropriate sector. If we order the sectors manufacturing, agriculture, and services, then the consumption matrix is
.10 C = .30 .30
.60 .20 .10
.60 0 .10
The intermediate demands created by the production vector x are given by Cx. Thus in this case the intermediate demand is .10 .60 .60 0 60 Cx = .30 .20 .00 100 = 20 .30 .10 .10 0 10
2. Solve the equation x = Cx + d for d:
x1 .10 d = x - Cx = x2 - .30 x3 .30
.60 .20 .10
.60 x1 .9 x1 .00 x2 = -.3 x1 .10 x3 -.3 x1 0 1 0 0 0 1
-.6 x2 +.8 x2 -.1x2
-.6 x3 0 = 18 +.9 x3 0
This system of equations has the augmented matrix 0 1 -.90 -.60 -.60 -.30 .80 .00 18 ~ 0 -.30 -.10 .90 0 0 so x = (33.33, 35.00, 15.00). 33.33 35.00 15.00
2.6
Solutions
129
3. Solving as in Exercise 2:
x1 .10 .60 .60 x1 .9 x1 -.6 x2 d = x - C x = x2 - .30 .20 .00 x2 = -.3 x1 +.8 x2 x3 .30 .10 .10 x3 -.3 x1 -.1x2 This system of equations has the augmented matrix .90 -.60 -.60 18 1 -.30 .80 .00 0 ~ 0 -.30 -.10 .90 0 0 so x = (40.00, 15.00, 15.00).
4. Solving as in Exercise 2:
-.6 x3 18 =0 +.9 x3 0
0 1 0
0 0 1
40.00 15.00 15.00
x1 .10 .60 .60 x1 .9 x1 d = x - Cx = x2 - .30 .20 .00 x2 = -.3 x1 x3 .30 .10 .10 x3 -.3 x1 This system of equations has the augmented matrix -.90 -.60 -.60 18 1 -.30 .80 .00 18 ~ 0 -.30 -.10 .90 0 0 so x = (73.33, 50.00, 30.00). 0 1 0 0 0 1 73.33 50.00 30.00
-.6 x2 +.8 x2 -.1x2
-.6 x3 18 = 18 +.9 x3 0
Note: Exercises 24 may be used by students to discover the linearity of the Leontief model.
1 5. x = ( I - C ) -1 d = -.6 .9 6. x = ( I - C ) -1 d = -.5
7. a. From Exercise 5,
-.5 50 1.6 = .8 20 1.2
-1
-1
1 50 110 = 2 20 120 30 / 21 18 50 = 45 / 21 11 45
-.6 18 40 / 21 = .8 11 25 / 21 1 2
1.6 ( I - C ) -1 = 1.2 so
1.6 x1 = ( I - C ) -1 d1 = 1.2 1.6 x 2 = ( I - C ) -1 d 2 = 1.2
1 1 1.6 = 2 0 1.2 1 51 111.6 = 2 30 121.2
which is the first column of ( I - C ) -1.
b.
130 CHAPTER 2 Matrix Algebra
50 110 c. From Exercise 5, the production x corressponding to d = is x = . 20 120 Note that d 2 = d + d1. Thus
x 2 = ( I - C ) -1 d 2
= ( I - C ) -1 (d + d1 ) = ( I - C ) -1 d + ( I - C ) -1 d1 = x + x1
8. a. Given ( I - C )x = d and ( I - C ) x = d,
( I - C )(x + x) = ( I - C )x + ( I - C ) x = d + d Thus x + x is the production level corresponding to a demand of d + d .
b. Since x = ( I - C ) -1 d and d is the first column of I, x will be the first column of ( I - C ) -1 . 9. In this case
.8 I - C = -.3 -.1 Row reduce .8 -.3 -.1
-.2 .9 .0
.0 -.3 .8 to find 40.0 1 60.0 ~ 0 80.0 0 0 1 0 0 82.8 0 131.0 1 110.3
[ I - C d] -.2 .0 .9 -.3 .0 .8
So x = (82.8, 131.0, 110.3).
10. From Exercise 8, the (i, j) entry in (I C)1 corresponds to the effect on production of sector i when the final demand for the output of sector j increases by one unit. Since these entries are all positive, an increase in the final demand for any sector will cause the production of all sectors to increase. Thus an increase in the demand for any sector will lead to an increase in the demand for all sectors. 11. (Solution in study Guide) Following the hint in the text, compute pTx in two ways. First, take the transpose of both sides of the price equation, p = C Tp + v, to obtain pT = (C T p + v)T = (C T p)T + vT = pT C + vT
and right-multiply by x to get
pTx = (pT C + vT )x = pT Cx + vT x
Another way to compute pTx starts with the production equation x = Cx + d. Left multiply by pT to get
pT x = pT (Cx + d) = pT C x + pT d
The two expression for pTx show that
pT C x + v T x = pT C x + pT d
so vTx = pTd. The Study Guide also provides a slightly different solution.
12. Since
Dm +1 = I + C + C 2 + ... + C m +1 = I + C ( I + C + ... + C m ) = I + CDm Dm +1 may be found iteratively by Dm +1 = I + CDm .
2.6
Solutions
131
13. [M] The matrix I C is
0.8412 -0.0057 -0.0264 -0.3299 -0.0089 -0.1190 -0.0063 0.8412 -0.0057 -0.0264 -0.3299 -0.0089 -0.1190 -0.0063
-0.0064 0.7355 -0.1506 -0.0565 -0.0081 -0.0901 -0.0126 -0.0064 0.7355 -0.1506 -0.0565 -0.0081 -0.0901 -0.0126
-0.0025 -0.0436 0.6443 -0.0495 -0.0333 -0.0996 -0.0196 -0.0025 -0.0436 0.6443 -0.0495 -0.0333 -0.0996 -0.0196 1 0 0 ~ 0 0 0 0 0 1 0 0 0 0 0
-0.0304 -0.0099 -0.0139 0.6364 -0.0295 -0.1260 -0.0098 -0.0304 -0.0099 -0.0139 0.6364 -0.0295 -0.1260 -0.0098 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0
-0.0014 -0.0083 -0.0142 -0.0204 0.6588 -0.1722 -0.0064 -0.0014 -0.0083 -0.0142 -0.0204 0.6588 -0.1722 -0.0064 0 0 0 0 0 1 0 0 0 0 0 0 0 1
-0.0083 -0.0201 -0.0070 -0.0483 -0.0237 0.7632 -0.0132 -0.0083 -0.0201 -0.0070 -0.0483 -0.0237 0.7632 -0.0132 99576 97703 51231 131570 49488 329554 13835
-0.1594 -0.3413 -0.0236 -0.0649 -0.0020 -0.3369 0.9988 74000 -0.1594 -0.3413 56000 10500 -0.0236 25000 -0.0649 17500 -0.0020 -0.3369 196000 0.9988 5000
so the augmented matrix [ I - C d ] may be row reduced to find
so x = (99576, 97703, 51321, 131570, 49488, 329554, 13835). Since the entries in d seem to be accurate to the nearest thousand, a more realistic answer would be x = (100000, 98000, 51000, 132000, 49000, 330000, 14000).
14. [M] The augmented matrix [ I - C d ] in this case may be row reduced to find
0.8412 -0.0057 -0.0264 -0.3299 -0.0089 -0.1190 -0.0063
-0.0064 0.7355 -0.1506 -0.0565 -0.0081 -0.0901 -0.0126
-0.0025 -0.0436 0.6443 -0.0495 -0.0333 -0.0996 -0.0196
-0.0304 -0.0099 -0.0139 0.6364 -0.0295 -0.1260 -0.0098
-0.0014 -0.0083 -0.0142 -0.0204 0.6588 -0.1722 -0.0064
-0.0083 -0.0201 -0.0070 -0.0483 -0.0237 0.7632 -0.0132
-0.1594 -0.3413 -0.0236 -0.0649 -0.0020 -0.3369 0.9988
99640 75548 14444 33501 23527 263985 6526
132 CHAPTER 2 Matrix Algebra
1 0 0 ~ 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
134034 131687 69472 176912 66596 443773 18431
so x = (134034, 131687, 69472, 176912, 66596, 443773, 18431). To the nearest thousand, x = (134000, 132000, 69000, 177000, 67000, 444000, 18000).
15. [M] Here are the iterations rounded to the nearest tenth: x(0) = (74000.0, 56000.0, 10500.0, 25000.0, 17500.0, 196000.0, 5000.0) x(1) = (89344.2, 77730.5, 26708.1, 72334.7, 30325.6, 265158.2, 9327.8) x(2) = (94681.2, 87714.5, 37577.3, 100520.5, 38598.0, 296563.8, 11480.0) x(3) = (97091.9, 92573.1, 43867.8, 115457.0, 43491.0, 312319.0, 12598.8) x(4) = (98291.6, 95033.2, 47314.5, 123202.5, 46247.0, 320502.4, 13185.5) x(5) = (98907.2, 96305.3, 49160.6, 127213.7, 47756.4, 324796.1, 13493.8) x(6) = (99226.6, 96969.6, 50139.6, 129296.7, 48569.3, 327053.8, 13655.9) x(7) = (99393.1, 97317.8, 50656.4, 130381.6, 49002.8, 328240.9, 13741.1) x(8) = (99480.0, 97500.7, 50928.7, 130948.0, 49232.5, 328864.7, 13785.9) x(9) = (99525.5, 97596.8, 51071.9, 131244.1, 49353.8, 329192.3, 13809.4) x(10) = (99549.4, 97647.2, 51147.2, 131399.2, 49417.7, 329364.4, 13821.7) x(11) = (99561.9, 97673.7, 51186.8, 131480.4, 49451.3, 329454.7, 13828.2) x(12) = (99568.4, 97687.6, 51207.5, 131523.0, 49469.0, 329502.1, 13831.6)
so x(12) is the first vector whose entries are accurate to the nearest thousand. The calculation of x(12) takes about 1260 flops, while the row reduction above takes about 550 flops. If C is larger than 20 20, then fewer flops are required to compute x(12) by iteration than by row reduction. The advantage of the iterative method increases with the size of C. The matrix C also becomes more sparse for larger models, so fewer iterations are needed for good accuracy.
2.7
SOLUTIONS
Notes: The content of this section seems to have universal appeal with students. It also provides practice with composition of linear transformations. The case study for Chapter 2 concerns computer graphics see this case study (available as a project on the website) for more examples of computer graphics in action. The Study Guide encourages the student to examine the book by Foley referenced in the text. This section could form the beginning of an independent study on computer graphics with an interested student.
2.7
Solutions
133
1. Refer to Example 5. The representation in homogenous coordinates can be written as a partitioned matrix A 0 of the form T , where A is the matrix of the linear transformation. Since in this case 1 0 1 .25 , the representation of the transformation with respect to homogenous coordinates is A= 1 0
1 .25 0 1 0 0
0 0 1
Note: The Study Guide shows the student why the action of
action of A on x.
A 0
T
0 x on the vector corresponds to the 1 1
-1 2. The matrix of the transformation is A = 0 -1 AD = 0
0 5 1 0 2 2 4 -5 = 3 0 -2 2
0 , so the transformed data matrix is 1 - 4 3
Both the original triangle and the transformed triangle are shown in the following sketch.
x2
2 5 5 x1
3. Following Examples 46,
2/2 2/2 0 .8 4. 0 0
0 1.2 0
- 2/2 2/2 0 0 1 0
0 1 0 0 1 0 -2 .8 3 = 0 1 0 0 -1 0
0 1 0
3 2 / 2 1 = 2 / 2 1 0 0 1.2 0 -1.6 3.6 1
- 2/2 2/2 0
2 2 2 1
0 1 0 0 1 0 -1/ 2 3/2 0
3/2 5. 1/ 2 0 1 6. 0 0
0 -1 0
0 1 0 0 1 0
0 3 / 2 0 = 1/ 2 1 0 0 3 / 2 0 = -1/ 2 1 0
1/ 2 - 3/2 0 -1/ 2 - 3/2 0
0 0 1 0 0 1
0 3 / 2 0 1/ 2 1 0
-1/ 2 3/2 0
134 CHAPTER 2 Matrix Algebra
7. A 60 rotation about the origin is given in homogeneous coordinates by the matrix 1/ 2 - 3 / 2 0 1/ 2 0 . To rotate about the point (6, 8), first translate by (6, 8), then rotate about the 3/2 0 0 1 origin, then translate back by (6, 8) (see the Practice Problem in this section). A 60 rotation about (6, 8) is thus is given in homogeneous coordinates by the matrix
1 0 0
0 1 0
6 1/ 2 8 3 / 2 1 0
- 3/2 1/ 2 0
0 1 0 0 1 0
0 1 0
-6 1/ 2 -8 = 3 / 2 1 0
- 3/2 1/ 2 0
3 + 4 3 4 - 3 3 1
8. A 45 rotation about the origin is given in homogeneous coordinates by the matrix 2 / 2 - 2 / 2 0 2 / 2 0 . To rotate about the point (3, 7), first translate by (3, 7), then rotate about the 2/2 0 0 1
origin, then translate back by (3, 7) (see the Practice Problem in this section). A 45 rotation about (3, 7) is thus is given in homogeneous coordinates by the matrix
1 0 0
0 1 0
3 2 / 2 7 2 / 2 1 0
- 2/2 2/2 0
0 1 0 0 1 0
0 1 0
-3 2 / 2 -7 = 2 / 2 1 0
- 2/2 2/2 0
3+ 2 2 7 - 5 2 1
9. To produce each entry in BD two multiplications are necessary. Since BD is a 2 200 matrix, it will take 2 2 200 = 800 multiplications to compute BD. By the same reasoning it will take 2 2 200 = 800 multiplications to compute A(BD). Thus to compute A(BD) from the beginning will take 800 + 800 = 1600 multiplications. To compute the 2 2 matrix AB it will take 2 2 2 = 8 multiplications, and to compute (AB)D it will take 2 2 200 = 800 multiplications. Thus to compute (AB)D from the beginning will take 8 + 800 = 808 multiplications. For computer graphics calculations that require applying multiple transformations to data matrices, it is thus more efficient to compute the product of the transformation matrices before applying the result to the data matrix. 10. Let the transformation matrices in homogeneous coordinates for the dilation, rotation, and translation be called respectively D, and R, and T. Then for some value of s, , h, and k,
s D = 0 0
0 s 0
0 cos , R = sin 0 1 0 - s sin s cos 0
- sin cos 0
0 1 , T = 0 0 0 1
0 1 0
h k 1
Compute the products of these matrices:
s cos DR = s sin 0
0 s cos , RD = s sin 0 0 1
- s sin s cos 0
0 0 1
2.7
Solutions
135
s DT = 0 0
0 s 0
sh s , TD = 0 sk 0 1
0 s 0
h k 1
cos RT = sin 0
- sin cos 0
h cos - k sin cos , TR = sin h sin + k cos 0 1
- sin cos 0
h k 1
Since DR = RD, DT TD and RT TR, D and R commute, D and T do not commute and R and T do not commute.
11. To simplify A 2 A 1 completely, the following trigonometric identities will be needed:
sin 1. - tan cos = - cos cos = - sin
sin sin 1 2. sec - tan sin = cos - cos sin = 1-cos = cos = cos cos
2 2
Using these identities,
sec A2 A1 = 0 0
- tan 1 0
0 1 0 sin 1 0
0 cos 0
0 0 1 0 0 1
sec - tan sin sin = 0 cos = sin 0
- sin cos 0 0 0 1
- tan cos cos 0
which is the transformation matrix in homogeneous coordinates for a rotation in 1 - cos sin = sin 1 + cos 1 - cos sin = cos sin
2
.
12. To simplify this product completely, the following trigonometric identity will be needed:
tan / 2 =
This identity has two important consequences: 1 - (tan / 2)(sin ) = 1 -
(cos )(- tan / 2) - tan / 2 = -(cos + 1) tan / 2 = -(cos + 1) The product may be computed and simplified using these results:
sin = - sin 1 + cos
1 0 0
- tan / 2 1 0
0 1 0 sin 1 0
0 1 0
0 1 0 0 1 0
- tan / 2 1 0
0 0 1 0 0 1
1 - (tan / 2)(sin ) sin = 0
- tan / 2 1 0
0 1 0 0 1 0
- tan / 2 1 0
136 CHAPTER 2 Matrix Algebra
cos = sin 0 cos = sin 0 cos = sin 0
- tan / 2 1 0
0 1 0 0 1 0
- tan / 2 1 0
(cos )(- tan / 2) - tan / 2 -(sin )(tan / 2) + 1 0 - sin cos 0 0 0 1
0 0 1 0 0 1
which is the transformation matrix in homogeneous coordinates for a rotation in
2
.
13. Consider first applying the linear transformation on 2 whose matrix is A, then applying a translation by the vector p to the result. The matrix representation in homogeneous coordinates of the linear A 0 transformation is T , while the matrix representation in homogeneous coordinates of the 1 0 p I translation is T . Applying these transformations in order leads to a transformation whose matrix 1 0 representation in homogeneous coordinates is p A 0 A I T = 1 0T 1 0T 0 which is the desired matrix. p 1
14. The matrix for the transformation in Exercise 7 was found to be
1/ 2 3/2 0
- 3/2 1/ 2 0
3 + 4 3 4 - 3 3 1
A This matrix is of the form T 0
p , where 1
1/ 2 A= 3/2
3 + 4 3 - 3 / 2 ,p = 1/ 2 4 - 3 3
By Exercise 13, this matrix may be written as p A 0 I T 1 0 T 1 0 that is, the composition of a linear transformation on 2 and a translation. The matrix A is the matrix of a rotation about the origin in 2. Thus the transformation in Exercise 7 is the composition of a rotation 3 + 4 3 about the origin and a translation by p = . 4 - 3 3
2.7
1 15. Since ( X , Y , Z , H ) = ( 1 , - 1 , 1 , 24 ), the corresponding point in 2 4 8
3
Solutions
137
has coordinates
X Y Z ( x, y , z ) = , , H H H
=
1 2 1 24
,
-1 4
1 24
,
1 8 1 24
= (12, -6,3)
16. The homogeneous coordinates (1, 2, 3, 4) represent the point (1/ 4, - 2 / 4, 3/ 4) = (1/ 4, - 1/ 2, 3/ 4) while the homogeneous coordinates (10, 20, 30, 40) represent the point (10 / 40, - 20 / 40, 30 / 40) = (1/ 4, - 1/ 2, 3/ 4) so the two sets of homogeneous coordinates represent the same point in
3
.
17. Follow Example 7a by first constructing that 3 3 matrix for this rotation. The vector e1 is not changed by this rotation. The vector e2 is rotated 60 toward the positive z-axis, ending up at the point (0, cos 60, sin 60) = (0, 1/ 2, 3 / 2). The vector e3 is rotated 60 toward the negative y-axis, stopping at the point (0, cos 150, sin 150) = (0, - 3 / 2, 1/ 2). The matrix A for this rotation is thus
1 A = 0 0
0 1/ 2 3/2
0 - 3 / 2 1/ 2 0 1/ 2 3/2 0 0 - 3/2 1/ 2 0 0 0 0 1
so in homogeneous coordinates the transformation is represented by the matrix
A T 0 1 0 0 = 1 0 0
18. First construct the 3 3 matrix for the rotation. The vector e1 is rotated 30 toward the negative y-axis, ending up at the point (cos(30), sin (30), 0) = ( 3 / 2, -1/ 2, 0). The vector e2 is rotated 60 toward the positive x-axis, ending up at the point (cos 60, sin 60, 0) = (1/ 2, 3 / 2, 0). The vector e3 is not changed by the rotation. The matrix A for the rotation is thus
3/2 1/ 2 0 A = -1/ 2 3 / 2 0 0 0 1 so in homogeneous coordinates the rotation is represented by the matrix A T 0 3/2 0 -1/ 2 = 1 0 0
1/ 2 3/2 0 0
0 0 1 0
0 0 0 1
Following Example 7b, in homogeneous coordinates the translation by the vector (5, 2, 1) is represented by the matrix 5 1 0 0 0 1 0 -2 0 0 1 1 1 0 0 0
138 CHAPTER 2 Matrix Algebra
Thus the complete transformation is represented in homogeneous coordinates by the matrix
1 0 0 0
0 1 0 0
0 0 1 0
5 3 / 2 -2 -1/ 2 1 0 1 0
1/ 2 3/2 0 0
0 0 1 0
0 3 / 2 0 -1/ 2 = 0 0 1 0
1/ 2 3/2 0 0
0 0 1 0
5 -2 1 1
19. Referring to the material preceding Example 8 in the text, we find that the matrix P that performs a perspective projection with center of projection (0, 0, 10) is
1 0 0 0
0 1 0 0
0 0 0 -.1
0 0 0 1
The homogeneous coordinates of the vertices of the triangle may be written as (4.2, 1.2, 4, 1), (6, 4, 2, 1), and (2, 2, 6, 1), so the data matrix for S is
4.2 6 2 1.2 4 2 4 2 6 1 1 1 and the data matrix for the transformed triangle is
0 0 4.2 6 2 4.2 6 2 1 0 0 1 0 0 1.2 4 2 1.2 4 2 = 0 0 0 0 4 2 6 0 0 0 0 0 -.1 1 1 1 1 .6 .8 .4 Finally, the columns of this matrix may be converted from homogeneous coordinates by dividing by the final coordinate: (4.2, 1.2, 0, .6) (4.2 /.6, 1.2 /.6, 0 /.6) = (7, 2, 0) (6, 4, 0, .8) (2, 2, 0, .4) (6 /.8, 2 /.8, 0 /.8) = (7.5, 5, 0) (2 /.4, 2 /.4, 0 /.4) = (5, 5, 0)
So the coordinates of the vertices of the transformed triangle are (7, 2, 0), (7.5, 5, 0), and (5, 5, 0). 20. As in the previous exercise, the matrix P that performs the perspective projection is
1 0 0 0
0 1 0 0
0 0 0 -.1
0 0 0 1
The homogeneous coordinates of the vertices of the triangle may be written as (9, 3, 5, 1), (12, 8, 2, 1), and (1.8, 2.7, 1, 1), so the data matrix for S is
9 12 3 8 -5 2 1 1
1.8 2.7 1 1
2.8
Solutions
139
and the data matrix for the transformed triangle is 0 0 9 12 1.8 9 12 1.8 1 0 0 1 0 0 3 8 2.7 3 8 2.7 = 0 0 0 0 -5 2 1 0 0 0 1 1 1.5 .8 .9 0 0 -.1 1 1 Finally, the columns of this matrix may be converted from homogeneous coordinates by dividing by the final coordinate: (9, 3, 0, 1.5) (9 /1.5, 3/1.5, 0 /1.5) = (6, 2, 0) (12, 8, 0, .8) (12 /.8, 8 /.8, 0 /.8) = (15, 10, 0) (1.8, 2.7, 0, .9) (1.8 /.9, 2.7 /.9, 0 /.9) = (2, 3, 0) So the coordinates of the vertices of the transformed triangle are (6, 2, 0), (15, 10, 0), and (2, 3, 0). 21. [M] Solve the given equation for the vector (R, G, B), giving
R .61 .29 G = .35 .59 B .04 .12
-1
.15 .063 .787
X 2.2586 Y = -1.3495 Z .0910
-1
-1.0395 2.3441 -.3046
-.3473 X .0696 Y 1.2777 Z
22. [M] Solve the given equation for the vector (R, G, B), giving
R .299 G = .596 B .212
.587 -.275 -.528
.114 -.321 .311
Y 1.0031 I = .9968 Q 1.0085
.9548 -.2707 -1.1105
.6179 Y -.6448 I 1.6996 Q
2.8
SOLUTIONS
Notes: Cover this section only if you plan to skip most or all of Chapter 4. This section and the next cover
everything you need from Sections 4.14.6 to discuss the topics in Section 4.9 and Chapters 57 (except for the general inner product spaces in Sections 6.7 and 6.8). Students may use Section 4.2 for review, particularly the Table near the end of the section. (The final subsection on linear transformations should be omitted.) Example 6 and the associated exercises are critical for work with eigenspaces in Chapters 5 and 7. Exercises 3136 review the Invertible Matrix Theorem. New statements will be added to this theorem in Section 2.9. Key Exercises: 520 and 2326. 1. The set is closed under sums but not under multiplication by a negative scalar. A counterexample to the subspace condition is shown at the right.
u
(1)u
Note: Most students prefer to give a geometric counterexample, but some may choose an algebraic calculation. The four exercises here should help students develop an understanding of subspaces, but they may be insufficient if you want students to be able to analyze an unfamiliar set on an exam. Developing that skill seems more appropriate for classes covering Sections 4.14.6.
