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 explicitl
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,
26
CHAPTER 1
1
~ 0
0
3
7
0
Linear Equations in Linear Algebra
1
= 0
6
b2 + 3b1
0 b3 5b1 + 2(b2 + 3b1 ) 0
4
b1
3
7
0
6
b2 + 3b1
0 b1 + 2b2 + b3
4
b1
The equation Ax = b is consistent if and only if b1 + 2b2 + b3 = 0. The set of such b is a plane th
28
CHAPTER 1
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 ha
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 import
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
5
6
5 12 15 3
2 = 2 4 3 3 = 8 + 9 = 1 , and
3
3
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.
7
8
4
7
4
11
5
6
38. [M]
4
9
9
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
1.6
1.6
Solutions
39
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
Services
input
.7
.
38
CHAPTER 1
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.)
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
5
4
7
8
9
x1
+ 4 x3 = 5
x2 3 x3 =
4
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 syst
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
T
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,
20
CHAPTER 1
Linear Equations in Linear Algebra
25. a. There are only three vectors in the set cfw_a1, a2, a3, and b is not one of them.
b. There are infinitely many vectors in W = Spancfw_a1, a2, a3. To determine if b is in W, use the method
of Exercise
8
CHAPTER 1
1.2
Linear Equations in Linear Algebra
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
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
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 o
1.1
SOLUTIONS
Notes: The 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
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
Replace R2 by R2 + (1)R1 