140 CHAPTER 2 Matrix Algebra
2. The set is closed under scalar multiples but not sums. For example, the sum of the vectors u and v shown here is not in H.
u v u+v
3. No. The set is not closed under sums or scalar multiples. The subset consisting of the points on the line x2 = x1 is a subspace, so any "counterexample" must use at least one point not on this line. Here are two counterexamples to the subspace conditions:
v
u u+v
3u
4. No. The set is closed under sums, but not under multiplication by a negative scalar.
(1)u
u
5. The vector w is in the subspace generated by v1 and v2 if and only if the vector equation x1v1 + x2v2 = w is consistent. The row operations below show that w is not in the subspace generated by v1 and v2.
2 [ v1 v 2 w ] ~ 3 -5
-4 -5 8
8 2 2 ~ 0 -9 0
-4 1 -2
8 2 -10 ~ 0 11 0
-4 1 0
8 -10 -9
6. The vector u is in the subspace generated by {v1, v2, v3} if and only if the vector equation x1v1 + x2v2 + x3v3 = u is consistent. The row operations below show that u is not in the subspace generated by {v1, v2, v3}.
1 -2 v 3 u] ~ 4 3
[ v1 v 2
4 -7 9 7
5 -8 6 5
-4 1 10 0 ~ -7 0 -5 0
4 1 -7 -5
5 2 -14 -10
-4 1 2 0 ~ 9 0 7 0
4 1 0 0
5 2 0 0
-4 2 23 17
Note: For a quiz, you could use w = (1, 3, 11, 8), which is in Span{v1, v2, v3}.
7. a. There are three vectors: v1, v2, and v3 in the set {v1, v2, v3}. b. There are infinitely many vectors in Span{v1, v2, v3} = Col A. c. Deciding whether p is in Col A requires calculation:
2 [ A p] ~ -8 6
-3 8 -7
-4 6 -7
6 2 -10 ~ 0 11 0
-3 -4 2
-4 -10 5
6 2 14 ~ 0 -7 0
-3 -4 0
-4 -10 0
6 14 0
The equation Ax = p has a solution, so p is in Col A.
2.8
Solutions
141
-3 8. [ A p] = 0 6
-2 2 3
0 -6 3
1 -3 14 ~ 0 -9 0
-2 2 -1
0 -6 3
1 -3 14 ~ 0 -7 0
-2 2 0
0 -6 0
1 14 0
Yes, the augmented matrix [A p] corresponds to a consistent system, so p is in Col A. 9. To determine whether p is in Nul A, simply compute Ap. Using A and p as in Exercise 7,
2 Ap = -8 6 -3 Au = 0 6
-3 8 -7
-4 6 -2 6 -10 = -62 . Since Ap 0, p is not in Nul A. -7 11 29 0 -2 0 -6 3 = 0 . Yes, u is in Nul A. 3 1 0
10. To determine whether u is in Nul A, simply compute Au. Using A as in Exercise 7 and u = (2, 3, 1), -2 2 3
11. p = 4 and q = 3. Nul A is a subspace of R4 because solutions of Ax = 0 must have 4 entries, to match the columns of A. Col A is a subspace of R3 because each column vector has 3 entries. 12. p = 3 and q = 4. Nul A is a subspace of R3 because solutions of Ax = 0 must have 3 entries, to match the columns of A. Col A is a subspace of R4 because each column vector has 4 entries. 13. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0. (Include an augmented column of zeros, to avoid errors.)
3 -9 9 3 ~ 0 0
2 -4 2 2 1 0
1 1 -5 1 2 0
-5 7 1 -5 -4 0
0 3 0 ~ 0 0 0 0 1 0 ~ 0 0 0
2 2 -4 0 1 0
1 4 -8 -1 2 0
-5 -8 16 1 -4 0
0 3 0 ~ 0 0 0 0 0 , 0 x1
2 2 0
1 4 0
-5 -8 0
0 0 0
- x3 + x4 = 0 x2 + 2 x3 - 4 x4 = 0 0 = 0
The general solution is x1 = x3 x4, and x2 = 2x3 + 4x4, with x3 and x4 free. The general solution in parametric vector form is not needed. All that is required here is one nonzero vector. So choose any values for x3 and x4 (not both zero). For instance, set x3 = 1 and x4 = 0 to obtain the vector (1, 2, 1, 0) in Nul A.
Note: Section 2.8 of Study Guide introduces the ref command (or rref, depending on the technology), which produces the reduced echelon form of a matrix. This will greatly speed up homework for students who have a matrix program available.
14. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0:
1 4 -5 2
2 3 5 7 -1 0 7 11
0 1 0 0 ~ 0 0 0 0
2 -3 9 3
3 -5 15 5
0 1 0 0 ~ 0 0 0 0
2 1 0 0
3 5/3 0 0
0 1 0 0 ~ 0 0 0 0
0 1 0 0
-1/ 3 5/3 0 0
0 0 0 0
The general solution is x1 = (1/3)x3 and x2 = (5/3) x3, with x3 free. The general solution in parametric vector form is not needed. All that is required here is one nonzero vector. So choose any values of x3 and x4 (not both zero). For instance, set x3 = 3 to obtain the vector (1, 5, 3) in Nul A.
142 CHAPTER 2 Matrix Algebra
15. Yes. Let A be the matrix whose columns are the vectors given. Then A is invertible because its determinant is nonzero, and so its columns form a basis for R2, by the Invertible Matrix Theorem (or by Example 5). (Other reasons for the invertibility of A could be given.) 16. No. One vector is a multiple of the other, so they are linearly dependent and hence cannot be a basis for any subspace. 17. No. Place the three vectors into a 33 matrix A and determine whether A is invertible:
0 A= 1 -2
5 -7 4
6 1 3 ~ 0 5 -2
-7 5 4
3 1 6 ~ 0 5 0
-7 5 -10
3 1 6 ~ 0 11 0
-7 5 0
3 6 23
The matrix A has three pivots, so A is invertible by the IMT and its columns form a basis for R3 (as pointed out in Example 5). 18. Yes. Place the three vectors into a 33 matrix A and determine whether A is invertible:
1 A= 1 -2
-5 -1 2
7 1 0 ~ 0 -5 0
-5 4 -8
7 1 -7 ~ 0 9 0
-5 4 0
7 -7 -5
The matrix A has three pivots, so A is invertible by the IMT and its columns form a basis for R3 (as pointed out in Example 5). 19. No. The vectors cannot be a basis for R3 because they only span a plan in R3. Or, point out that the 1 -5 columns of the matrix 1 -1 cannot possibly span R3 because the matrix cannot have a pivot in -2 2 every row. So the columns are not a basis for R3.
Note: The Study Guide warns students not to say that the two vectors here are a basis for R2.
20. No. The vectors are linearly dependent because there are more vectors in the set than entries in each vector. (Theorem 8 in Section 1.7.) So the vectors cannot be a basis for any subspace. 21. a. b. c. d. e. False. See the definition at the beginning of the section. The critical phrases "for each" are missing. True. See the paragraph before Example 4. False. See Theorem 12. The null space is a subspace of Rn, not Rm. True. See Example 5. True. See the first part of the solution of Example 8.
22. a. False. See the definition at the beginning of the section. The condition about the zero vector is only one of the conditions for a subspace. b. True. See Example 3. c. True. See Theorem 12. d. False. See the paragraph after Example 4. e. False. See the Warning that follows Theorem 13.
2.8
Solutions
143
4 23. (Solution in Study Guide) A = 6 3
5 5 4
9 1 8
-2 1 12 ~ 0 -3 0
2 1 0
6 5 0
-5 -6 . The echelon form identifies 0
4 5 columns 1 and 2 as the pivot columns. A basis for Col A uses columns 1 and 2 of A: 6 , 5 . This is not 3 4 the only choice, but it is the "standard" choice. A wrong choice is to select columns 1 and 2 of the echelon form. These columns have zero in the third entry and could not possibly generate the columns displayed in A.
24. For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:
1 0 0
0 1 0
-4 5 0
7 -6 0
0 x1 - 4 x3 + 7 x4 = 0 x2 + 5 x3 - 6 x4 = 0 . 0 . This corresponds to: 0 0 = 0
4 -7 -5 6 , 1 0 0 1
Solve for the basic variables and write the solution of Ax = 0 in parametric vector form:
x1 4 x3 - 7 x4 4 -7 x -5 x + 6 x -5 3 4 2 = + x 6 . Basis for Nul A: =x x3 3 1 4 0 x3 x4 0 1 x4
Notes: (1) A basis is a set of vectors. For simplicity, the answers here and in the text list the vectors without
enclosing the list inside set brackets. This style is also easier for students. I am careful, however, to distinguish between a matrix and the set or list whose elements are the columns of the matrix. (2) Recall from Chapter 1 that students are encouraged to use the augmented matrix when solving Ax = 0, to avoid the common error of misinterpreting the reduced echelon form of A as itself the augmented matrix for a nonhomogeneous system. (3) Because the concept of a basis is just being introduced, I insist that my students write the parametric vector form of the solution of Ax = 0. They see how the basis vectors span the solution space and are obviously linearly independent. A shortcut, which some instructors might introduce later in the course, is only to solve for the basic variables and to produce each basis vector one at a time. Namely, set all free variables equal to zero except for one free variable, and set that variable equal to a suitable nonzero number.
-3 24. A = 2 3 -3 -2 2 , 4 . 3 -2 For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:
9 -6 -9
-2 4 -2
-7 1 8 ~ 0 2 0
-3 0 0
6 4 0
9 5 . Basis for Col A: 0
x1 - 3x2 + 1.50 x4 = 0 1 -3 0 1.50 0 0 . This corresponds to: x3 + 1.25 x4 = 0 . 0 1 1.25 0 0 0 0 0 0 0 = 0 Solve for the basic variables and write the solution of Ax = 0 in parametric vector form:
x1 3 x2 - 1.5 x4 3 -1.5 x 1 x2 2 = = x + x 0 . Basis for Nul A: 2 x3 -1.25 x4 0 4 -1.25 x4 0 1 x4 3 -1.5 1 0 , . 0 -1.25 0 1
144 CHAPTER 2 Matrix Algebra
1 -1 25. A = -2 3
4 2 2 6
8 7 9 9
-3 3 5 -5 0 1 0 0
-7 1 4 0 ~ 5 0 -2 0 -2 2.5 0 0 0 0 1 0
4 2 0 0 7 -.5 4 0
8 5 0 0 0 0 . 0 0
0 0 1 0
5 -1 . Basis for Col A: 4 0 x1 - 2 x3 x2 + 2.5 x3
1 -1 , -2 3
4 -3 2 3 , . 2 5 6 -5
For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:
1 0 [ A 0] ~ 0 0
+ 7 x5 = - .5 x5 = x4 + 4 x5 = 0 =
0 0 . 0 0
x1 2 x3 - 7 x5 2 -7 x -2.5 x + .5 x -2.5 .5 3 5 2 = x3 1 + x5 0 . Thesolution of Ax = 0in parametric vector form : x3 = x3 -4 x5 x4 0 -4 x5 0 1 x5
u Basis for Nul A: {u, v}.
v
Note: The solution above illustrates how students could write a solution on an exam, when time is precious, namely, describe the basis by giving names to appropriate vectors found in the calculations.
3 -2 26. A = -5 -2
-1 2 9 6
7 -2 3 6 0 1 0 0
3 7 3 3 3 2 0 0
9 3 5 0 ~ 4 0 7 0 0 0 1 0 2.5 1.5 1 0
-1 2 0 0 0 0 . 0 0
7 4 0 0
0 0 1 0 x1
6 3 . Basis for Col A: 1 0 + 3x x2 + 2 x3
3 -1 3 -2 2 7 , , . -5 9 3 -2 6 3
For Nul A,
1 0 [ A 0] ~ 0 0
+ 2.5 x5 = + 1.5 x5 = x4 + x5 = 0 =
0 0 0 0
The solution of Ax = 0 in parametric vector form:
x1 -3 x3 - 2.5 x5 -3 -2.5 x -2 x - 1.5 x -2 -1.5 3 5 2 x3 = = x3 1 + x5 0 x3 - x5 x4 0 -1 . Basis for Nul A: {u, v}. x5 0 1 x5
u
v
27. Construct a nonzero 33 matrix A and construct b to be almost any convenient linear combination of the columns of A.
2.8
Solutions
145
28. The easiest construction is to write a 33 matrix in echelon form that has only 2 pivots, and let b be any vector in R3 whose third entry is nonzero. 29. (Solution in Study Guide) A simple construction is to write any nonzero 33 matrix whose columns are obviously linearly dependent, and then make b a vector of weights from a linear dependence relation among the columns. For instance, if the first two columns of A are equal, then b could be (1, 1, 0). 30. Since Col A is the set of all linear combinations of a1, ... , ap, the set {a1, ... , ap} spans Col A. Because {a1, ... , ap} is also linearly independent, it is a basis for Col A. (There is no need to discuss pivot columns and Theorem 13, though a proof could be given using this information.) 31. If Col F R5, then the columns of F do not span R5. Since F is square, the IMT shows that F is not invertible and the equation Fx = 0 has a nontrivial solution. That is, Nul F contains a nonzero vector. Another way to describe this is to write Nul F {0}. 32. If Nul R contains nonzero vectors, then the equation Rx = 0 has nontrivial solutions. Since R is square, the IMT shows that R is not invertible and the columns of R do not span R6. So Col R is a subspace of R6, but Col R R6. 33. If Col Q = R4, then the columns of Q span R4. Since Q is square, the IMT shows that Q is invertible and the equation Qx = b has a solution for each b in R4. Also, each solution is unique, by Theorem 5 in Section 2.2. 34. If Nul P = {0}, then the equation Px = 0 has only the trivial solution. Since P is square, the IMT shows that P is invertible and the equation Px = b has a solution for each b in R5. Also, each solution is unique, by Theorem 5 in Section 2.2. 35. If the columns of B are linearly independent, then the equation Bx = 0 has only the trivial (zero) solution. That is, Nul B = {0}. 36. If the columns of A form a basis, they are linearly independent. This means that A cannot have more columns than rows. Since the columns also span Rm, A must have a pivot in each row, which means that A cannot have more rows than columns. As a result, A must be a square matrix. 37. [M] Use the command that produces the reduced echelon form in one step (ref or rref depending on the program). See the Section 2.8 in the Study Guide for details. By Theorem 13, the pivot columns of A form a basis for Col A.
3 -7 A= -5 3
-5 9 7 -7
0 -4 -2 -3
-1 9 5 4
3 1 -11 0 ~ -7 0 0 0
0 1 0 0
2.5 1.5 0 0
-4.5 -2.5 0 0
3.5 1.5 Basis for Col A: 0 0
3 -5 -7 9 , -5 7 3 -7
For Nul A, obtain the solution of Ax = 0 in parametric vector form: x1 + 2.5 x3 - 4.5 x4 + 3.5 x5 = 0 x2 + 1.5 x3 - 2.5 x4 + 1.5 x5 = 0
x1 = -2.5 x3 + 4.5 x4 - 3.5 x5 Solution: x2 = -1.5 x3 + 2.5 x4 - 1.5 x5 x , x , and x are free 5 3 4
146 CHAPTER 2 Matrix Algebra
x1 -2.5 x3 + 4.5 x4 - 3.5 x5 -2.5 4.5 -3.5 x -1.5 x + 2.5 x - 1.5 x -1.5 2.5 -1.5 3 4 5 2 = x3 1 + x4 0 + x5 0 = x3u + x4v + x5w x = x3 = x3 x4 x4 0 1 0 x5 0 0 1 x5 By the argument in Example 6, a basis for Nul A is {u, v, w}. 5 4 38. [M] A = 5 -8 2 1 1 -5 0 2 3 6 -8 -8 5 8 -8 1 -9 0 ~ 19 0 5 0 0 1 0 0 0 0 1 0 60 -154 -47 0 122 -309 . -94 0
5 2 0 4 1 2 The pivot columns of A form a basis for Col A: , , . 5 1 3 -8 -5 6 x1 For Nul A, solve Ax = 0: x1 x 2 Solution: x3 x4 x2 + 60 x4 + 122 x5 = 0 - 154 x4 - 309 x5 = 0 x3 - 47 x4 - 94 x5 = 0
= -60 x4 - 122 x5 = 154 x4 + 309 x5 = 47 x4 + 94 x5 and x5 are free
x1 -60 x4 - 122 x5 -60 -122 x 154 x + 309 x 154 309 4 5 2 x = x3 = 47 x4 + 94 x5 = x4 47 + x5 94 = x4u + x5v x4 x4 1 0 x5 0 1 x5 By the method of Example 6, a basis for Nul A is {u, v}
Note: The Study Guide for Section 2.8 gives directions for students to construct a review sheet for the
concept of a subspace and the two main types of subspaces, Col A and Nul A, and a review sheet for the concept of a basis. I encourage you to consider making this an assignment for your class.
2.9
SOLUTIONS
Notes: This section contains the ideas from Sections 4.44.6 that are needed for later work in Chapters 57.
If you have time, you can enrich the geometric content of "coordinate systems" by discussing crystal lattices (Example 3 and Exercises 35 and 36 in Section 4.4.) Some students might profit from reading Examples 13 from Section 4.4 and Examples 2, 4, and 5 from Section 4.6. Section 4.5 is probably not a good reference for students who have not considered general vector spaces. Coordinate vectors are important mainly to give an intuitive and geometric feeling for the isomorphism between a k-dimensional subspace and Rk. If you plan to omit Sections 5.4, 5.6, 5.7 and 7.2, you can safely omit Exercises 18 here.
2.9
Solutions
147
Exercises 116 may be assigned after students have read as far as Example 2. Exercises 19 and 20 use the Rank Theorem, but they can also be assigned before the Rank Theorem is discussed. The Rank Theorem in this section omits the nontrivial fact about Row A which is included in the Rank Theorem of Section 4.6, but that is used only in Section 7.4. The row space itself can be introduced in Section 6.2, for use in Chapter 6 and Section 7.4. Exercises 916 include important review of techniques taught in Section 2.8 (and in Sections 1.2 and 2.5). They make good test questions because they require little arithmetic. My students need the practice here. Nearly every time I teach the course and start Chapter 5, I find that at least one or two students cannot find a basis for a two-dimensional eigenspace! 3 1. If [x]B = , then x is formed from b1 and b2 using 2 weights 3 and 2: 1 2 7 x = 3b1 + 2b2 = 3 + 2 = 1 -1 1
x2 3b 1 2b 1 b1 x x1 b2 2b 2
-1 2. If [x]B = , then x is formed from b1 and b2 using weights 1 and 3: 3 -2 3 11 x = (1)b1 + 3b2 = (-1) + 3 = 1 1 2
b1 x2 3b 2 2b 2 b2 x x1 b1
3. To find c1 and c2 that satisfy x = c1b1 + c2b2, row reduce the augmented matrix: 1 -2 -3 1 -2 -3 1 0 7 [b1 b 2 x] = ~ ~ . Or, one can write a matrix equation as -4 7 7 0 -1 -5 0 1 5 suggested by Exercise 7 and solve using the matrix inverse. In either case, c 7 [x]B = 1 = . c2 5
1 4. As in Exercise 3, [b1 b 2 x] = -3 c 5 [x]B = 1 = . c2 4 1 x] = 5 -3 -3 -7 5 4 1 10 ~ 0 -7 0 -3 8 -4 4 1 -10 ~ 0 5 0 0 1 0 1/ 4 -5 / 4 . [x]B = 0 c1 1/ 4 c = -5 / 4 . 2 -3 5 -7 1 ~ 5 0 -3 -4 -7 1 ~ -16 0 0 1 5 , and 4
5. [b1 b 2
148 CHAPTER 2 Matrix Algebra
6. [b1 b 2
-3 x] = 1 -4
7 5 -6
11 1 0 ~ 0 7 0
5 22 14
0 1 11 ~ 0 7 0
0 1 0
-5 / 2 1/ 2 0
c -5 / 2 [x]B = 1 = . c2 1/ 2
7. Fig. 1 suggests that w = 2b1 b2 and x = 1.5b1 + .5b2, in which case,
2 [w]B = and [x]B = -1 1.5 .5 . To confirm [x]B, compute
3 - 1 4 1.5 b1 + .5 b 2 = 1.5 + .5 = = x 0 2 1
y x
b2 0 b1
b1
x
0 z
b2
w
Figure 1
Figure 2
Note: Figures 1 and 2 display what Section 4.4 calls B-graph paper.
8. Fig. 2 suggests that x = 2b1 b2, y = 1.5b1 + b2, and z = b1 .5b2. If so, then
2 [x]B = , [y]B = -1 1.5 1.0 , and [z]B = -1 -.5 . To confirm [y]B and [z]B, compute
0 2 2 0 2 -1 1.5 b1 + b 2 = 1.5 + = = y and - b1 - .5 b 2 = -1 - .5 = =z. 2 1 4 2 1 -2.5 1 -3 2 -4 1 -3 -3 9 -1 5 0 0 ~ 9. The information A = 2 -6 4 -3 0 0 -4 12 2 7 0 0 1 2 -4 -3 -1 5 A form a basis for Col A: , , . 2 4 -3 -4 2 7 Columns 1, 2 and 4, of the echelon form certain cannot span Col A since those vectors all have zero in their fourth entries. For Nul A, use the reduced echelon form, augmented with a zero column to insure that the equation Ax = 0 is kept in mind: 2 -4 5 -7 is enough to see that columns 1, 3, and 4 of 0 5 0 0
2.9
Solutions
149
1 0 0 0
-3 0 0 0
0 1 0 0
0 0 1 0
0 0 . 0 0
x1 - 3x2
= 0 x3 = 0 , x4 = 0
x2 is the free variable
x1 3x2 3 3 x x 1 1 x = 2 = 2 = x2 . So is x3 0 0 0 0 0 x4 0
a basis for Nul A. From this information, dim Col A = 3 (because A has three pivot columns) and dim Nul A = 1 (because the equation Ax = 0 has only one free variable). 1 1 10. The information A = -2 4 4 1 -2 9 0 -3 1 -3 ~ -2 0 0 0 -9 0 0 0 1 -2 5 1 -1 5 and 4 of A form a basis for Col A: , , . For Nul A, -2 0 1 4 1 1 9 6 -6 9 5 5 1 1 1 0 [ A 0] ~ 0 0 0 1 0 0 3 -3 0 0 0 0 1 0 0 -7 -2 0 0 0 . 0 0 x1 + 3x3 x2 - 3x3 -2 -1 0 1 5 0 1 0 4 -7 shows that columns 1, 2, -2 0
= 0 - 7 x5 = 0 x4 - 2 x5 = 0 -3 0 3 7 1 , 0 . 0 2 0 1 -1 5 shows that columns 1, 2, 2 0
x3 and x5 are free variables
x1 -3x3 -3 0 x 3 x + 7 x 3 7 5 2 3 = x3 1 + x5 0 . Basis for Nul A: x = x3 = x3 x4 2 x5 0 2 x5 x5 0 1 From this, dim Col A = 3 and dim Nul A = 2. 1 2 11. The information A = -3 3 2 5 -9 10 -5 -8 9 -7 0 4 -7 11 -1 1 3 0 ~ -2 0 7 0 2 1 0 0 -5 2 0 0
0 4 1 0
1 2 0 2 5 4 and 4 of A form a basis for Col A: , , . For Nul A, -3 -9 -7 3 10 11 1 0 A 0] ~ [ 0 0 0 1 0 0 -9 2 0 0 0 0 1 0 5 -3 2 0 - 9 x3 + 5 x5 = 0 0 x1 x2 + 2 x3 - 3x5 = 0 0 . x4 + 2 x5 = 0 0 0 x3 and x5 are free variables
150 CHAPTER 2 Matrix Algebra
x1 9 x3 - 5 x5 9 -5 x -2 x + 3 x -2 3 3 5 2 = x3 1 + x5 0 . Basis for Nul A: x = x3 = x3 x4 -2 x5 0 -2 x5 0 1 x5 From this, dim Col A = 3 and dim Nul A = 2. 1 5 12. The information A = 4 -2 2 10 8 -4 -4 -9 -9 5 3 -7 -2 0 3 1 8 0 ~ 7 0 -6 0 2 0 0 0
9 -2 1 , 0 0
-5 3 0 . -2 1 3 0 shows that columns 1, 3, -5 0
-4 1 0 0
3 -2 0 0
1 -4 3 5 -9 8 and 5 of A form a basis for Col A: , , . For Nul A 4 -9 7 -2 5 -6 1 0 [ A 0] ~ 0 0 2 0 0 0 0 -5 1 -2 0 0 0 0 0 0 1 0 0 0 . 0 0 x1 + 2 x2 - 5 x4 x3 - 2 x4 = 0 = 0 x5 = 0 -2 1 0 , 0 0 5 0 2 . 1 0
x2 and x4 are free variables 5 0 2 . Basis for Nul A: 1 0
x1 -2 x2 + 5 x4 -2 x 1 x2 2 = x2 0 + x4 x = x3 = 2 x4 x4 x4 0 x5 0 0
From this, dim Col A = 3 and dim Nul A = 2.
13. The four vectors span the column space H of a matrix that can be reduced to echelon form:
1 -3 2 -4 -3 9 -6 12 2 -1 4 2 -4 1 5 0 ~ -3 0 7 0 -3 2 0 5 0 0 0 10 -4 1 -7 0 ~ 5 0 -9 0 -3 0 0 0 2 5 0 0 -4 1 -7 0 ~ 5 0 5 0 -3 0 0 0 2 5 0 0 -4 -7 5 0
Columns 1, 3, and 4 of the original matrix form a basis for H, so dim H = 3.
Note: Either Exercise 13 or 14 should be assigned because there are always one or two students who confuse
Col A with Nul A. Or, they wrongly connect "set of linear combinations" with "parametric vector form" (of the general solution of Ax = 0).
14. The five vectors span the column space H of a matrix that can be reduced to echelon form:
1 -1 -2 5 2 -3 -1 6 0 2 -6 8 -1 4 -7 7 3 1 -8 0 ~ 9 0 -5 0 2 -1 3 -4 0 2 -6 8 -1 3 -9 12 3 1 -5 0 ~ 15 0 -20 0 2 -1 0 0 0 2 0 0 -1 3 0 0 3 -5 0 0
Columns 1 and 2 of the original matrix form a basis for H, so dim H = 2.
2.9
Solutions
151
15. Col A = R3, because A has a pivot in each row and so the columns of A span R3. Nul A cannot equal R2, because Nul A is a subspace of R5. It is true, however, that Nul A is two-dimensional. Reason: the equation Ax = 0 has two free variables, because A has five columns and only three of them are pivot columns. 16. Col A cannot be R3 because the columns of A have four entries. (In fact, Col A is a 3-dimensional subspace of R4, because the 3 pivot columns of A form a basis for Col A.) Since A has 7 columns and 3 pivot columns, the equation Ax = 0 has 4 free variables. So, dim Nul A = 4. 17. a. True. This is the definition of a B-coordinate vector. b. False. Dimension is defined only for a subspace. A line must be through the origin in Rn to be a subspace of Rn. c. True. The sentence before Example 1 concludes that the number of pivot columns of A is the rank of A, which is the dimension of Col A by definition. d. True. This is equivalent to the Rank Theorem because rank A is the dimension of Col A. e. True, by the Basis Theorem. In this case, the spanning set is automatically a linearly independent set. 18. a. True. This fact is justified in the second paragraph of this section. b. True. See the second paragraph after Fig. 1. c. False. The dimension of Nul A is the number of free variables in the equation Ax = 0. See Example 2. d. True, by the definition of rank. e. True, by the Basis Theorem. In this case, the linearly independent set is automatically a spanning set. 19. The fact that the solution space of Ax = 0 has a basis of three vectors means that dim Nul A = 3. Since a 57 matrix A has 7 columns, the Rank Theorem shows that rank A = 7 dim Nul A = 4.
Note: One can solve Exercises 1922 without explicit reference to the Rank Theorem. For instance, in
Exercise 19, if the null space of a matrix A is three-dimensional, then the equation Ax = 0 has three free variables, and three of the columns of A are nonpivot columns. Since a 57 matrix has seven columns, A must have four pivot columns (which form a basis of Col A). So rank A = dim Col A = 4.
20. A 45 matrix A has 5 columns. By the Rank Theorem, rank A = 5 dim Nul A. Since the null space is three-dimensional, rank A = 2. 21. A 76 matrix has 6 columns. By the Rank Theorem, dim Nul A = 6 rank A. Since the rank is four, dim Nul A = 2. That is, the dimension of the solution space of Ax = 0 is two. 22. The wording of this problem was poor in the first printing, because the phrase "it spans a fourdimensional subspace" was never defined. Here is a revision that I will put in later printings of the third edition: Show that a set {v1, ..., v5} in Rn is linearly dependent if dim Span{v1, ..., v5} = 4. Solution: Suppose that the subspace H = Span{v1, ..., v5} is four-dimensional. If {v1, ..., v5} were linearly independent, it would be a basis for H. This is impossible, by the statement just before the definition of dimension in Section 2.9, which essentially says that every basis of a p-dimensional subspace consists of p vectors. Thus, {v1, ..., v5} must be linearly dependent. 23. A 34 matrix A with a two-dimensional column space has two pivot columns. The remaining two columns will correspond to free variables in the equation Ax = 0. So the desired construction is possible.
152 CHAPTER 2 Matrix Algebra
* There are six possible locations for the two pivot columns, one of which is 0 0 0
* * 0
* * . A simple 0
construction is to take two vectors in R3 that are obviously not linearly dependent, and put two copies of these two vectors in any order. The resulting matrix will obviously have a two-dimensional column space. There is no need to worry about whether Nul A has the correct dimension, since this is guaranteed by the Rank Theorem: dim Nul A = 4 rank A.
24. A rank 1 matrix has a one-dimensional column space. Every column is a multiple of some fixed vector. To construct a 43 matrix, choose any nonzero vector in R4, and use it for one column. Choose any multiples of the vector for the other two columns. 25. The p columns of A span Col A by definition. If dim Col A = p, then the spanning set of p columns is automatically a basis for Col A, by the Basis Theorem. In particular, the columns are linearly independent. 26. If columns a1, a3, a5, and a6 of A are linearly independent and if dim Col A = 4, then {a1, a3, a5, a6} is a linearly independent set in a 4-dimensional column space. By the Basis Theorem, this set of four vectors is a basis for the column space. 27. a. Start with B = [b1 bp] and A = [a1 aq], where q > p. For j = 1, ..., q, the vector aj is in W. Since the columns of B span W, the vector aj is in the column space of B. That is, aj = Bcj for some vector cj of weights. Note that cj is in Rp because B has p columns. b. Let C = [c1 cq]. Then C is a pq matrix because each of the q columns is in Rp. By hypothesis, q is larger than p, so C has more columns than rows. By a theorem, the columns of C are linearly dependent and there exists a nonzero vector u in Rq such that Cu = 0. c. From part (a) and the definition of matrix multiplication A = [a1 aq] = [Bc1 Bcq] = BC From part (b), Au = (BC) u = B(Cu) = B0 = 0. Since u is nonzero, the columns of A are linearly dependent. 28. If A contained more vectors than B, then A would be linearly dependent, by Exercise 27, because B spans W. Repeat the argument with B and A interchanged to conclude that B cannot contain more vectors than A. 29. [M] Apply the matrix command ref or rref to the matrix [v1 v2 x]:
11 -5 10 7 14 -8 13 10 19 1 -13 0 ~ 18 0 15 0 0 1 0 0 -1.667 2.667 0 0
The equation c1v1 + c2v2 = x is consistent, so x is in the subspace H. The decimal approximations suggest c1 = 5/3 and c2 = 8/3, and it can be checked that these values are precise. Thus, the B-coordinate of x is (5/3, 8/3).
30. [M] Apply the matrix command ref or rref to the matrix [v1 v2 v3 x]:
-6 4 -9 4 8 -3 7 -3 -9 5 -8 3 4 1 7 0 ~ -8 0 3 0 0 1 0 0 0 0 1 0 3 5 2 0
Chapter 2
Supplementary Exercises
153
The first three columns of [v1 v2 v3 x] are pivot columns, so v1, v2 and v3 are linearly independent. Thus v1, v2 and v3 form a basis B for the subspace H which they span. View [v1 v2 v3 x] as an augmented matrix for c1v1 + c2v2 + c3v3 = x. The reduced echelon form shows that x is in H and 3 [x]B = 5 . 2
Notes: The Study Guide for Section 2.9 contains a complete list of the statements in the Invertible Matrix
Theorem that have been given so far. The format is the same as that used in Section 2.3, with three columns: statements that are logically equivalent for any mn matrix and are related to existence concepts, those that are equivalent only for any nn matrix, and those that are equivalent for any np matrix and are related to uniqueness concepts. Four statements are included that are not in the text's official list of statements, to give more symmetry to the three columns. The Study Guide section also contains directions for making a review sheet for "dimension" and "rank."
Chapter 2
SUPPLEMENTARY EXERCISES
1. a. True. If A and B are mn matrices, then BT has as many rows as A has columns, so ABT is defined. Also, ATB is defined because AT has m columns and B has m rows. b. False. B must have 2 columns. A has as many columns as B has rows. c. True. The ith row of A has the form (0, ..., di, ..., 0). So the ith row of AB is (0, ..., di, ..., 0)B, which is di times the ith row of B. d. False. Take the zero matrix for B. Or, construct a matrix B such that the equation Bx = 0 has nontrivial solutions, and construct C and D so that C D and the columns of C D satisfy the equation Bx = 0. Then B(C D) = 0 and BC = BD. 1 0 0 0 e. False. Counterexample: A = and C = 0 1 . 0 0 f. False. (A + B)(A B) = A2 AB + BA B2. This equals A2 B2 if and only if A commutes with B. g. True. An nn replacement matrix has n + 1 nonzero entries. The nn scale and interchange matrices have n nonzero entries. h. True. The transpose of an elementary matrix is an elementary matrix of the same type. i. True. An nn elementary matrix is obtained by a row operation on In. j. False. Elementary matrices are invertible, so a product of such matrices is invertible. But not every square matrix is invertible. k. True. If A is 33 with three pivot positions, then A is row equivalent to I3. l. False. A must be square in order to conclude from the equation AB = I that A is invertible. m. False. AB is invertible, but (AB)1 = B1A1, and this product is not always equal to A1B1. n. True. Given AB = BA, left-multiply by A1 to get B = A1BA, and then right-multiply by A1 to obtain BA1 = A1B. o. False. The correct equation is (rA)1 = r1A1, because
(rA)(r1A1) = (rr1)(AA1) = 1I = I. 1 p. True. If the equation Ax = 0 has a unique solution, then there are no free variables in this equation, 0 which means that A must have three pivot positions (since A is 33). By the Invertible Matrix Theorem, A is invertible.
154 CHAPTER 2 Matrix Algebra
2. C = (C 1 ) 1 =
0 3. A = 1 0
3
1 7 -2 -6
2
-5 -7 / 2 = 4 3 0 0 1 0 0 0 0 0 0 1 0 0
5 / 2 -2 0 0 1 0 0 0 = 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1
2
0 0 , 0
0 A = 1 0 0 0 1 0 0 0 0 0 1
0 A = A A = 1 0
0 0 0 = 0 0 0
Next, ( I - A)( I + A + A2 ) = I + A + A2 - A( I + A + A2 ) = I + A + A2 - A - A2 - A3 = I - A3 . Since A3 = 0, ( I - A)( I + A + A2 ) = I .
4. From Exercise 3, the inverse of I A is probably I + A + A2 + + An1. To verify this, compute
( I - A)( I + A + " + An -1 ) = I + A + " + An -1 - A( I + A + " + An -1 ) = I - AAn -1 = I - An
If An = 0, then the matrix B = I + A + A2 + + An1 satisfies (I A)B = I. Since I A and B are square, they are invertible by the Invertible Matrix Theorem, and B is the inverse of I A.
5. A2 = 2A I. Multiply by A: A3 = 2A2 A. Substitute A2 = 2A I: A3 = 2(2A I) A = 3A 2I. 4 2 2 Multiply by A again: A = A(3A 2I) = 3A 2A. Substitute the identity A = 2A I again:
Finally, A4 = 3(2A I) 2A = 4A 3I.
1 6. Let A = 0 0 0 and B = 1 -1 1 0 2 2 . By direct computation, A = I, B = I, and AB = -1 0 1 = BA. 0
7. (Partial answer in Study Guide) Since A1B is the solution of AX = B, row reduction of [A B] to [I X] will produce X = A1B. See Exercise 12 in Section 2.2.
[A
1 ~ 0 0
1 B] = 2 1 3 1 0 8 3 1
3 8 4 11 2 5 -3 -6 -5 -1 10 . -3
-3 1 3
5 1 5 ~ 0 4 0 3 1 0 0 0 1
3 -2 -1 37 9 -5
8 -5 -3
-3 7 6
5 1 -5 ~ 0 -1 0 0 1 0 0 0 1
3 1 -2 10 9 -5
8 3 -5 -1 10 -3
-3 -6 7
5 1 -5
5 1 1 ~ 0 -3 0
29 1 10 ~ 0 -3 0
10 Thus, A1B = 9 -5
8. By definition of matrix multiplication, the matrix A satisfies
1 A 3 2 1 = 7 1 3 1
Chapter 2
Supplementary Exercises
155
1 Right-multiply both sides by the inverse of 3 1 A= 1 3 7 1 -3 -2 -2 = 1 4 1 -1
2 . The left side becomes A. Thus, 7
5 9. Given AB = -2 1 B -1 = -2
4 7 and B = 2 3
3 , notice that ABB1 = A. Since det B = 7 6 =1, 1 4 1 3 -2 -3 -3 = 7 -8 13 27
-3 5 -1 and A = ( AB ) B = -2 7
Note: Variants of this question make simple exam questions.
10. Since A is invertible, so is AT, by the Invertible Matrix Theorem. Then ATA is the product of invertible matrices and so is invertible. Thus, the formula (ATA)1AT makes sense. By Theorem 6 in Section 2.2,
(ATA)1AT = A1(AT)1AT = A1I = A1 An alternative calculation: (ATA)1ATA = (ATA)1(ATA) = I. Since A is invertible, this equation shows that its inverse is (ATA)1AT. c0 11. a. For i = 1,..., n, p(xi) = c0 + c1xi + + cn-1 x in-1 = row i (V ) # = row i (V )c . cn -1 By a property of matrix multiplication, shown after Example 6 in Section 2.1, and the fact that c was chosen to satisfy Vc= y, row i (V )c = row i (Vc) = row i (y ) = yi Thus, p(xi) = yi. To summarize, the entries in Vc are the values of the polynomial p(x) at x1, ..., xn. b. Suppose x1, ..., xn are distinct, and suppose Vc = 0 for some vector c. Then the entries in c are the coefficients of a polynomial whose value is zero at the distinct points x1, ..., xn. However, a nonzero polynomial of degree n 1 cannot have n zeros, so the polynomial must be identically zero. That is, the entries in c must all be zero. This shows that the columns of V are linearly independent. c. (Solution in Study Guide) When x1, ..., xn are distinct, the columns of V are linearly independent, by (b). By the Invertible Matrix Theorem, V is invertible and its columns span Rn. So, for every y = (y1, ..., yn) in Rn, there is a vector c such that Vc = y. Let p be the polynomial whose coefficients are listed in c. Then, by (a), p is an interpolating polynomial for (x1, y1), ..., (xn, yn).
12. If A = LU, then col1(A) = Lcol1(U). Since col1(U) has a zero in every entry except possibly the first,
Lcol1(U) is a linear combination of the columns of L in which all weights except possibly the first are zero. So col1(A) is a multiple of col1(L). Similarly, col2(A) = Lcol2(U), which is a linear combination of the columns of L using the first two entries in col2(U) as weights, because the other entries in col2(U) are zero. Thus col2(A) is a linear combination of the first two columns of L.
13. a. P2 = (uuT)(uuT) = u(uTu)uT = u(1)uT = P, because u satisfies uTu = 1. b. PT = (uuT)T = uTTuT = uuT = P c. Q2 = (I 2P)(I 2P) = I I(2P) 2PI + 2P(2P) = I 4P + 4P2 = I, because of part (a).
156 CHAPTER 2 Matrix Algebra
0 14. Given u = 0 , define P and Q as in Exercise 13 by 1 0 P = uu = 0 [ 0 1
T
0 0 0 0
0 1] = 0 0
0 0 0
0 1 , Q = I - 2 P = 0 0 0 1 1 and Qx = 0 0 0 1 0
0 1 0
0 0 - 2 0 0 1 0
0 0 0
0 1 0 = 0 1 0
0 1 0
0 0 -1
1 0 5 , then Px = 0 If x = 3 0
0 1 0 0 5 = 0 1 3 3
0 1 1 0 5 = 5 . -1 3 -3
15. Left-multiplication by an elementary matrix produces an elementary row operation:
B ~ E1 B ~ E2 E1 B ~ E3 E2 E1 B = C so B is row equivalent to C. Since row operations are reversible, C is row equivalent to B. (Alternatively, show C being changed into B by row operations using the inverse of the Ei .)
16. Since A is not invertible, there is a nonzero vector v in Rn such that Av = 0. Place n copies of v into an nn matrix B. Then AB = A[v v] = [Av Av] = 0. 17. Let A be a 64 matrix and B a 46 matrix. Since B has more columns than rows, its six columns are linearly dependent and there is a nonzero x such that Bx = 0. Thus ABx = A0 = 0. This shows that the matrix AB is not invertible, by the IMT. (Basically the same argument was used to solve Exercise 22 in Section 2.1.)
Note: (In the Study Guide) It is possible that BA is invertible. For example, let C be an invertible 44 matrix
C and construct A = and B = [C -1 0]. Then BA = I4, which is invertible. 0
18. By hypothesis, A is 53, C is 35, and AC = I3. Suppose x satisfies Ax = b. Then CAx = Cb. Since CA = I, x must be Cb. This shows that Cb is the only solution of Ax = b.
.4 .2 .3 .31 .26 .3 .6 .3 . Then A2 = .39 .48 19. [M] Let A = .3 .2 .4 .30 .26 calculations by computing .2875 .2834 A = A A = .4251 .4332 .2874 .2834
4 2 2
.30 .39 . Instead of computing A3 next, speed up the .31 .2857 .4286 .2857 .2857 .4285 .2857 2/7 3/ 7 2/7 2 / 7 3/ 7 . 2 / 7
.2874 .4251 , .2875
.2857 A = A A = .4285 .2857
8 4 4
To four decimal places, as k increases, .2857 A .4286 .2857
k
.2857 .4286 .2857
.2857 2 / 7 , or, in rational format, Ak 3/ 7 .4286 2 / 7 .2857
Chapter 2 Supplementary Exercises
157
0 .2 If B = .1 .6 .9 .2
4
.3 .3 , then .4 .1998 .3764 .4218
.29 B 2 = .33 .38
.18 .44 .38
.18 .33 , .49 .2022 .3709 .4269 .2022 .3707 .4271 18 / 89 33/ 89 38 / 89 18 / 89 33/ 89 . 38 / 89
.1998 .2024 , B8 = .3707 .3663 .4269 .4339 To four decimal places, as k increases, .2022 B .3708 .4270
k
.2119 B = .3663 .4218
.2022 .3708 .4270
.2022 18 / 89 , or, in rational format, B k 33/ 89 .3708 38 / 89 .4270
20. [M] The 44 matrix A4 is the 44 matrix of ones, minus the 44 identity matrix. The MATLAB command is A4 = ones(4) eye(4). For the inverse, use inv(A4).
0 1 A4 = 1 1 0 1 A5 = 1 1 1 0 1 1 A6 = 1 1 1
1 0 1 1 1 0 1 1 1 1 0 1 1 1 1
1 1 0 1 1 1 0 1 1 1 1 0 1 1 1
1 1 , 1 0 1 1 1 0 1 1 1 1 0 1 1
-2 / 3 1/ 3 - A4 1 = 1/ 3 1/ 3 1 1 1 , 1 0 1 1 1 1 0 1
1/ 3 -2 / 3 1/ 3 1/ 3
1/ 3 1/ 3 -2 / 3 1/ 3
1/ 3 1/ 3 1/ 3 -2 / 3 1/ 4 1/ 4 1/ 4 -3/ 4 1/ 4 1/ 4 1/ 4 1/ 4 1/ 4 -3/ 4 1/ 5 1/ 5 1/ 5 1/ 5 -4 / 5 1/ 5 1/ 5 1/ 5 1/ 5 1/ 5 1/ 5 -4 / 5
-3/ 4 1/ 4 -1 A5 = 1/ 4 1/ 4 1/ 4 1 1 1 , 1 1 0
1/ 4 -3/ 4 1/ 4 1/ 4 1/ 4
1/ 4 1/ 4 -3/ 4 1/ 4 1/ 4
-4 / 5 1/ 5 1/ 5 -1 A6 = 1/ 5 1/ 5 1/ 5
1/ 5 -4 / 5 1/ 5 1/ 5 1/ 5 1/ 5
1/ 5 1/ 5 -4 / 5 1/ 5 1/ 5 1/ 5
1/ 5 1/ 5 1/ 5 -4 / 5 1/ 5 1/ 5
The construction of A6 and the appearance of its inverse suggest that the inverse is related to I6. In fact, - A6 1 + I 6 is 1/5 times the 66 matrix of ones. Let J denotes the nn matrix of ones. The conjecture is: 1 J - In n -1 Proof: (Not required) Observe that J 2 = nJ and An J = (J I ) J = J 2 J = (n 1) J. Now compute An((n 1)1J I) = (n 1)1A n J An = J (J I) = I Since An is square, An is invertible and its inverse is (n 1)1J I. An = J In and
- An 1 =
3.1
SOLUTIONS
Notes: Some exercises in this section provide practice in computing determinants, while others allow the
student to discover the properties of determinants which will be studied in the next section. Determinants are developed through the cofactor expansion, which is given in Theorem 1. Exercises 3336 in this section provide the first step in the inductive proof of Theorem 3 in the next section. 1. Expanding along the first row:
3 2 0 3 2 0
0 3 5 0 3 5
4 3 2 =3 5 -1
2 2 -0 -1 0
2 2 +4 -1 0
3 = 3( -13) + 4(10) = 1 5
Expanding along the second column: 4 2 2 = (-1)1+ 2 0 0 -1 2 3 + (-1) 2+ 2 3 0 -1 4 3 + (-1)3+ 2 5 2 -1 4 = 3(-3) - 5(-2) = 1 2
2. Expanding along the first row:
0 4 2 0 4 2
5 -3 4 5 -3 4
1 -3 0 =0 4 1
0 4 -5 1 2
0 4 +1 1 2
-3 = -5(4) + 1(22) = 2 4
Expanding along the second column: 1 4 0 = (-1)1+ 2 5 2 1 0 0 + (-1) 2 + 2 (-3) 1 2 1 0 + ( -1)3+ 2 4 1 4 1 = -5(4) - 3(-2) - 4(-4) = 2 0
3. Expanding along the first row:
2 3 1
-4 1 4
3 1 2 =2 4 -1
2 3 - ( -4) -1 1
2 3 +3 -1 1
1 = 2( -9) + 4(-5) + (3)(11) = -5 4
159
160
CHAPTER 3
Determinants
Expanding along the second column: 2 3 1 -4 1 4 3 2 = ( -1)1+ 2 ( -4) 1 -1 3 2 -1 + (-1) 2+ 2 1 2 1 3 -1 + (-1)3+ 2 4 2 3 3 2 = 4(-5) + 1( -5) - 4(-5) = -5
4. Expanding along the first row:
1 2 3 1 2 3
3 1 4 3 1 4
1 1 =1 4 2 5
5
1 2
-3
2 3
1 2
+5
2 3
1 4
= 1(-2) - 3(1) + 5(5) = 20
Expanding along the second column: 2 1 = ( -1)1+ 2 3 3 2 1 1 + (-1) 2+ 2 1 2 3 5 1 + (-1)3+ 2 4 2 2 5 = -3(1) + 1(-13) - 4(-9) = 20 1
5. Expanding along the first row:
2 4 5
3 0 1
-4 0 5 =2 1 6
5 4 -3 6 5
5 4 + ( -4) 6 5
0 = 2( -5) - 3(-1) - 4(4) = -23 1
6. Expanding along the first row:
5 0 2
-2 3 -4
4 3 -5 = 5 -4 7
-5 0 - (-2) 7 2
-5 0 +4 7 2
3 = 5(1) + 2(10) + 4(-6) = 1 -4
7. Expanding along the first row:
4 6 9
3 5 7
0 5 2 =4 7 3
2 6 -3 3 9
2 6 +0 3 9
5 = 4(1) - 3(0) = 4 7
8. Expanding along the first row:
8 4 3
1 0 -2
6
0 3 =8 -2 5
3 4 -1 5 3
3 4 +6 5 3
0 = 8(6) - 1(11) + 6(-8) = -11 -2
9. First expand along the third row, then expand along the first row of the remaining matrix:
6 1 2 8
0 7 0 3
0 2 0 1
5 0 -5 3+1 = (-1) 2 7 0 3 8
0 2 1
5 7 -5 = 2 (-1)1+3 5 3 8
2 = 10(1) = 10 1
3.1
Solutions
161
10. First expand along the second row, then expand along either the third row or the second column of the remaining matrix.
1 0 2 5
-2 0 -6 0
5 3 -7 4
2 1 0 2+3 = (-1) 3 2 5 5 4
-2 -6 0
2 5 4 -2 = (-3) (5(2) + 4(-2)) = -6 -6
-2 = (-3) (-1)3+1 5 -6
2 1 + (-1)3+3 4 5 2
or 1 0 2 5 -2 0 -6 0 5 3 -7 4 2 1 0 2+3 = (-1) 3 2 5 5 4 -2 -6 0 2 5 4 2 = (-3) ( 2(-17) - 6(-6) ) = -6 4
2 = (-3) (-1)1+ 2 (-2) 5
5 1 + (-1) 2+ 2 ( -6) 4 5
11. There are many ways to do this determinant efficiently. One strategy is to always expand along the first column of each matrix:
3 0 0 0
5 -2 0 0
-8 3 1 0
4 -2 -7 1+1 = (-1) 3 0 5 0 2
3 1 0
-7 1 5 = 3 (-1)1+1 (-2) 0 2
5 = 3(2)(2) = 12 2
12. There are many ways to do this determinant efficiently. One strategy is to always expand along the first row of each matrix:
4 7 2 5
0 -1 6 -8
0 0 3 4
0 -1 0 1+1 = (-1) 4 6 0 -8 -3
0 3 4
0 3 0 = 4 (-1)1+1 (-1) 4 -3
0 = 4(1)( 9) = 36 -3
13. First expand along either the second row or the second column. Using the second row,
4 0 7 5 0
0 0 3 0 0
-7 2 -6 5 9
3 0 4 2 -1 0 3 0 0
-5 0 -8 = (-1) 2 [phone]
4 7 5 0
0 3 0 0
3 4 2 -1
-5 -8 -3 2
Now expand along the second column to find: 4 7 (-1) 2+3 2 5 0 -5 4 -8 2+ 2 = -2 (-1) 3 5 -3 0 2 3 2 -1 -5 -3 2
162
CHAPTER 3
Determinants
Now expand along either the first column or third row. The first column is used below.
4 2+ 2 -2 (-1) 3 5 0 3 2 -1 -5 2 -3 = -6 (-1)1+1 4 -1 2 3 -3 + (-1) 2+1 5 2 -1 -5 = (-6)(4(1) - 5(1)) = 6 2
14. First expand along either the fourth row or the fifth column. Using the fifth column,
6 9 8 3 4
3 0 -5 0 2 6
2 -4 6 0 3 3 0 0 2
4 1 7 0 2
0
6 0 9 1 = (-1)3+5 1 3 0 4 0 4
3 0 0 2
2 -4 0 3
4 1 0 2
Now expand along the third row to find: 2 -4 0 3 (-1)3+5 1 9 3 4 3 3+1 1 (-1) 3 0 2 3
15. 2 0
3 1 = 1 (-1)3+1 3 0 0 2 2
2 -4 3
4 1 2
Now expand along either the first column or second row. The first column is used below. 2 -4 3 4 -4 1 = 3 ( -1)1+1 3 3 2 1 2 + (-1)3+1 2 2 -4 4 = (3)(3( -11) + 2(18)) = 9 1
0 3 5
4 2 = (3)(3)(1) + (0)(2)(0) + (4)(2)(5) (0)(3)(4) (5)(2)(3) (1)(2)(0) = -1 9 + 0 + 40 0 30 0 = 1
0
16. 4 2
5 -3 4
1 0 = (0)(3)(1) + (5)(0)(2) + (1)(4)(4) (2)(3)(1) (4)(0)(0) (1)(4)(5) = 1 0 + 0 + 16 (6) 0 20 = 2
2
17. 3 1
-4 1 4
3 2 = (2)(1)(1) + (4)(2)(1) + (3)(3)(4) (1)(1)(3) (4)(2)(2) (1)(3)(4) = -1 2 + (8) + 36 3 16 12 = 5
1
18. 2 3
3 1 4
5 1 = (1)(1)(2) + (3)(1)(3) + (5)(2)(4) (3)(1)(5) (4)(1)(1) (2)(2)(3) = 2 2 + 9 + 40 15 4 12 = 20
3.1
Solutions
163
19.
a c
b c = ad - bc, d a
d = cb - da = -(ad - bc) b
The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed.
20.
a c
b a = ad - bc, d kc
b = a(kd ) - (kc)b = kad - kbc = k (ad - bc) kd
The row operation scales row 2 by k, and the determinant is multiplied by k.
21.
3 5
4 3 = 18 - 20 = -2, 6 5 + 3k
4 = 3(6 + 4k ) - (5 + 3k )4 = -2 6 + 4k
The row operation replaces row 2 with k times row 1 plus row 2, and the determinant is unchanged.
22. a c b d
= ad - bc,
a + kc c
b + kd d
= (a + kc)d - c(b + kd ) = ad + kcd - bc - kcd = ad - bc
The row operation replaces row 1 with k times row 2 plus row 1, and the determinant is unchanged. 1 23. -3 2 1 8 -3 1 k -4 = 1(4) - 1(2) + 1(-7) = -5, -3 2 2
k k
8 -3
-4 = k (4) - k (2) + k (-7) = -5k 2
The row operation scales row 1 by k, and the determinant is multiplied by k.
a
24. 3 6
b
c
2 5 2 b 5
2 = a (2) - b(6) + c(3) = 2a - 6b + 3c, 6 2 c = 3(6b - 5c) - 2(6a - 6c) + 2(5a - 6b) = -2a + 6b - 3c 6
3 a 6
The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed.
25. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
1 0 0
0 1 k
0 0 = (1)(1)(1) = 1 1
26. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
1 0 k
0 1 0
0 0 = (1)(1)(1) = 1 1
27. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
k 0 0
0 1 0
0 0 = (k )(1)(1) = k 1
164
CHAPTER 3
Determinants
28. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:
1 0 0
0 k 0
0 0 = (1)(k )(1) = k 1
29. A cofactor expansion along row 1 gives
0 1 0
1 0 0
0 1 0 = -1 0 1
0 = -1 1
30. A cofactor expansion along row 1 gives
0 0 1
0 1 0
1 0 0 =1 1 0
1 = -1 0
31. A 3 3 elementary row replacement matrix looks like one of the six matrices
1 k 0
0 1 0
0 1 0 , 0 1 k
0 1 0
0 1 0 , 0 1 0
0 1 k
0 1 0 , 0 1 0
0 1 0
0 1 k , 0 1 0
0 1 0
k 1 0 , 0 1 0
k
1 0
0 0 1
In each of these cases, the matrix is triangular and its determinant is the product of its diagonal entries, which is 1. Thus the determinant of a 3 3 elementary row replacement matrix is 1.
32. A 3 3 elementary scaling matrix with k on the diagonal looks like one of the three matrices
k 0 0
0 1 0
0 1 0 , 0 1 0
0 k 0
0 1 0 , 0 1 0
0 1 0
0 0 k
In each of these cases, the matrix is triangular and its determinant is the product of its diagonal entries, which is k. Thus the determinant of a 3 3 elementary scaling matrix with k on the diagonal is k. 0 33. E = 1 1 a b c d , A = c d , EA = a b 0 det E = 1, det A = ad bc, det EA = cb da = 1(ad bc) = (det E)(det A)
1 34. E = 0
b 0 a b a , A = c d , EA = kc kd k det E = k, det A = ad bc, det EA = a(kd) (kc)b = k(ad bc) = (det E)(det A) k a , A = c 1 b a + kc , EA = c d b + kd d
1 35. E = 0
det E = 1, det A = ad bc, det EA = (a + kc)d c(b + kd) = ad + kcd bc kcd = 1(ad bc) = (det E)(det A)
3.1
Solutions
165
1 36. E = k
b 0 a b a , A = c d , EA = ka + c kb + d 1 det E = 1, det A = ad bc, det EA = a(kb + d) (ka + c)b = kab + ad kab bc = 1(ad bc) = (det E)(det A)
3 37. A = 4
1 15 , 5A = 2 20
5 , det A = 2, det 5A = 50 5det A 10
a b ka kb 38. A = , kA = kc kd , det A = ad bc, c d det kA = ( ka)( kd ) - ( kb)( kc) = k 2 ( ad - bc) = k 2 det A
39. a. True. See the paragraph preceding the definition of the determinant. b. False. See the definition of cofactor, which precedes Theorem 1. 40. a. False. See Theorem 1. b. False. See Theorem 2.
1 3 41. The area of the parallelogram determined by u = , v = , u + v, and 0 is 6, since the base of the 0 2 parallelogram has length 3 and the height of the parallelogram is 2. By the same reasoning, the area of x 3 the parallelogram determined by u = , x = , u + x, and 0 is also 6. 0 2
X2 V 2 1 1 2 U 4 X1 X2 2 1 1 2 U X
X1
3 x 3 1 v ] = det = 6, and det [u x ] = det 0 2 = 6. The determinant of the 0 2 matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 is equal to the area of the parallelogram. Also note that det [u a c 42. The area of the parallelogram determined by u = , v = , u + v, and 0 is cb, since the base of the b 0 parallelogram has length c and the height of the parallelogram is b.
X2 U b
V a c
X1
166
CHAPTER 3
Determinants
a c c a v ] = det = -cb , and det [ v u ] = det 0 b = cb. The determinant of the b 0 matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 either is equal to the area of the parallelogram or is equal to the negative of the area of the parallelogram. Also note that det [u 43. [M] Answers will vary. The conclusion should be that det (A + B) det A + det B. 44. [M] Answers will vary. The conclusion should be that det (AB) = (det A)(det B). 45. [M] Answers will vary. For 4 4 matrices, the conclusions should be that det AT = det A, det(A) = det A, det(2A) = 16det A, and det (10 A) = 104 det A . For 5 5 matrices, the conclusions should be that det AT = det A, det(A) = det A, det(2A) = 32det A, and det (10 A) = 105 det A. For 6 6 matrices, the conclusions should be that det AT = det A , det(A) = det A, det(2A) = 64det A, and det (10 A) = 106 det A. 46. [M] Answers will vary. The conclusion should be that det A-1 = 1/ det A.
3.2
SOLUTIONS
Notes: This section presents the main properties of the determinant, including the effects of row operations
on the determinant of a matrix. These properties are first studied by examples in Exercises 120. The properties are treated in a more theoretical manner in later exercises. An efficient method for computing the determinant using row reduction and selective cofactor expansion is presented in this section and used in Exercises 1114. Theorems 4 and 6 are used extensively in Chapter 5. The linearity property of the determinant studied in the text is optional, but is used in more advanced courses. 1. Rows 1 and 2 are interchanged, so the determinant changes sign (Theorem 3b.). 2. The constant 2 may be factored out of the Row 1 (Theorem 3c.). 3. The row replacement operation does not change the determinant (Theorem 3a.). 4. The row replacement operation does not change the determinant (Theorem 3a.). 1 5.
-1 -2
5
-4 -7
-6
1 4 = 0 9 0
5 1 3 5 -18 3
-6
1
-2 = 0 -3 0 -3
5 1 0
-6 -2 = 3 3
1 6. 3 2 1 -2 7. 3 1
5 -3 13 3 -5 5 -1
1 3 = 0 0 -7 0 7 2 2
-3
1 12 = 6 0 0 -1 0 7 2 2
5 -3 3
-3
1 2 =6 0 0 -1 3 1 0 0 0 7 30 30
5 -3 0
-3
2 = (6)(-3) = -18 1 3 1 0 0 0 7 30 0 2 8 =0 27 0
2 1 4 0 = 1 0 -3 0
3 1 -4 -4
2 1 8 0 = -5 0 -5 0
2 1 8 0 = 27 0 27 0
3.2
Solutions
167
1 0 8. 2 -3 1 0 9. -1 3 1 0 10. -2 3 3 1 0 - 0 0 0
3 1 5 -7
-1
3 2 4 -5
-3
-4
1 0 -5 = -3 0 2 0
3 1 -1 2
-1
3 2 -2 4
-3
1 0 -5 = 5 0 0 -10 0 1 4 0 = 5 0 3 0
-1 -4 0 -1
-4
3 1 0 0
-3
3 2 0 0
-4 -5 =0 0
0 0 4 =- 1 -5 3 2 0 0 0 1 0 0 0
-1 -4 0 -4 0 -1 -3
1 2 -1 3 2 -6 7 5 3 2 0 0 0
5 8 -2
-1 -4 2 -3 5 -1 -4 -4
0 1 4 0 = 5 0 3 0 0 -1 3 8 2 0 -1 7 3 0
-2
1 1 2
5 5 7 3 2 0 -2 -4
1 0 0 0 -1 3 8 2
5 0 -3
-2
1 0 0 0 -1 3 7 0
5 -3 0
-2
0 4 = -(-3) = 3 -5 1
1 0 -6 9 = 0 0 -7 7 0
-2
0 8
1 0 -6 5 = 0 0 -1 13 0
-6 5 = -7 1
-6 -7 = -(-24) = 24
0 0
5 1
11. First use a row replacement to create zeros in the second column, and then expand down the second column: 2 5 3 0 -6 0 4 10 3
-5 -6 0 -3 -1
2
1 -4 -4 1
-4 2
3 -3 = -6 9
-1 -3
0
5 0 0 0 1
-2 2
-3
1 -4 2
-3
-1 3 -3 = -5 -6 9 0 1
1 -4 2
-3 9
1
Now use a row replacement to create zeros in the first column, and then expand down the first column: 3 9 = -5 0 1 0 3 = ( -5)(3) 1
-2
2
3 = (-5)(3)( -8) = 120 1
12. First use a row replacement to create zeros in the fourth column, and then expand down the fourth column:
-1
3 5 4
2 4 4 2
3 3 6 4
0 -1 0 3 = 6 -3 3 4
2 4 0 2
3 3 -2 4
0 -1 0 =3 3 0 -3 3
2 4 0
3 3 -2
Now use a row replacement to create zeros in the first column, and then expand down the first column: -1 2 3 -1 2 3 10 12 3 3 4 3 = 3 0 10 12 = 3(-1) = 3(-1)(-38) = 114 -6 -11 -3 0 -2 0 -6 -11
168
CHAPTER 3
Determinants
13. First use a row replacement to create zeros in the fourth column, and then expand down the fourth column: 2 4 6 -6 5 7 -2 7 4 6 -4 7 1 2 2 0 = 0 6 -6 0 5
-3 -2
4
-2 -4
7
7
1 0 0 = -1 6 0 -6 0
-3 -2
-2 -4
7
7
Now use a row replacement to create zeros in the first column, and then expand down the first column: 0 -3 -2 0 -3 -2 -3 -2 -1 6 -2 -4 = -1 6 -2 -4 = (-1)(-6) = (-1)(-6)(1) = 6 5 3 7 7 0 5 3 -6 14. First use a row replacement to create zeros in the third column, and then expand down the third column:
-3 -2
1 -3 3 1 1 -9 3
a 15. d 5g a 16. 3d g a 17. g d
g 18. a d a
3 4 -4 3 0 -4
b e 5h b 3e h b h e
h b e
1 0 -2 0
-3
-4
-3
-2
1 -3 = -9 8 4 3
3 0 -4
1 0 0 0
-4
1 -3 = 1 -9 0 3 4
3 0 -4
-3 0
4
Now expand along the second row: 3 0 = 1(-(-9)) -4 4
b e h b e h b e h
b h e c
-3 = (1)(9)(0) = 0
4
c a f =5 d 5i g c a 3f =3 d i g c a i =- d f g
i a c =- g f b d
c f = 5(7) = 35 i c f = 3(7) = 21 i c f = -7 i
a c i = - - d g f
a b
b e h c
c f = -(-7) = 7 i a b e h c f = 2(7) = 14 i
19. 2d + a g
2e + b h
2 f + c = 2d i g
2e h
2f =2 d i g
3.2
Solutions
169
a+d
b+e e h
c+ f f i
a
= d g
b e h
c f =7 i
20.
d g
2 21. Since 1 1 5 22. Since 1 0 2 1 23. Since 3 0 4 24. Since 6 -7 7 25. Since -4 -6 3 5 26. Since -6 4 27. a. b. c. d. 28. a. b. c. d.
3 3 2 0 -3 5 0 -7 8 7
-7
0 4 = -1 0 , the matrix is invertible. 1
-1 -2 = 0 , the matrix is not invertible. 3
0 -5 6 5
-3
8 0 = 0 , the matrix is not invertible. 0 4
0 2
-8
-5 = 11 0 , the columns of the matrix form a linearly independent set. 6
5 7 2 -6 0 7
7 0 = -1 0 , the columns of the matrix form a linearly independent set. -5
-2 -1 3
0
0 0 = 0 , the columns of the matrix form a linearly dependent set. 0 -3
True. See Theorem 3. True. See the paragraph following Example 2. True. See the paragraph following Theorem 4. False. See the warning following Example 5. True. See Theorem 3. False. See the paragraphs following Example 2. False. See Example 3. False. See Theorem 5.
29. By Theorem 6, det B 5 = (det B )5 = ( -2)5 = -32 . 30. Suppose the two rows of a square matrix A are equal. By swapping these two rows, the matrix A is not changed so its determinant should not change. But since swapping rows changes the sign of the determinant, det A = det A. This is only possible if det A = 0. The same may be proven true for columns by applying the above result to AT and using Theorem 5.
170
CHAPTER 3
Determinants
31. By Theorem 6, (det A)(det A-1 ) = det I = 1 , so det A-1 = 1/ det A. 32. By factoring an r out of each of the n rows, det (rA) = r n det A. 33. By Theorem 6, det AB = (det A)(det B) = (det B)(det A) = det BA. 34. By Theorem 6 and Exercise 31, det ( PAP -1 ) = (det P)(det A)(det P -1 ) = (det P )(det P -1 )(det A)
1 = (det P) (det A) = 1det A det P = det A
35. By Theorem 6 and Theorem 5, det U T U = (det U T )(det U ) = (det U ) 2 . Since U T U = I ,
det U T U = det I = 1 , so (det U ) 2 = 1. Thus det U = 1.
36. By Theorem 6 det A4 = (det A) 4 . Since det A4 = 0 , then (det A) 4 = 0 . Thus det A = 0, and A is not invertible by Theorem 4.
6 37. One may compute using Theorem 2 that det A = 3 and det B = 8, while AB = 17
det AB = 24 = 3 8 = (det A)(det B). 0 . Thus 4
6 38. One may compute that det A = 0 and det B = 2, while AB = -2 0 2 = (det A)(det B).
0 . Thus det AB = 0 = 0
39. a. By Theorem 6, det AB = (det A)(det B) = 4 3 = 12. b. By Exercise 32, det 5 A = 53 det A = 125 4 = 500 . c. By Theorem 5, det BT = det B = -3 . d. By Exercise 31, det A-1 = 1/ det A = 1/ 4 . e. By Theorem 6, det A3 = (det A)3 = 43 = 64 . 40. a. By Theorem 6, det AB = (det A)(det B) = 1 2 = 2. b. By Theorem 6, det B 5 = (det B )5 = 25 = 32 . c. By Exercise 32, det 2 A = 24 det A = 16 -1 = -16 . d. By Theorems 5 and 6, det AT A = (det AT )(det A) = (det A)(det A) = -1 -1 = 1 . e. By Theorem 6 and Exercise 31, det B -1 AB = (det B -1 )(det A)(det B ) = (1/ det B)(det A)(det B ) = det A = -1 . 41. det A = (a + e)d c(b + f) = ad + ed bc cf = (ad bc) + (ed cf) = det B + det C. 42. det ( A + B) =
= (1 + a)(1 + d ) - cb = 1 + a + d + ad - cb = det A + a + d + det B , so c 1+ d det (A + B) = det A + det B if and only if a + d = 0. 1+ a b
3.3
Solutions
171
43. Compute det A by using a cofactor expansion down the third column: det A = (u1 + v1 )det A13 - (u2 + v2 )det A23 + (u3 + v3 )det A33
= u1det A13 - u2 det A23 + u3det A33 + v1det A13 - v2 det A23 + v3det A33 = det B + det C
44. By Theorem 5, det AE = det ( AE )T . Since ( AE )T = E T AT , det AE = det( E T AT ). Now E T is itself an
elementary matrix, so by the proof of Theorem 3, det ( E T AT ) = (det E T )(det AT ). Thus it is true that det AE = (det E T )(det AT ), and by applying Theorem 5, det AE = (det E)(det A).
45. [M] Answers will vary, but will show that det AT A always equals 0 while det AAT should seldom be
zero. To see why AT A should not be invertible (and thus det AT A = 0 ), let A be a matrix with more columns than rows. Then the columns of A must be linearly dependent, so the equation Ax = 0 must have a non-trivial solution x. Thus ( AT A)x = AT ( Ax) = AT 0 = 0, and the equation ( AT A) x = 0 has a non-trivial solution. Since AT A is a square matrix, the Invertible Matrix Theorem now says that AT A is not invertible. Notice that the same argument will not work in general for AAT , since AT has more rows than columns, so its columns are not automatically linearly dependent.
46. [M] One may compute for this matrix that det A = 1 and cond A 23683. Note that this is the A 2 condition number, which is used in Section 2.3. Since det A 0, it is invertible and
-19 -549 A-1 = 267 -278
-14 -401 195 -203
0 -2 1 -1
7 196 -95 99
The determinant is very sensitive to scaling, as det10 A = 104 det A = 10,000 and det 0.1A = (0.1) 4 det A = 0.0001. The condition number is not changed at all by scaling: cond(10A) = cond(0.1A) = condA 23683. When A = I 4 , det A=1 and cond A = 1. As before the determinant is sensitive to scaling: det10 A = 104 det A = 10,000 and det 0.1A = (0.1) 4 det A = 0.0001. Yet the condition number is not changed by scaling: cond(10A) = cond(0.1A) = cond A = 1.
3.3
SOLUTIONS
Notes: This section features several independent topics from which to choose. The geometric interpretation of the determinant (Theorem 10) provides the key to changes of variables in multiple integrals. Students of economics and engineering are likely to need Cramer's Rule in later courses. Exercises 110 concern Cramer's Rule, exercises 1118 deal with the adjugate, and exercises 1932 cover the geometric interpretation of the determinant. In particular, Exercise 25 examines students' understanding of linear independence and requires a careful explanation, which is discussed in the Study Guide. The Study Guide also contains a heuristic proof of Theorem 9 for 2 2 matrices.
172
CHAPTER 3
Determinants
5 7 3 1. The system is equivalent to Ax = b, where A = and b = 1 . We compute 2 4 3 7 5 3 A1 (b) = , A2 (b) = 2 1 , det A = 6, det A1 (b) = 5, det A2 (b) = -1, 1 4 det A1 (b) 5 det A2 (b) 1 x1 = = , x2 = =- . det A 6 det A 6 4 1 6 and b = . We compute 2. The system is equivalent to Ax = b, where A = 5 2 7 6 1 4 6 A1 (b) = , A2 (b) = 5 7 , det A = 3, det A1 (b) = 5, det A2 (b) = -2, 7 2
x1 = det A1 (b) 5 det A2 (b) 2 = , x2 = =- . det A 3 det A 3
3 -2 7 3. The system is equivalent to Ax = b, where A = and b = -5 . We compute 6 -5 7 7 -2 3 A1 (b) = , A2 (b) = -5 -5 , det A = 8, det A1 (b) = 32, det A2 (b) = 20, 6 -5
x1 = det A1 (b) 32 det A2 (b) 20 5 = = 4, x2 = = = . det A 8 det A 8 2
3 -5 9 4. The system is equivalent to Ax = b, where A = and b = -5 . We compute 3 -1 9 3 -5 9 A1 (b) = , A2 (b) = , det A = -4, det A1 (b) = 6, det A2 (b) = -2, -5 -1 3 -5 x1 = det A1 (b) 6 det A2 (b) -2 1 3 = = - , x2 = = = . -4 -4 2 det A 2 det A
2 1 0 5. The system is equivalent to Ax = b, where A = -3 0 1 0 1 2 7 0 7 1 0 2 2 A1 (b) = -8 0 1 , A2 (b) = -3 -8 1 , A3 (b) = -3 -3 1 2 0 -3 2 0
det A = 4,det A1 (b) = 6,det A2 (b) = 16,det A3 (b) = -14, x1 =
7 and b = -8 . We compute -3 1 7 0 -8 , 1 -3
det A3 (b) -14 det A1 (b) 6 3 det A2 (b) 16 7 = = , x2 = = = 4, x3 = = =- . det A 4 2 det A 4 det A 4 2
3.3
Solutions
173
2 6. The system is equivalent to Ax = b, where A = -1 3 4 A1 (b) = 2 -2
1 0 1 1 2 , A (b) = -1 2 2 3 3 4 2 -2
1 4 and b = 2 . We compute 0 2 -2 1 3 1 4 2 1 , A (b) = -1 0 2 3 2 , 3 1 -2 3 1
det A = 4, det A1 (b) = -16, det A2 (b) = 52, det A3 (b) = -4, x1 = det A3 (b) -4 det A1 (b) -16 det A2 (b) 52 = = -4, x2 = = = 13, x3 = = = -1. det A 4 det A 4 det A 4
4 6s 5 7. The system is equivalent to Ax = b, where A = and b = -2 . We compute 9 2s 4 5 5 6s A1 (b) = , A2 (b) = , det A1 (b) = 10s + 8, det A2 (b) = -12s - 45. 9 -2 -2 2 s Since det A = 12 s 2 - 36 = 12( s 2 - 3) 0 for s 3 , the system will have a unique solution when s 3 . For such a system, the solution will be det A1 (b) det A2 (b) -12 s - 45 -4s - 15 10 s + 8 5s + 4 x1 = = = , x2 = = = . 2 2 det A det A 12( s - 3) 6( s - 3) 12( s 2 - 3) 4( s 2 - 3)
3s -5 3 8. The system is equivalent to Ax = b, where A = and b = 2 . We compute 9 5s 3 -5 3s 3 A1 (b) = , A2 (b) = 9 2 , det A1 (b) = 15s + 10, det A2 (b) = 6s - 27. 2 5s
Since det A = 15s 2 + 45 = 15( s 2 + 3) 0 for all values of s, the system will have a unique solution for all values of s. For such a system, the solution will be det A1 (b) 15s + 10 det A2 (b) 3s + 2 6 s - 27 2s - 9 x1 = = = , x2 = = = . 2 2 2 det A det A 15( s + 3) 3( s + 3) 15( s + 3) 5( s 2 + 3)
s -2s -1 9. The system is equivalent to Ax = b, where A = and b = 4 . We compute 6s 3 -1 -2 s s -1 A1 (b) = , det A1 (b) = 2s, det A2 (b) = 4 s + 3. , A2 (b) = 3 6s 4 4
Since det A = 6s 2 + 6s = 6 s ( s + 1) = 0 for s = 0, 1, the system will have a unique solution when s 0, 1. For such a system, the solution will be det A1 (b) det A2 (b) 2s 1 4s + 3 = = = x1 = , x2 = . det A 6s ( s + 1) 3( s + 1) det A 6s ( s + 1) 1 2s 1 10. The system is equivalent to Ax = b, where A = and b = . We compute 3s 6s 2 1 1 2s 1 A1 (b) = , A2 (b) = , det A1 (b) = 6s - 2, det A2 (b) = s. 2 6s 3s 2
174
CHAPTER 3
Determinants
Since det A = 12 s 2 - 3s = 3s (4s - 1) = 0 for s = 0,1/4, the system will have a unique solution when s 0,1/4. For such a system, the solution will be det A1 (b) det A2 (b) 6s - 2 s 1 = = = x1 = , x2 = . det A 3s (4 s - 1) det A 3s (4s - 1) 3(4 s - 1)
11. Since det A = 3 and the cofactors of the given matrix are
C11 = 0 1 0 1 -2 1 -1 0 1 -1 2 0 = 0, -1 1 = 1, C12 = - C22 = 3 -1 0 -1 0 3 0 1 -1 1 -1 0 = -3, = -1, = -3, C13 = 3 -1 0 -1 -2 0 0 1 = 3, -2 1 = 2,
C21 = - C31 =
C23 = - C33 = 1/ 3 -1/ 3 2/3 0 3
-2
= 0,
C32 = -
= 6,
0 adj A = -3 3
0 0 and A-1 = 1 adj A = -1 -3 det A 1 6
0 -1 . 2 -2 1 1 1 0 1 1 -2
12. Since det A = 5 and the cofactors of the given matrix are
C11 = -2 1 1 1 1 -2 3 0 -1 1 0 3 0 3 1 = -1, = 3, = 7, C12 = - C22 = 1 0 1 2 2 0 3 0 1 0 = 0, C13 = 2 0 = 2, = -1, = -4,
C21 = - C31 =
= 0, 3 1 = 5,
C23 = - C33 = 1 2
C32 = -
-1 adj A = 0 2
7 -1/ 5 and A-1 = 1 adj A = 5 0 det A 2/5 -4
3/ 5 0 -1/ 5
7 / 5 1 . -4 / 5
13. Since det A = 6 and the cofactors of the given matrix are
C11 = 0 5 1 4 1 -1 -5 7 1 1 1 = -1, 4 1 = -1, C12 = - C22 = 3 2 3 1 1 1 2 1 4 1 = 1, = -5, 4 1 = 1, C13 = 1 2 3 2 5 0 0 1 = 1, 5 1 = 7,
C21 = - C31 = 5 0
C23 = - C33 = 3 1
= 5,
C32 = -
= -5, 5 / 6 1/ 6 . -5 / 6
-1 adj A = 1 1
5 -1/ 6 1 1 and A-1 = adj A = 1/ 6 det A 1/ 6 -5
-1/ 6 -5 / 6 7/6
3.3
Solutions
175
14. Since det A = 1 and the cofactors of the given matrix are
C11 = 2 3 6 3 7 1 -3 -2 3 1 4 = 5, 7 3 = -3, C12 = - C22 = 3 2 3 0 0 2 7 4 1 4 = 2, C13 = 0 2 3 2 6 2 2 3 = -4, 6 3 = 3,
C21 = - C31 = 6 2
= -2, 7 1 = -3,
C23 = - C33 = 3 2 -3 3 0
= -8,
C32 = -
= 6,
5 adj A = 2 -4
-8 -5 and A-1 = 1 adj A = -2 -3 det A 4 6 -1 -2 3 -2 0 1 0 0 0 2
8 3 . -6 -1 -2 3 -2 3 -1 0 0 1
15. Since det A = 6 and the cofactors of the given matrix are
C11 = 1 3 0 3 0 0 0 6 -9 0 2 = 2, 0 2 = 0, C12 = - C22 = C31 = 0 2 = 2, C13 = 1 3 = -1, 0 3 = -9,
C21 = - C31 = 0 1
= 6,
C23 = - C33 =
= 0,
= 0,
= 3,
2 adj A = 2 -1
0 1/ 3 and A-1 = 1 adj A = 1/ 3 0 det A -1/ 6 3
0 1 0 . -3/ 2 1/ 2 -3 0 1 0 2 0 2 -3
16. Since det A = 9 and the cofactors of the given matrix are
C11 = -3 0 2 0 2 -3 -6 3 0 1 3 4 3 4 1 = -9, = -6, = 14, C12 = - C22 = 1 0 1 0 0 0 4 3 1 3 = 0, C13 = 0 0 = 0, = 0, = -3,
C21 = - C31 =
= 3, 4 1 = -1,
C23 = - C33 = 2/3 -1/ 3 0 1 0
C32 = -
-9 adj A = 0 0 a 17. Let A = c
14 1 and A-1 = 1 adj A = 0 -1 det A 0 -3
-14 / 9 1/ 9 . 1/ 3
b . Then the cofactors of A are C11 = d = d , C12 = - c = -c, d d -b C21 = - b = -b , and C22 = a = a . Thus adj A = . Since det A = ad bc, Theorem 8 gives that a -c 1 1 d -b . This result is identical to that of Theorem 4 in Section 2.2. A-1 = adj A = a det A ad - bc -c
176
CHAPTER 3
Determinants
18. Each cofactor of A is an integer since it is a sum of products of entries in A. Hence all entries in adj A will be integers. Since det A = 1, the inverse formula in Theorem 8 shows that all the entries in A-1 will be integers.
5 19. The parallelogram is determined by the columns of A = 2 |det A| = |8| = 8. -1 20. The parallelogram is determined by the columns of A = 3 |det A| = |7| = 7.
6 , so the area of the parallelogram is 4 4 , so the area of the parallelogram is -5
21. First translate one vertex to the origin. For example, subtract (1, 0) from each vertex to get a new parallelogram with vertices (0, 0),(1, 5),(2, 4), and (3, 1). This parallelogram has the same area as the 2 1 original, and is determined by the columns of A = , so the area of the parallelogram is 5 -4 |det A| = |14| = 14. 22. First translate one vertex to the origin. For example, subtract (0, 2) from each vertex to get a new parallelogram with vertices (0, 0),(6, 1),(3, 3), and (3, 4). This parallelogram has the same area as 6 -3 the original, and is determined by the columns of A = , so the area of the parallelogram is 3 1 |det A| = |21| = 21.
1 23. The parallelepiped is determined by the columns of A = 0 -2 parallelepiped is |det A| = |22| = 22. 1 24. The parallelepiped is determined by the columns of A = 4 0 parallelepiped is |det A| = |15| = 15.
1 2 4 7 1 , so the volume of the 0 -1 2 , so the volume of the -1
-2 -5 2
25. The Invertible Matrix Theorem says that a 3 3 matrix A is not invertible if and only if its columns are linearly dependent. This will happen if and only if one of the columns is a linear combination of the others; that is, if one of the vectors is in the plane spanned by the other two vectors. This is equivalent to the condition that the parallelepiped determined by the three vectors has zero volume, which is in turn equivalent to the condition that det A = 0. 26. By definition, p + S is the set of all vectors of the form p + v, where v is in S. Applying T to a typical vector in p + S, we have T(p + v) = T(p) + T(v). This vector is in the set denoted by T(p) + T(S). This proves that T maps the set p + S into the set T(p) + T(S). Conversely, any vector in T(p) + T(S) has the form T(p) + T(v) for some v in S. This vector may be written as T(p + v). This shows that every vector in T(p) + T(S) is the image under T of some point p + v in p + S.
3.3
Solutions
177
-2 -2 27. Since the parallelogram S is determined by the columns of , the area of S is 5 3 6 -2 -2 -2 = 6 . By Theorem 10, the area of T(S) is det = | -4 | = 4. The matrix A has det A = -3 2 5 3
|det A|{area of S} = 6 4 = 24. Alternatively, one may compute the vectors that determine the image, namely, the columns of
6 -2 -2 -2 -18 -22 b2 ] = = 2 3 5 12 16 -3 The determinant of this matrix is 24, so the area of the image is 24.
A[b1
4 0 28. Since the parallelogram S is determined by the columns of , the area of S is -7 1 7 2 4 0 det = | 4 | = 4 . The matrix A has det A = 1 1 = 5 . By Theorem 10, the area of T(S) is -7 1
|det A|{area of S} =5 4 = 20. Alternatively, one may compute the vectors that determine the image, namely, the columns of
7 2 4 0 14 2 b2 ] = = 1 1 -7 1 -3 1 The determinant of this matrix is 20, so the area of the image is 20.
A[b1
29. The area of the triangle will be one half of the area of the parallelogram determined by v1 and v 2 . By
Theorem 9, the area of the triangle will be (1/2)|det A|, where A = [ v1
v 2 ].
30. Translate R to a new triangle of equal area by subtracting ( x3 , y3 ) from each vertex. The new triangle has vertices (0, 0), ( x1 - x3 , y1 - y3 ) , and ( x2 - x3 , y2 - y3 ). By Exercise 29, the area of the triangle will be
x - x 1 det 1 3 2 y1 - y3 x1 det x2 x3
By Theorem 5, y1 - y3 x2 - x3 x - x x - x det 1 3 = det 1 3 x2 - x3 y2 - y3 y1 - y3 y2 - y3 So the above observation allows us to state that the area of the triangle will be y1 y2 y3 x2 - x3 . y2 - y3 1 x1 - x3 = det x - x 1 2 3 x3 1 y1 - y3 y2 - y3 y3 0 x - x 0 = det 1 3 x2 - x3 1
Now consider using row operations and a cofactor expansion to compute the determinant in the formula: y1 - y3 y2 - y3
x - x 1 det 1 3 2 y1 - y3
x1 x2 - x3 1 = det x2 y2 - y3 2 x3
y1 y2 y3
1 1 1
178
CHAPTER 3
Determinants
u1 2 2 x12 x2 x3 31. a. To show that T(S) is bounded by the ellipsoid with equation 2 + 2 + 2 = 1 , let u = u2 and let a b c u3 x1 2 2 x = x2 = Au . Then u1 = x1 / a , u2 = x2 / b , and u3 = x3 / c , and u lies inside S (or u12 + u2 + u3 1 ) if x3
2 2 x12 x2 x3 + 2 + 2 1 ). a2 b c b. By the generalization of Theorem 10, {volume of ellipsoid} = {volume of T ( S )}
and only if x lies inside T(S) (or
= | det A | {volume of S } = abc
4 4abc = 3 3
32. a. A linear transformation T that maps S onto S will map e1 to v1 , e 2 to v 2 , and e3 to v 3 ; that is, T (e1 ) = v1 , T (e 2 ) = v 2 , and T (e3 ) = v 3 . The standard matrix for this transformation will be
A = [T (e1 ) T (e 2 ) T (e3 ) ] = [ v1
v2
v 3 ].
b. The area of the base of S is (1/2)(1)(1) = 1/2, so the volume of S is (1/3)(1/2)(1) = 1/6. By part a. T(S) = S , so the generalization of Theorem 10 gives that the volume of S is |det A|{volume of S} = (1/6)|det A|. 33. [M] Answers will vary. In MATLAB, entries in B inv(A) are approximately 10-15 or smaller. 34. [M] Answers will vary, as will the commands which produce the second entry of x. For example, the MATLAB command is x2 = det([A(:,1) b A(:,3:4)])/det(A) while the Mathematica command is x2 = Det[{Transpose[A][[1]],b,Transpose[A][[3]], Transpose[A][[4]]}]/Det[A]. 35. [M] MATLAB Student Version 4.0 uses 57,771 flops for inv A and 14,269,045 flops for the inverse formula. The inv(A) command requires only about 0.4% of the operations for the inverse formula.
Chapter 3
SUPPLEMENTARY EXERCISES
1. a. True. The columns of A are linearly dependent. b. True. See Exercise 30 in Section 3.2. c. False. See Theorem 3(c); in this case det 5 A = 53 det A .
2 d. False. Consider A = 0
0 1 , B = 0 1 0 3 , and A + B = 0 3 0 . 4
e. f. g. h. i. j. k.
False. By Theorem 6, det A3 = 23 . False. See Theorem 3(b). True. See Theorem 3(c). True. See Theorem 3(a). False. See Theorem 5. False. See Theorem 3(c); this statement is false for n n invertible matrices with n an even integer. True. See Theorems 6 and 5; det AT A = (det A) 2 .
Chapter 3
Supplementary Exercises
179
l. m. n. o. p.
12
False. The coefficient matrix must be invertible. False. The area of the triangle is 5. True. See Theorem 6; det A3 = (det A)3 . False. See Exercise 31 in Section 3.2. True. See Theorem 6. 13 16 19
a b c
14
12
13 14 3 6
a b-a c-a a
2. 15 18
1 3. 1 1
17 = 3 20 6
b+c
3 =0 6
b+c
1
1
a
b+c -1 = 0 -1
a+c = 0 0 a+b b b+ x b+ y c
a - b = (b - a)(c - a) 0 0 a-c b x y c a b c
1 1
a 4. a + x a+ y
c+x = x c+ y y
x = xy 1 1 y
1 1
1 =0 1
9 9 5. 4 9 6
1 0 0 0 0
9 9 0 3 0
9 9 5 9 7 4 6
9
9 2 4 0 = (-1) 9 0 6 0 5 7
9 0 3 0
9 5 9 7
4 0 = (-1)(-2) 9 0 6 0
2
0 3 0
5 9 7
= (-1)(-2)(3)
= (-1)(-2)(3)(-2) = -12
4 0 6. 6 0 0
8 1 8 8 8
8 0 8 8 2
8 0 8 3 0
4 0 6 7 = (1) 0 0 0 0
5
8 8 8 2
8 8 3 0
4 7 = (1)(2) 6 0 0 0
5
8 8 3
5 4 7 = (1)(2)(-3) 6 0
5 = (1)(2)(-3)(-2) = 12 7
7. Expand along the first row to obtain 1 x y 1 y1 1 x y1 1 x1 y1 = 1 1 -x +y x2 y2 1 y2 1 1 x2 y2
x1 = 0. This is an equation of the form ax + by + c = 0, x2
and since the points ( x1 , y1 ) and ( x2 , y2 ) are distinct, at least one of a and b is not zero. Thus the equation is the equation of a line. The points ( x1 , y1 ) and ( x2 , y2 ) are on the line, because when the coordinates of one of the points are substituted for x and y, two rows of the matrix are equal and so the determinant is zero.
180
CHAPTER 3
Determinants
8. Expand along the first row to obtain 1 x y x y1 1 y1 1 -x +y 1 x1 y1 = 1 1 1 m 0 m 0 0 1 m
x1
1
= 1( mx1 - y1 ) - x(m) + y (1) = 0. This equation may be
rewritten as mx1 - y1 - mx + y = 0, or y - y1 = m( x - x1 ).
1
a b c
a2 b c
2 2
1 = 0 0
a b-a c-a
2
a2 b -a
2
9. det T = 1
1
1 = 0 0
a b-a c-a
a2 (b - a)(b + a) (c - a)(c + a)
c2 - a2
1 = (b - a )(c - a ) 0 0
a 1 1
a2 1 b + a = (b - a )(c - a ) 0 c+a 0
a 1 0
a2 b + a = (b - a )(c - a )(c - b) c-b
10. Expanding along the first row will show that f (t ) = det V = c0 + c1t + c2t 2 + c3t 3 . By Exercise 9,
1 c3 = 1 1 x1 x2 x3 x12
2 x2 = ( x2 - x1 )( x3 - x1 )( x3 - x2 ) 0 2 x3
since x1 , x2 , and x3 are distinct. Thus f (t) is a cubic polynomial. The points ( x1 ,0) , ( x2 ,0) , and ( x3 ,0) are on the graph of f, since when any of x1 , x2 or x3 are substituted for t, the matrix has two equal rows and thus its determinant (which is f (t)) is zero. Thus f ( xi ) = 0 for i = 1, 2, 3.
11. To tell if a quadrilateral determined by four points is a parallelogram, first translate one of the vertices to the origin. If we label the vertices of this new quadrilateral as 0, v1 , v 2 , and v 3 , then they will be the vertices of a parallelogram if one of v1 , v 2 , or v 3 is the sum of the other two. In this example, subtract (1, 4) from each vertex to get a new parallelogram with vertices 0 = (0, 0), v1 = (-2,1) , v 2 = (2,5) , and v 3 = (4, 4) . Since v 2 = v 3 + v1 , the quadrilateral is a parallelogram as stated. The translated parallelogram has the same area as the original, and is determined by the columns of -2 4 A = [ v1 v 3 ] = , so the area of the parallelogram is |det A| = |12| = 12. 1 4 12. A 2 2 matrix A is invertible if and only if the parallelogram determined by the columns of A has nonzero area. 13. By Theorem 8, (adj A)
(adj A) -1 = 1 A. det A 1 A = A-1 A = I . By the Invertible Matrix Theorem, adj A is invertible and det A
A O 14. a. Consider the matrix Ak = , where 1 k n and O is an appropriately sized zero matrix. We O I k will show that det Ak = det A for all 1 k n by mathematical induction.
Chapter 3
Supplementary Exercises
181
First let k = 1. Expand along the last row to obtain A O ( n +1) + ( n +1) det A1 = det 1 det A = det A. = ( -1) O 1 Now let 1 < k n and assume that det Ak -1 = det A. Expand along the last row of Ak to obtain
A O det Ak = det = ( -1)( n + k ) + ( n + k ) 1 det Ak -1 = det Ak -1 = det A. Thus we have proven the result, O Ik and the determinant of the matrix in question is det A.
O I b. Consider the matrix Ak = k , where 1 k n, Ck is an n k matrix and O is an appropriately Ck D sized zero matrix. We will show that det Ak = det D for all 1 k n by mathematical induction. First let k = 1. Expand along the first row to obtain 1 O = ( -1)1+1 1 det D = det D. det A1 = det C1 D Now let 1 < k n and assume that det Ak -1 = det D. Expand along the first row of Ak to obtain O I det Ak = det k = ( -1)1+1 1 det Ak -1 = det Ak -1 = det D. Thus we have proven the result, and the Ck D determinant of the matrix in question is det D. c. By combining parts a. and b., we have shown that A O A O I O det = det O I det C D = (det A)(det D). C D From this result and Theorem 5, we have T AT O A B A B T T det = (det A )(det D ) = (det A)(det D). = det O D = det T T O D D B
15. a. Compute the right side of the equation:
I X A C O A I O B A = Y XA B XB + Y B
Set this equal to the left side of the equation: B A = D XA so that XA = C XB + Y XB + Y = D
Since XA = C and A is invertible, X = CA-1. Since XB + Y = D, Y = D - XB = D - CA-1B . Thus by Exercise 14(c), A det C I B = det -1 D CA O A det I O D - CA B B
-1
= (det A)(det ( D - CA-1 B ))
b. From part a.,
A det C B = (det A)(det ( D - CA-1 B)) = det[ A( D - CA-1 B)] D = det[ AD - ACA-1 B ] = det[ AD - CAA-1 B ] = det[ AD - CB]
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16. a. Doing the given operations does not change the determinant of A since the given operations are all row replacement operations. The resulting matrix is
a - b 0 0 # b -a + b a -b 0 0 -a + b a -b ... ... ... 0 0 0 # a
# b
# b
% ...
b. Since column replacement operations are equivalent to row operations on AT and det AT = det A , the given operations do not change the determinant of the matrix. The resulting matrix is
0 0 0 ... a - b 0 0 0 a -b ... 0 0 0 a -b ... # # % # # b 2b 3b ... a + (n - 1)b c. Since the preceding matrix is a triangular matrix with the same determinant as A, det A = (a - b) n -1 (a + (n - 1)b).
17. First consider the case n = 2. In this case a -b b b det B = = a( a - b),det C = 0 a b
b a
= ab - b 2 ,
so det A = det B + det C = a (a - b) + ab - b 2 = a 2 - b 2 = (a - b)( a + b) = (a - b) 2 -1 (a + (2 - 1)b) , and the formula holds for n = 2. Now assume that the formula holds for all (k 1) (k 1) matrices, and let A, B, and C be k k matrices. By a cofactor expansion along the first column, a det B = (a - b) b # b b a # b ... ... % ... b b = (a - b)(a - b) k - 2 (a + (k - 2)b) = (a - b) k -1 (a + (k - 2)b) # a
since the matrix in the above formula is a (k 1) (k 1) matrix. We can perform a series of row operations on C to "zero out" below the first pivot, and produce the following matrix whose determinant is det C: b 0 # 0 b a -b # 0 ... ... % ... b 0 . # a - b
Since this is a triangular matrix, we have found that det C = b(a - b) k -1 . Thus det A = det B + det C = (a - b) k -1 (a + (k - 2)b) + b(a - b) k -1 = (a - b) k -1 (a + (k - 1)b), which is what was to be shown. Thus the formula has been proven by mathematical induction.
18. [M] Since the first matrix has a = 3, b = 8, and n = 4, its determinant is (3 - 8) 4 -1 (3 + (4 - 1)8) = (-5)3 (3 + 24) = (-125)(27) = -3375. Since the second matrix has a = 8, b = 3,
and n = 5, its determinant is (8 - 3)5-1 (8 + (5 - 1)3) = (5) 4 (8 + 12) = (625)(20) = 12,500.
Chapter 3
Supplementary Exercises
183
19. [M] We find that
1 1 1 1 2 2 1 1 2 = 1, 1 3 1 1 1 2 2 2 1 2 3 3 1 1 1 2 2 2 2 1 2 3 3 3 1 2 3 4 4 1 2 3 = 1. 4 5 1 2 = 1, 1 3 1 4 1
Our conjecture then is that 1 1 1 1 2 2 1 2 3 ... ... ... 1 2 3 = 1.
# 1
# 2
# 3
% ...
# n
To show this, consider using row replacement operations to "zero out" below the first pivot. The resulting matrix is 1 0 0 # 0 1 1 1 1 1 2 ... ... ... 1 1 2 . # n - 1
# 1
# 2
% ...
Now use row replacement operations to "zero out" below the second pivot, and so on. The final matrix which results from this process is 1 0 0 # 0 1 1 0 1 1 1 ... 1 ... 1 ... 1 , % # ... 1
# 0
# 0
which is an upper triangular matrix with determinant 1.
20. [M] We find that
1 1 1 1 3 3 1 1 1 3 = 6, 1 6 1 1 3 3 3 1 3 6 6 1 1 1 3 3 3 3 1 3 6 6 6 1 3 6 9 9 1 3 6 = 54. 9 12 1 3 = 18, 1 6 1 9 1
Our conjecture then is that 1 1 1 1 3 3 1 3 6 ... ... ... 1 3 6 = 2 3n - 2.
# 1
# 3
# 6
% ...
# 3(n - 1)
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Determinants
To show this, consider using row replacement operations to "zero out" below the first pivot. The resulting matrix is 1 0 0 # 0 1 2 2 1 2 5 ... ... ... 2 . 5 # 3(n - 1) - 1 1
# 2
# 5
% ...
Now use row replacement operations to "zero out" below the second pivot. The matrix which results from this process is 1 0 0 0 0 # 0 1 2 0 0 0 1 2 3 3 3 1 2 3 6 6 1 2 3 6 9 1 2 3 6 9 ... ... ... ... ... 6 . 9 # 3(n - 2) 1 2 3
# 0
# 3
# 6
# 9
# 12
% ...
This matrix has the same determinant as the original matrix, and is recognizable as a block matrix of the form A O where 3 3 1 and D = 3 2 # 3 3 6 6 3 6 9 3 6 9 ... ... ... 1 1 = 3 1 # # 1 3(n - 2) 3 6 9 1 2 2 1 2 3 1 2 3 ... ... ... 1 2 3 . # n - 2 B , D
1 A= 0
# 6
# 9
# 12
% ...
# 2
# 3
# 4
% !
A As in Exercise 14(c), the determinant of the matrix O Since D is an (n 2) (n 2) matrix, 1 det D = 3
n-2
B is (det A)(det D) = 2 det D. D
1 2 2
1 2 3
1 2 3
... ... ...
1 2 3 = 3n - 2 (1) = 3n - 2
1 1
# 1
# 2
# 3
# 4
% ...
# n-2
B is 2det D = 2 3n - 2. D
A by Exercise 19. Thus the determinant of the matrix O
4.1
SOLUTIONS
Notes: This section is designed to avoid the standard exercises in which a student is asked to check ten axioms on an array of sets. Theorem 1 provides the main homework tool in this section for showing that a set is a subspace. Students should be taught how to check the closure axioms. The exercises in this section (and the next few sections) emphasize n, to give students time to absorb the abstract concepts. Other vectors do appear later in the chapter: the space of signals is used in Section 4.8, and the spaces n of polynomials are used in many sections of Chapters 4 and 6.
1. a. If u and v are in V, then their entries are nonnegative. Since a sum of nonnegative numbers is nonnegative, the vector u + v has nonnegative entries. Thus u + v is in V.
2 b. Example: If u = and c = 1, then u is in V but cu is not in V. 2 x cx x 2. a. If u = is in W, then the vector cu = c = is in W because (cx)(cy ) = c 2 ( xy ) 0 y cy y since xy 0. -1 2 b. Example: If u = and v = , then u and v are in W but u + v is not in W. 3 -7 .5 3. Example: If u = and c = 4, then u is in H but cu is not in H. Since H is not closed under scalar .5 multiplication, H is not a subspace of 2. 4. Note that u and v are on the line L, but u + v is not.
u u+v
v L
5. Yes. Since the set is Span{t 2 } , the set is a subspace by Theorem 1.
185
186
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6. No. The zero vector is not in the set. 7. No. The set is not closed under multiplication by scalars which are not integers. 8. Yes. The zero vector is in the set H. If p and q are in H, then (p + q)(0) = p(0) + q(0) = 0 + 0 = 0, so p + q is in H. For any scalar c, (cp)(0) = c p(0) = c 0 = 0, so cp is in H. Thus H is a subspace by Theorem 1.
1 9. The set H = Span {v}, where v = 3 . Thus H is a subspace of 2 2 10. The set H = Span {v}, where v = 0 . Thus H is a subspace of -1
3
by Theorem 1.
3
by Theorem 1.
2 5 11. The set W = Span {u, v}, where u = 1 and v = 0 . Thus W is a subspace of 0 1
3
by Theorem 1.
3 1 -1 1 12. The set W = Span {u, v}, where u = and v = . Thus W is a subspace of 2 -1 0 4
4
by Theorem 1.
13. a. The vector w is not in the set {v1 , v 2 , v 3 } . There are 3 vectors in the set {v1 , v 2 , v 3 }.
b. The set Span{v1 , v 2 , v 3 } contains infinitely many vectors. c. The vector w is in the subspace spanned by {v1 , v 2 , v 3 } if and only if the equation x1 v1 + x2 v 2 + x3 v 3 = w has a solution. Row reducing the augmented matrix for this system of linear equations gives
1 0 -1
2 1 3
4 2 6
3 1 1 0 2 0
0 1 0
0 2 0
1 1 , 0
so the equation has a solution and w is in the subspace spanned by {v1 , v 2 , v 3 } . 14. The augmented matrix is found as in Exercise 13c. Since
1 0 -1
2 1 3
4 2 6
8 1 4 0 7 0
0 1 0
0 2 0
0 0 , 1
the equation x1 v1 + x2 v 2 + x3 v 3 = w has no solution, and w is not in the subspace spanned by {v1 , v 2 , v 3 }. 15. Since the zero vector is not in W, W is not a vector space. 16. Since the zero vector is not in W, W is not a vector space.
4.1
Solutions
187
17. Since a vector w in W may be written as
1 -1 0 0 1 -1 w = a +b + c -1 0 1 0 1 0 S = 1 -1 0 0 1 -1 , , -1 0 1 0 1 0
is a set that spans W. 18. Since a vector w in W may be written as
4 3 0 0 0 0 w = a +b + c 1 1 1 -2 0 1 S = 4 3 0 0 0 0 , , 1 1 1 -2 0 1
is a set that spans W. 19. Let H be the set of all functions described by y (t ) = c1cos t + c2sin t. Then H is a subset of the vector space V of all real-valued functions, and may be written as H = Span {cos t, sin t}. By Theorem 1, H is a subspace of V and is hence a vector space. 20. a. The following facts about continuous functions must be shown. 1. The constant function f(t) = 0 is continuous. 2. The sum of two continuous functions is continuous. 3. A constant multiple of a continuous function is continuous. b. Let H = {f in C[a, b]: f(a) = f(b)}. 1. Let g(t) = 0 for all t in [a, b]. Then g(a) = g(b) = 0, so g is in H. 2. Let g and h be in H. Then g(a) = g(b) and h(a) = h(b), and (g + h)(a) = g(a) + h(a) = g(b) + h(b) = (g + h)(b), so g + h is in H. 3. Let g be in H. Then g(a) = g(b), and (cg)(a) = cg(a) = cg(b) = (cg)(b), so cg is in H. Thus H is a subspace of C[a, b]. 21. The set H is a subspace of M 22 . The zero matrix is in H, the sum of two upper triangular matrices is upper triangular, and a scalar multiple of an upper triangular matrix is upper triangular. 22. The set H is a subspace of M 24 . The 2 4 zero matrix 0 is in H because F0 = 0. If A and B are matrices in H, then F(A + B) = FA + FB = 0 + 0 = 0, so A + B is in H. If A is in H and c is a scalar, then F(cA) = c(FA) = c0 = 0, so cA is in H.
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Vector Spaces
23. a. False. The zero vector in V is the function f whose values f(t) are zero for all t in . b. False. An arrow in three-dimensional space is an example of a vector, but not every arrow is a vector. c. False. See Exercises 1, 2, and 3 for examples of subsets which contain the zero vector but are not subspaces. d. True. See the paragraph before Example 6. e. False. Digital signals are used. See Example 3. 24. a. b. c. d. e. True. See the definition of a vector space. True. See statement (3) in the box before Example 1. True. See the paragraph before Example 6. False. See Example 8. False. The second and third parts of the conditions are stated incorrectly. For example, part (ii) does not state that u and v represent all possible elements of H.
25. 2, 4 26. a. 3 b. 5 c. 4 27. a. b. c. d. 28. a. b. c. d. e. 8 3 5 4 4 7 3 5 4
29. Consider u + (1)u. By Axiom 10, u + (1)u = 1u + (1)u. By Axiom 8, 1u + (1)u = (1 + (1))u = 0u. By Exercise 27, 0u = 0. Thus u + (1)u = 0, and by Exercise 26 (1)u = u. 30. By Axiom 10 u = 1u. Since c is nonzero, c -1c = 1 , and u = (c -1c)u . By Axiom 9, (c -1c)u = c -1 (cu) = c -1 0 since cu = 0. Thus u = c -1 0 = 0 by Property (2), proven in Exercise 28. 31. Any subspace H that contains u and v must also contain all scalar multiples of u and v, and hence must also contain all sums of scalar multiples of u and v. Thus H must contain all linear combinations of u and v, or Span {u, v}.
Note: Exercises 3234 provide good practice for mathematics majors because these arguments involve simple symbol manipulation typical of mathematical proofs. Most students outside mathematics might profit more from other types of exercises.
32. Both H and K contain the zero vector of V because they are subspaces of V. Thus the zero vector of V is in H K. Let u and v be in H K. Then u and v are in H. Since H is a subspace u + v is in H. Likewise u and v are in K. Since K is a subspace u + v is in K. Thus u + v is in H K. Let u be in H K. Then u is in H. Since H is a subspace cu is in H. Likewise v is in K. Since K is a subspace cu is in K. Thus cu is in H K for any scalar c, and H K is a subspace of V.
4.1
Solutions
189
The union of two subspaces is not in general a subspace. For an example in 2 let H be the x-axis and let K be the y-axis. Then both H and K are subspaces of 2, but H K is not closed under vector addition. The subset H K is thus not a subspace of 2. 33. a. Given subspaces H and K of a vector space V, the zero vector of V belongs to H + K, because 0 is in both H and K (since they are subspaces) and 0 = 0 + 0. Next, take two vectors in H + K, say w1 = u1 + v1 and w 2 = u 2 + v 2 where u1 and u 2 are in H, and v1 and v 2 are in K. Then w1 + w 2 = u1 + v1 + u 2 + v 2 = (u1 + u 2 ) + ( v1 + v 2 ) because vector addition in V is commutative and associative. Now u1 + u 2 is in H and v1 + v 2 is in K because H and K are subspaces. This shows that w1 + w 2 is in H + K. Thus H + K is closed under addition of vectors. Finally, for any scalar c, cw1 = c(u1 + v1 ) = cu1 + cv1 The vector cu1 belongs to H and cv1 belongs to K, because H and K are subspaces. Thus, cw1 belongs to H + K, so H + K is closed under multiplication by scalars. These arguments show that H + K satisfies all three conditions necessary to be a subspace of V. b. Certainly H is a subset of H + K because every vector u in H may be written as u + 0, where the zero vector 0 is in K (and also in H, of course). Since H contains the zero vector of H + K, and H is closed under vector addition and multiplication by scalars (because H is a subspace of V ), H is a subspace of H + K. The same argument applies when H is replaced by K, so K is also a subspace of H + K. 34. A proof that H + K = Span{u1 ,..., u p , v1 ,..., v q } has two parts. First, one must show that H + K is a subset of Span{u1 ,..., u p , v1 ,..., v q }. Second, one must show that Span{u1 ,..., u p , v1 ,..., v q } is a subset of H + K. (1) A typical vector H has the form c1u1 + ...+ c p u p and a typical vector in K has the form
d1 v1 + ...+ d q v q . The sum of these two vectors is a linear combination of u1 ,..., u p , v1 ,..., v q and so
belongs to Span{u1 ,..., u p , v1 ,..., v q }. Thus H + K is a subset of Span{u1 ,..., u p , v1 ,..., v q }. (2) Each of the vectors u1 ,..., u p , v1 ,..., v q belongs to H + K, by Exercise 33(b), and so any linear combination of these vectors belongs to H + K, since H + K is a subspace, by Exercise 33(a). Thus, Span{u1 ,..., u p , v1 ,..., v q } is a subset of H + K. 35. [M] Since
7 -4 -2 9
-4 5 -1 -7
-9 4 4 -7
-9 1 7 0 4 0 8 0
0 1 0 0
0 15 / 2 0 3 , 1 11/ 2 0 0
w is in the subspace spanned by {v1 , v 2 , v 3 }. 36. [M] Since
[A
5 8 y] = -5 3
-5 8 -9 -2
-9 -6 3 -7
6 1 7 0 1 0 -4 0
0 1 0 0
0 11/ 2 0 -2 , 1 7 / 2 0 0
y is in the subspace spanned by the columns of A.
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Vector Spaces
37. [M] The graph of f(t) is given below. A conjecture is that f(t) = cos 4t.
1 0.5 1 0.5 1 2 3 4 5 6
The graph of g(t) is given below. A conjecture is that g(t) = cos 6t.
1 0.5 1 0.5 1 2 3 4 5 6
38. [M] The graph of f(t) is given below. A conjecture is that f(t) = sin 3t.
1 0.5 1 0.5 1 2 3 4 5 6
The graph of g(t) is given below. A conjecture is that g(t) = cos 4t.
1 0.5 1 0.5 1 2 3 4 5 6
The graph of h(t) is given below. A conjecture is that h(t) = sin 5t.
1 0.5 1 0.5 1 2 3 4 5 6
4.2
Solutions
191
4.2
SOLUTIONS
Notes: This section provides a review of Chapter 1 using the new terminology. Linear tranformations are introduced quickly since students are already comfortable with the idea from n. The key exercises are 1726, which are straightforward but help to solidify the notions of null spaces and column spaces. Exercises 3036 deal with the kernel and range of a linear transformation and are progressively more advanced theoretically. The idea in Exercises 714 is for the student to use Theorems 1, 2, or 3 to determine whether a given set is a subspace.
1. One calculates that
3 Aw = 6 -8
-5 -2 4
-3 1 0 0 3 = 0 , 1 -4 0
so w is in Nul A. 2. One calculates that
5 Aw = 13 8
21 19 5 0 23 2 -3 = 0 , 14 1 2 0
so w is in Nul A. 3. First find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0] 0
0 1
-7 4
6 -2
0 , 0
the general solution is x1 = 7 x3 - 6 x4 , x2 = -4 x3 + 2 x4 , with x3 and x4 free. So
x1 7 -6 x -4 2 x = 2 = x3 + x4 , x3 1 0 0 1 x4
and a spanning set for Nul A is
7 -6 -4 , 2 1 0 0 1 .
4. First find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0] 0
-6 0
0 1
0 0
0 , 0
the general solution is x1 = 6 x2 , x3 = 0 , with x2 and x4 free. So
x1 6 0 x 1 0 x = 2 = x2 + x4 , x3 0 0 0 1 x4
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CHAPTER 4
Vector Spaces
and a spanning set for Nul A is
6 0 1 0 , 0 0 0 1 .
5. First find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0] 0 0
-2 0 0
0 1 0
4 -9 0
0 0 1
0 0 , 0
the general solution is x1 = 2 x2 - 4 x4 , x3 = 9 x4 , x5 = 0 , with x2 and x4 free. So
x1 2 -4 x 1 0 2 x = x3 = x2 0 + x4 9 , x4 0 1 x5 0 0
and a spanning set for Nul A is
2 -4 1 0 0 , 9 0 1 0 0 . -8 1 0
6. First find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0] 0 0
0 1 0
6 -2 0
1 0 0
0 0 , 0
the general solution is x1 = -6 x3 + 8 x4 - x5 , x2 = 2 x3 - x4 , with x3 , x4 , and x5 free. So x1 -6 8 -1 x 2 -1 0 2 x = x3 = x3 1 + x4 0 + x5 0 , x4 0 1 0 x5 0 0 1 and a spanning set for Nul A is -6 8 -1 2 -1 0 1 , 0 , 0 0 1 0 0 0 1 .
4.2
Solutions
193
7. The set W is a subset of would be a subspace of not a vector space. 8. The set W is a subset of would be a subspace of not a vector space.
3 3
. If W were a vector space (under the standard operations in 3), then it . But W is not a subspace of 3 since the zero vector is not in W. Thus W is . If W were a vector space (under the standard operations in 3), then it . But W is not a subspace of 3 since the zero vector is not in W. Thus W is
3 3
9. The set W is the set of all solutions to the homogeneous system of equations a 2b 4c = 0, 0 1 -2 -4 4 2a c 3d = 0. Thus W = Nul A, where A = . Thus W is a subspace of by 0 -1 -3 2 Theorem 2, and is a vector space. 10. The set W is the set of all solutions to the homogeneous system of equations a + 3b c = 0, 0 1 3 -1 a + b + c d = 0. Thus W = Nul A, where A = . Thus W is a subspace of 4 by 1 -1 1 1 Theorem 2, and is a vector space. 11. The set W is a subset of would be a subspace of a vector space.
4 4
. If W were a vector space (under the standard operations in 4), then it . But W is not a subspace of 4 since the zero vector is not in W. Thus W is not
12. The set W is a subset of 4. If W were a vector space (under the standard operations in 4), then it would be a subspace of 4. But W is not a subspace of 4 since the zero vector is not in W. Thus W is not a vector space. 13. An element w on W may be written as
1 -6 1 w = c 0 + d 1 = 0 1 0 1 -6 c 1 d 0 -6 1 . Thus W is a subspace of 0
3
1 where c and d are any real numbers. So W = Col A where A = 0 1 Theorem 3, and is a vector space.
by
14. An element w on W may be written as
-1 2 -1 1 + b -2 = 1 w = a 3 -6 3 2 a -2 b -6 2 -2 . Thus W is a subspace of -6
3
-1 where a and b are any real numbers. So W = Col A where A = 1 3 Theorem 3, and is a vector space.
by
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CHAPTER 4
Vector Spaces
15. An element in this set may be written as
0 2 3 0 1 1 -2 1 r + s +t = 4 1 0 4 3 -1 -1 3 2 1 1 -1 3 r -2 s 0 t -1 2 1 1 -1 3 -2 . 0 -1
0 1 where r, s and t are any real numbers. So the set is Col A where A = 4 3
16. An element in this set may be written as
1 -1 0 1 2 1 1 2 b + c + d = 0 5 -4 0 0 0 1 0 -1 1 5 0 0 b 1 c -4 d 1 -1 1 5 0 0 1 . -4 1
1 2 where b, c and d are any real numbers. So the set is Col A where A = 0 0
17. The matrix A is a 4 2 matrix. Thus (a) Nul A is a subspace of 2, and (b) Col A is a subspace of 4. 18. The matrix A is a 4 3 matrix. Thus (a) Nul A is a subspace of 3, and (b) Col A is a subspace of 4. 19. The matrix A is a 2 5 matrix. Thus (a) Nul A is a subspace of 5, and (b) Col A is a subspace of 2. 20. The matrix A is a 1 5 matrix. Thus (a) Nul A is a subspace of 5, and (b) Col A is a subspace of 1 = . 21. Either column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0 0] 0 0
-3 0 0 0
0 0 , 0 0
4.2
Solutions
195
the general solution is x1 = 3 x2 , with x2 free. Letting x2 be a nonzero value (say x2 = 1 ) gives the nonzero vector x 3 x = 1 = x2 1 which is in Nul A.
22. Any column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general solution of Ax = 0 in terms of the free variables. Since
[A
1 0] 0
0 1
-7 4
6 -2
0 , 0
the general solution is x1 = 7 x3 - 6 x4 , x2 = -4 x3 + 2 x4 , with x3 and x4 free. Letting x3 and x4 be nonzero values (say x3 = x4 = 1 ) gives the nonzero vector x1 1 x -2 x = 2 = x3 1 x4 1 which is in Nul A.
23. Consider the system with augmented matrix [ A
w ] . Since
1 -2 -1/ 3 w] , 0 0 0 the system is consistent and w is in Col A. Also, since
[A
-6 12 2 0 Aw = = -3 6 1 0
w is in Nul A. 24. Consider the system with augmented matrix [ A w ] . Since
[A
1 w ] 0 0 -2 4 0
0 1 1 1/ 2 0 0
-1/ 2 1 , 0
the system is consistent and w is in Col A. Also, since -8 Aw = 6 4 -9 2 0 8 1 = 0 4 -2 0
w is in Nul A. 25. a. b. c. d. e. f.
True. See the definition before Example 1. False. See Theorem 2. True. See the remark just before Example 4. False. The equation Ax = b must be consistent for every b. See #7 in the table on page 226. True. See Figure 2. True. See the remark after Theorem 3.
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26. a. b. c. d. e. f.
True. See Theorem 2. True. See Theorem 3. False. See the box after Theorem 3. True. See the paragraph after the definition of a linear transformation. True. See Figure 2. True. See the paragraph before Example 8.
27. Let A be the coefficient matrix of the given homogeneous system of equations. Since Ax = 0 for 3 x = 2 , x is in Nul A. Since Nul A is a subspace of 3, it is closed under scalar multiplication. Thus -1
30 10x = 20 is also in Nul A, and x1 = 30 , x2 = 20 , x3 = -10 is also a solution to the system of -10 equations.
28. Let A be the coefficient matrix of the given systems of equations. Since the first system has a solution, 0 the constant vector b = 1 is in Col A. Since Col A is a subspace of 3, it is closed under scalar 9
0 multiplication. Thus 5b = 5 is also in Col A, and the second system of equations must thus have a 45 solution.
29. a. Since A0 = 0, the zero vector is in Col A.
b. Since Ax + Aw = A(x + w ), A x + Aw is in Col A. c. Since c( Ax) = A(cx), cAx is in Col A. 30. Since T (0V ) = 0W , the zero vector 0W of W is in the range of T. Let T(x) and T(w) be typical elements in the range of T. Then since T (x) + T (w ) = T ( x + w ), T (x) + T (w ) is in the range of T and the range of T is closed under vector addition. Let c be any scalar. Then since cT (x) = T (cx), cT (x) is in the range of T and the range of T is closed under scalar multiplication. Hence the range of T is a subspace of W. 31. a. Let p and q be arbitary polynomials in
2,
and let c be any scalar. Then
(p + q)(0) p(0) + q(0) p(0) q(0) T (p + q ) = = = + = T (p) + T (q) (p + q)(1) p(1) + q(1) p(1) q(1)
and
(cp)(0) p(0) T (cp) = = c p(1) = cT (p) (cp)(1)
so T is a linear transformation.
4.2
Solutions
197
b. Any quadratic polynomial q for which q(0) = 0 and q(1) = 0 will be in the kernel of T. The
x polynomial q must then be a multiple of p(t ) = t (t - 1). Given any vector 1 in x2 p = x1 + ( x2 - x1 )t has p(0) = x1 and p(1) = x2 . Thus the range of T is all of 2.
2
, the polynomial
32. Any quadratic polynomial q for which q(0) = 0 will be in the kernel of T. The polynomial q must then be q = at + bt 2 . Thus the polynomials p1 (t ) = t and p 2 (t ) = t 2 span the kernel of T. If a vector is in the
a range of T, it must be of the form . If a vector is of this form, it is the image of the polynomial a a p(t ) = a in 2. Thus the range of T is : a real . a
33. a. For any A and B in M 22 and for any scalar c, T ( A + B ) = ( A + B ) + ( A + B )T = A + B + AT + BT = ( A + AT ) + ( B + BT ) = T ( A) + T ( B ) and T (cA) = (cA)T = c( AT ) = cT ( A) so T is a linear transformation. b. Let B be an element of M 22 with BT = B, and let A = 1 B. Then 2
T ( A) = A + AT =
1 1 1 1 1 1 B + ( B )T = B + BT = B + B = B 2 2 2 2 2 2
c. Part b. showed that the range of T contains the set of all B in M 22 with BT = B. It must also be shown that any B in the range of T has this property. Let B be in the range of T. Then B = T(A) for some A in M 22 . Then B = A + AT , and
BT = ( A + AT )T = AT + ( AT )T = AT + A = A + AT = B so B has the property that BT = B.
a d. Let A = c
b be in the kernel of T. Then T ( A) = A + AT = 0 , so d
c + b 0 0 a b a c 2a + = = A + AT = c d b d b + c 2d 0 0 Solving it is found that a = d = 0 and c = -b . Thus the kernel of T is 0 b : b real . -b 0
34. Let f and g be any elements in C[0, 1] and let c be any scalar. Then T(f) is the antiderivative F of f with F(0) = 0 and T(g) is the antiderivative G of g with G(0) = 0. By the rules for antidifferentiation F + G will be an antiderivative of f + g, and ( F + G )(0) = F (0) + G (0) = 0 + 0 = 0. Thus T (f + g ) = T ( f ) + T (g ). Likewise cF will be an antiderivative of cf, and (cF )(0) = cF (0) = c0 = 0. Thus T (cf ) = cT (f ), and T is a linear transformation. To find the kernel of T, we must find all functions f in C[0,1] with antiderivative equal to the zero function. The only function with this property is the zero function 0, so the kernel of T is {0}.
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35. Since U is a subspace of V, 0V is in U. Since T is linear, T (0V ) = 0W . So 0W is in T(U). Let T(x) and T(y) be typical elements in T(U). Then x and y are in U, and since U is a subspace of V, x + y is also in U. Since T is linear, T (x) + T (y ) = T (x + y ). So T ( x) + T (y ) is in T(U), and T(U) is closed under vector addition. Let c be any scalar. Then since x is in U and U is a subspace of V, cx is in U. Since T is linear, T (cx) = cT (x) and cT(x) is in T(U ). Thus T(U) is closed under scalar multiplication, and T(U) is a subspace of W. 36. Since Z is a subspace of W, 0W is in Z. Since T is linear, T (0V ) = 0W . So 0V is in U. Let x and y be typical elements in U. Then T(x) and T(y) are in Z, and since Z is a subspace of W, T ( x) + T (y ) is also in Z. Since T is linear, T (x) + T (y ) = T (x + y ). So T ( x + y ) is in Z, and x + y is in U. Thus U is closed under vector addition. Let c be any scalar. Then since x is in U, T(x) is in Z. Since Z is a subspace of W, cT(x) is also in Z. Since T is linear, cT (x) = T (cx) and T(cx) is in T(U). Thus cx is in U and U is closed under scalar multiplication. Hence U is a subspace of V. 37. [M] Consider the system with augmented matrix [ A w ]. Since
[A
1 0 w] 0 0
0 1 0 0
0 0 1 0 -4 0 7 7
-1/ 95 39 /19 267 / 95 0
1/ 95 -20 /19 , -172 / 95 0
the system is consistent and w is in Col A. Also, since
7 -5 Aw = 9 19
6 -1 -11 -9
1 1 14 -2 1 0 = -3 -1 0 1 -3 0 w ] . Since
w is not in Nul A. 38. [M] Consider the system with augmented matrix [ A
[A
1 0 w] 0 0
0 1 0 0
-1 -2 0 0 -2 1 6 1
0 0 1 0
-2 -3 , 1 0
the system is consistent and w is in Col A. Also, since
-8 -5 Aw = 10 3
5 2 -8 -2
0 1 0 -2 2 0 = -3 1 0 0 0 0
w is in Nul A.
4.2
Solutions
199
39. [M] a. To show that a3 and a5 are in the column space of B, we can row reduce the matrices [ B
[B
a3 ] :
a3 ] and
1 0 [ B a3 ] 0 0
0 1 0 0 0 1 0 0
0 1/ 3 0 1/ 3 1 0 0 0 0 0 1 0 10 / 3 -26 / 3 -4 0
[B
1 0 a5 ] 0 0
Since both these systems are consistent, a3 and a5 are in the column space of B. Notice that the same conclusions can be drawn by observing the reduced row echelon form for A:
1 0 A 0 0
0 1/ 3 1 1/ 3 0 0 0 0
0 0 1 0
10 / 3 -26 / 3 -4 0
b. We find the general solution of Ax = 0 in terms of the free variables by using the reduced row echelon form of A given above: x1 = (-1/ 3) x3 - (10 / 3) x5 , x2 = (-1/ 3) x3 + (26 / 3) x5 , x4 = 4 x5 with x3 and x5 free. So
x1 -1/ 3 -10 / 3 x -1/ 3 26 / 3 2 x = x3 = x3 1 + x5 0 , 0 4 x4 x5 0 1
and a spanning set for Nul A is
-1/ 3 -10 / 3 -1/ 3 26 / 3 1 , 0 0 4 0 1 .
c. The reduced row echelon form of A shows that the columns of A are linearly dependent and do not span 4. Thus by Theorem 12 in Section 1.9, T is neither one-to-one nor onto. 40. [M] Since the line lies both in H = Span{v1 , v 2 } and in K = Span{v 3 , v 4 } , w can be written both as c1 v1 + c2 v 2 and c3 v 3 + c4 v 4 . To find w we must find the cj's which solve c1 v1 + c2 v 2 - c3 v 3 - c4 v 4 = 0 . Row reduction of [ v1
5 3 8
v2
- v3
-v4 0 1 0 0 0 1
0] yields -10 / 3 26 / 3 -4
1 3 4
-2 1 -5
0 12 28
0 1 0 0 0 0
0 0 , 0
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so the vector of cj's must be a multiple of (10/3, 26/3, 4, 1). One simple choice is (10, 26, 12, 3), which gives w = 10 v1 - 26 v 2 = 12 v 3 + 3v 4 = (24, -48, -24) . Another choice for w is (1, 2, 1).
4.3
SOLUTIONS
Notes: The definition for basis is given initially for subspaces because this emphasizes that the basis elements must be in the subspace. Students often overlook this point when the definition is given for a vector space (see Exercise 25). The subsection on bases for Nul A and Col A is essential for Sections 4.5 and 4.6. The subsection on "Two Views of a Basis" is also fundamental to understanding the interplay between linearly independent sets, spanning sets, and bases. Key exercises in this section are Exercises 2125, which help to deepen students' understanding of these different subsets of a vector space.
1. Consider the matrix whose columns are the given set of vectors. This 3 3 matrix is in echelon form, and has 3 pivot positions. Thus by the Invertible Matrix Theorem, its columns are linearly independent and span 3. So the given set of vectors is a basis for 3. 2. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent and thus cannot be a basis for 3. Now consider the matrix whose columns are the given set of vectors. This 3 3 matrix has only 2 pivot positions. Thus by the Invertible Matrix Theorem, its columns do not span 3. 3. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this matrix is
1 3 -3 1 0 9 / 2 0 2 -5 0 1 -5 / 2 -2 -4 1 0 0 0
so the matrix has only two pivot positions. Thus its columns do not form a basis for is neither linearly independent nor does it span 3.
3
; the set of vectors
4. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this matrix is
2 -2 1
1 -3 2
-7 1 5 0 4 0
0 1 0
0 0 1
3
so the matrix has three pivot positions. Thus its columns form a basis for
.
5. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent and thus cannot be a basis for 3. Now consider the matrix whose columns are the given set of vectors. The reduced echelon form of this matrix is
1 -3 0
-2 9 0
0 0 0
0 1 -3 0 5 0
0 1 0
0 0 0
0 0 1
3
so the matrix has a pivot in each row. Thus the given set of vectors spans
.
4.3
Solutions
201
6. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot in each row, its columns cannot span 3; thus the given set of vectors is not a basis for 3. The reduced echelon form of the matrix is
1 2 -3
-4 1 -5 0 6 0
0 1 0
so the matrix has a pivot in each column. Thus the given set of vectors is linearly independent. 7. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot in each row, its columns cannot span 3; thus the given set of vectors is not a basis for 3. The reduced echelon form of the matrix is
-2 3 0
6 1 -1 0 5 0
0 1 0
so the matrix has a pivot in each column. Thus the given set of vectors is linearly independent. 8. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot in each column, the set cannot be linearly independent and thus cannot be a basis for 3. The reduced echelon form of this matrix is
1 -4 3
0 3 -1
3 -5 4
0 1 2 0 -2 0
0 1 0
0 0 1
-3/ 2 -1/ 2 1/ 2
3
so the matrix has a pivot in each row. Thus the given set of vectors spans
.
9. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon form of A:
1 0 3
0 1 -2
-3 -5 1
2 1 4 0 -2 0
0 1 0
-3 -5 0
2 4 . 0
So x1 = 3 x3 - 2 x4 , x2 = 5 x3 - 4 x4 , with x3 and x4 free. So
x1 3 -2 x 5 -4 x = 2 = x3 + x4 , x3 1 0 0 1 x4
and a basis for Nul A is
3 -2 5 -4 , 1 0 0 1 .
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10. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon form of A:
1 -2 0
0 1 2
-5 6 -8
1 -2 1
4 1 -2 0 9 0
0 1 0
-5 -4 0
0 0 1
7 6 . -3
So x1 = 5 x3 - 7 x5 , x2 = 4 x3 - 6 x5 , x4 = 3 x5 , with x3 and x5 free. So
x1 5 -7 x 4 -6 2 x = x3 = x3 1 + x5 0 , x4 0 3 x5 0 1
and a basis for Nul A is
5 -7 4 -6 1 , 0 0 3 0 1 .
11. Let A = [1
2
1] . Then we wish to find a basis for Nul A. We find the general solution of Ax = 0 in
-2 -1 y 1 + z 0 , 0 1
.
2
terms of the free variables: x = 2y z with y and z free. So
x x = y = z
and a basis for Nul A is
-2 -1 1 , 0 0 1
12. We want to find a basis for the set of vectors in
in the line 5x y = 0. Let A = [5
-1] . Then we wish
to find a basis for Nul A. We find the general solution of Ax = 0 in terms of the free variables: y = 5x with x free. So x 1 x = = x , y 5 and a basis for Nul A is 1 . 5
4.3
Solutions
203
13. Since B is a row echelon form of A, we see that the first and second columns of A are its pivot columns. Thus a basis for Col A is
-2 4 2 , -6 -3 8 .
To find a basis for Nul A, we find the general solution of Ax = 0 in terms of the free variables: x1 = -6 x3 - 5 x4 , x2 = (-5 / 2) x3 - (3/ 2) x4 , with x3 and x4 free. So x1 -6 -5 x -5 / 2 + x -3/ 2 , x = 2 = x3 4 x3 1 0 0 1 x4 and a basis for Nul A is -6 -5 -5 / 2 -3/ 2 , 1 0 0 1 .
14. Since B is a row echelon form of A, we see that the first, third, and fifth columns of A are its pivot columns. Thus a basis for Col A is
1 -5 -3 2 -5 2 , , 1 0 5 3 -5 -2 .
To find a basis for Nul A, we find the general solution of Ax = 0 in terms of the free variables, mentally completing the row reduction of B to get: x1 = -2 x2 - 4 x4 , x3 = (7 / 5) x4 , x5 = 0, with x2 and x4 free. So x1 -2 -4 x 1 0 2 x = x3 = x2 0 + x4 7 / 5 , x4 0 1 x5 0 0 and a basis for Nul A is -2 -4 1 0 0 , 7 / 5 0 1 0 0 .
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15. This problem is equivalent to finding a basis for Col A, where A = [ v1
reduced echelon form of A is 1 0 -3 2 0 1 2 -3 -3 -4 1 6 1 -3 -8 7 2 1 1 0 -6 0 9 0 0 1 0 0 -3 -4 0 0 0 0 1 0 4 -5 , -2 0
v2
v3
v4
v5 ] . Since the
we see that the first, second, and fourth columns of A are its pivot columns. Thus a basis for the space spanned by the given vectors is 1 0 1 0 1 -3 , , -3 2 -8 2 -3 7 .
16. This problem is equivalent to finding a basis for Col A, where A = [ v1
reduced echelon form of A is 1 0 0 1 -2 1 -1 1 6 -1 2 -1 5 -3 3 -4 0 1 3 0 -1 0 1 0 0 1 0 0 0 0 1 0 -1 -3 0 0 -2 5 , 2 0
v2
v3
v4
v5 ] . Since the
we see that the first, second, and third columns of A are its pivot columns. Thus a basis for the space spanned by the given vectors is 1 -2 6 0 1 -1 , , 0 -1 2 1 1 -1 .
17. [M] This problem is equivalent to finding a basis for Col A, where A = [ v1 the reduced echelon form of A is
8 9 -3 -6 0 4 5 1 -4 4 -1 -4 -9 6 -7 6 8 4 -7 10 -1 1 4 0 11 0 -8 0 -7 0 0 1 0 0 0 0 0 1 0 0 -1/ 2 5/ 2 0 0 0 3 -7 -3 , 0 0
v2
v3
v4
v5 ] . Since
we see that the first, second, and third columns of A are its pivot columns. Thus a basis for the space spanned by the given vectors is 8 4 -1 9 5 -4 -3 , 1 , -9 -6 -4 6 0 4 -7 .
4.3
Solutions
205
18. [M] This problem is equivalent to finding a basis for Col A, where A = [ v1
the reduced echelon form of A is -8 7 6 5 -7 8 -7 -9 -5 7 -8 7 4 5 -7 1 4 9 6 -7 -9 1 3 0 -4 0 -1 0 0 0 0 1 0 0 0 5/3 2/3 0 0 0 0 0 1 0 0 4 / 3 1/ 3 -1 , 0 0
v2
v3
v4
v 5 ]. Since
we see that the first, second, and fourth columns of A are its pivot columns. Thus a basis for the space spanned by the given vectors is -8 8 1 7 -7 4 6 , -9 , 9 5 -5 6 -7 7 -7 .
19. Since 4 v1 + 5 v 2 - 3v 3 = 0, we see that each of the vectors is a linear combination of the others. Thus the sets {v1 , v 2 }, {v1 , v 3 }, and {v 2 , v 3 } all span H. Since we may confirm that none of the three vectors is a multiple of any of the others, the sets {v1 , v 2 }, {v1 , v 3 }, and {v 2 , v 3 } are linearly independent and thus each forms a basis for H. 20. Since v1 - 3v 2 + 5v 3 = 0, we see that each of the vectors is a linear combination of the others. Thus the sets {v1 , v 2 }, {v1 , v 3 }, and {v 2 , v 3 } all span H. Since we may confirm that none of the three vectors is a multiple of any of the others, the sets {v1 , v 2 }, {v1 , v 3 }, and {v 2 , v 3 } are linearly independent and thus each forms a basis for H. 21. a. False. The zero vector by itself is linearly dependent. See the paragraph preceding Theorem 4. b. False. The set {b1 ,..., b p } must also be linearly independent. See the definition of a basis. c. True. See Example 3. d. False. See the subsection "Two Views of a Basis." e. False. See the box before Example 9. 22. a. False. The subspace spanned by the set must also coincide with H. See the definition of a basis. b. True. Apply the Spanning Set Theorem to V instead of H. The space V is nonzero because the spanning set uses nonzero vectors. c. True. See the subsection "Two Views of a Basis." d. False. See the two paragraphs before Example 8. e. False. See the warning after Theorem 6. 23. Let A = [ v1 v2 v3 v 4 ]. Then A is square and its columns span
4
since
4
= Span{v1 , v 2 , v3 , v 4 }.
So its columns are linearly independent by the Invertible Matrix Theorem, and {v1 , v 2 , v 3 , v 4 } is a basis for 4.
24. Let A = [ v1
n
...
v n ]. Then A is square and its columns are linearly independent, so its columns span
n
by the Invertible Matrix Theorem. Thus {v1 ,..., v n } is a basis for
.
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25. In order for the set to be a basis for H, {v1 , v 2 , v 3 } must be a spanning set for H; that is, H = Span{v1 , v 2 , v 3 }. The exercise shows that H is a subset of Span{v1 , v 2 , v 3 }. but there are vectors in Span{v1 , v 2 , v 3 } which are not in H ( v1 and v 3 , for example). So H Span{v1 , v 2 , v 3 }, and {v1 , v 2 , v 3 } is not a basis for H. 26. Since sin t cos t = (1/2) sin 2t, the set {sin t, sin 2t} spans the subspace. By inspection we note that this set is linearly independent, so {sin t, sin 2t} is a basis for the subspace. 27. The set {cos t, sin t} spans the subspace. By inspection we note that this set is linearly independent, so {cos t, sin t} is a basis for the subspace. 28. The set {e- bt , te-bt } spans the subspace. By inspection we note that this set is linearly independent, so
{e- bt , te-bt } is a basis for the subspace.
29. Let A be the n k matrix [ v1
...
v k ] . Since A has fewer columns than rows, there cannot be a pivot
n
position in each row of A. By Theorem 4 in Section 1.4, the columns of A do not span a basis for n.
and thus are not
30. Let A be the n k matrix [ v1
...
v k ] . Since A has fewer rows than columns rows, there cannot be a
pivot position in each column of A. By Theorem 8 in Section 1.6, the columns of A are not linearly independent and thus are not a basis for n.
31. Suppose that {v1 ,..., v p } is linearly dependent. Then there exist scalars c1 ,..., c p not all zero with c1 v1 + ... + c p v p = 0.
Since T is linear, T (c1 v1 + ...+ c p v p ) = c1T ( v1 ) + ...+ c pT ( v p ) and
T (c1 v1 + ...+ c p v p ) = T (0) = 0.
Thus
c1T ( v1 ) + ...+ c pT ( v p ) = 0
and since not all of the ci are zero, {T ( v1 ),..., T ( v p )} is linearly dependent.
32. Suppose that {T ( v1 ),..., T ( v p )} is linearly dependent. Then there exist scalars c1 ,..., c p not all zero with c1T ( v1 ) + ...+ c pT ( v p ) = 0.
Since T is linear,
T (c1 v1 + ...+ c p v p ) = c1T ( v1 ) + ... + c pT ( v p ) = 0 = T (0)
Since T is one-to-one
T (c1 v1 + ...+ c p v p ) = T (0)
implies that c1 v1 + ... + c p v p = 0. Since not all of the ci are zero, {v1 ,..., v p } is linearly dependent.
4.3
Solutions
207
3.
33. Neither polynomial is a multiple of the other polynomial. So {p1 , p 2 } is a linearly independent set in Note: {p1 , p 2 } is also a linearly independent set in 2 since p1 and p 2 both happen to be in 2. 34. By inspection, p3 = p1 + p 2 , or p1 + p 2 - p3 = 0 . By the Spanning Set Theorem, Span{p1 , p 2 , p3 } = Span{p1 , p 2 } . Since neither p1 nor p 2 is a multiple of the other, they are linearly independent and hence {p1 , p 2 } is a basis for Span{p1 , p 2 , p3 }. 35. Let {v1 , v 3 } be any linearly independent set in a vector space V, and let v 2 and v 4 each be linear combinations of v1 and v 3 . For instance, let v 2 = 5 v1 and v 4 = v1 + v 3 . Then {v1 , v 3 } is a basis for Span{v1 , v 2 , v 3 , v 4 }. 36. [M] Row reduce the following matrices to identify their pivot columns:
[u1
u2
1 2 u3 ] = 3 -1 1 0 v3 ] = 8 -4
0 2 -1 1 2 -2 9 -5
2 1 2 0 7 0 -3 0 -1 1 4 0 6 0 -2 0 1 2 v3 ] = 3 -1 1 0 0 0
0 1 0 0 0 1 0 0 0 2 -1 1 0 1 0 0
2 -1 , so {u1 , u 2 } is a basis for H. 0 0 3 -2 , so {v1 , v 2 } is a basis for K. 0 0 2 2 7 -3 2 -1 0 0 0 0 1 0 1 0 8 -4 2 -3 0 0 2 -2 9 -5 -1 4 6 -2
[ v1
v2
[u1
u2
u3
v1
v2
-4 6 , so {u1 , u 2 , v1} is a basis for H + K. 3 0
37. [M] For example, writing c1 t + c2 sin t + c3cos 2t + c4sin t cos t = 0
with t = 0, .1, .2, .3 gives the following coefficent matrix A for the homogeneous system Ac = 0 (to four decimal places): 0 .1 A= .2 .3 sin 0 sin .1 sin .2 sin .3 cos 0 cos .2 cos .4 cos .6 sin 0 cos 0 0 sin .1 cos .1 .1 = sin .2 cos .2 .2 sin .3 cos .3 .3 0 .0998 .1987 .2955 1 .9801 .9211 .8253 0 .0993 . .1947 .2823
This matrix is invertible, so the system Ac = 0 has only the trivial solution and {t, sin t, cos 2t, sin t cos t} is a linearly independent set of functions.
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38. [M] For example, writing c1 1 + c2 cos t + c3 cos 2t + c4 cos3t + c5 cos 4t + c6 cos5t + c7 cos 6t = 0
with t = 0, .1, .2, .3, .4, .5, .6 gives the following coefficent matrix A for the homogeneous system Ac = 0 (to four decimal places):
1 1 1 A = 1 1 1 1 1 1 1 = 1 1 1 1
cos 0 cos.1 cos.2 cos.3 cos.4 cos.5 cos.6 1 .9950 .9801 .9553 .9211 .8776 .8253
cos 2 0 cos 2 .1 cos 2 .2 cos 2 .3 cos 2 .4 cos 2 .5 cos 2 .6 1 .9900 .9605 .9127 .8484 .7702 .6812
cos3 0 cos3 .1 cos3 .2 cos3 .3 cos3 .4 cos3 .5 cos3 .6 1 .9851 .9414 .8719 .7814 .6759 .5622
cos 4 0 cos 4 .1 cos 4 .2 cos 4 .3 cos 4 .4 cos 4 .5 cos 4 .6
cos5 0 cos5 .1 cos5 .2 cos5 .3 cos5 .4 cos5 .5 cos5 .6 1 .9704 .8862 .7602 .6106 .4568 .3161
cos6 0 cos6 .1 cos6 .2 cos 6 .3 cos6 .4 cos6 .5 cos6 .6
1 .9802 .9226 .8330 .7197 .5931 .4640
1 .9753 .9042 .7958 .6629 .5205 .3830
This matrix is invertible, so the system Ac = 0 has only the trivial solution and {1, cos t, cos2t, cos3t, cos4t, cos5t, cos6t} is a linearly independent set of functions.
4.4
SOLUTIONS
n
Notes: Section 4.7 depends heavily on this section, as does Section 5.4. It is possible to cover the
parts of the two later sections, however, if the first half of Section 4.4 (and perhaps Example 7) is covered. The linearity of the coordinate mapping is used in Section 5.4 to find the matrix of a transformation relative to two bases. The change-of-coordinates matrix appears in Section 5.4, Theorem 8 and Exercise 27. The concept of an isomorphism is needed in the proof of Theorem 17 in Section 4.8. Exercise 25 is used in Section 4.7 to show that the change-of-coordinates matrix is invertible.
1. We calculate that
3 -4 3 x = 5 + 3 = . -5 6 -7
2. We calculate that
4 6 2 x = 8 + (-5) = . 5 7 5
3. We calculate that
1 5 4 -1 -4 + 0 2 + ( -1) -7 = -5 . x = 3 3 -2 0 9
4.4
Solutions
209
4. We calculate that
-1 3 4 0 2 + 8 -5 + (-7) -7 = 1 . x = (-4) 0 2 3 -5
5. The matrix [b1 6. The matrix [b1
b2
1 x ] row reduces to 0 1 x ] row reduces to 0
0 1 0 1
8 8 , so [x]B = -5 . -5 -6 -6 , so [x]B = 2 . 2 0 1 0 0 1 0 0 0 1 0 0 1 -1 -1 , so [x] = -1 . -1 B 3 3 -2 -2 0 , so [x]B = 0 . 5 5
2
b2
7. The matrix [b1
b2
b3
1 x ] row reduces to 0 0 1 x ] row reduces to 0 0
8. The matrix [b1
b2
b3
9. The change-of-coordinates matrix from B to the standard basis in 2 1 PB = [b1 b 2 ] = . -9 8 10. The change-of-coordinates matrix from B to the standard basis in 2 8 3 -1 PB = [b1 b 2 b3 ] = 0 -2 . 4 -5 7 11. Since PB
-1
is
3
is
converts x into its B-coordinate vector, we find that
-1
3 [x]B = PB x = -5
-4 2 -3 = 6 -6 -5 / 2
-1
-2 2 6 = . -3/ 2 -6 4
12. Since PB
-1
converts x into its B-coordinate vector, we find that 6 2 -7 / 2 = 7 0 5 / 2
-1
4 -1 [x]B = PB x = 5
3 2 -7 = . -2 0 5
13. We must find c1 , c2 , and c3 such that
c1 (1 + t 2 ) + c2 (t + t 2 ) + c3 (1 + 2t + t 2 ) = p(t ) = 1 + 4t + 7t 2 .
Equating the coefficients of the two polynomials produces the system of equations c1 + c3 = 1 c2 + 2c3 = 4 c1 + c2 + c3 = 7
210
CHAPTER 4
Vector Spaces
We row reduce the augmented matrix for the system of equations to find
1 0 1
0 1 1
1 2 1
1 1 4 0 7 0
0 1 0
0 0 1
2 2 , so [p] = 6 . 6 B -1 -1
One may also solve this problem using the coordinate vectors of the given polynomials relative to the standard basis {1, t , t 2 }; the same system of linear equations results.
14. We must find c1 , c2 , and c3 such that
c1 (1 - t 2 ) + c2 (t - t 2 ) + c3 (2 - 2t + t 2 ) = p(t ) = 3 + t - 6t 2 . Equating the coefficients of the two polynomials produces the system of equations c1 + 2c3 = 3 c2 - 2c3 = 1 -c1 - c2 + c3 = -6 We row reduce the augmented matrix for the system of equations to find
1 0 -1
0 1 -1
2 -2 1
3 1 1 0 -6 0
0 1 0
0 0 1
7 7 , so [p] = -3 . -3 B -2 -2
One may also solve this problem using the coordinate vectors of the given polynomials relative to the standard basis {1, t , t 2 }; the same system of linear equations results.
15. a. True. See the definition of the B-coordinate vector. b. False. See Equation (4). c. False. 3 is isomorphic to 4. See Example 5. 16. a. True. See Example 2. b. False. By definition, the coordinate mapping goes in the opposite direction. c. True. If the plane passes through the origin, as in Example 7, the plane is isomorphic to
2
.
1 2 -3 1 17. We must solve the vector equation x1 + x2 + x3 = . We row reduce the augmented -3 -8 7 1 matrix for the system of equations to find 1 -3
2 -8
-3 1 1 7 1 0
0 1
-5 1
5 . -2
Thus we can let x1 = 5 + 5 x3 and x2 = -2 - x3 , where x3 can be any real number. Letting x3 = 0 and
1 x3 = 1 produces two different ways to express as a linear combination of the other vectors: 1 5v1 - 2 v 2 and 10 v1 - 3v 2 + v 3 . There are infintely many correct answers to this problem.
18. For each k, b k = 0 b1 + + 1 b k + + 0 b n , so [b k ]B = (0,...,1,...,0) = e k . 19. The set S spans V because every x in V has a representation as a (unique) linear combination of elements in S. To show linear independence, suppose that S = {v1 ,..., v n } and that c1 v1 + + cn v n = 0 for some scalars c1 , ..., cn . The case when c1 = = cn = 0 is one possibility. By hypothesis, this is the unique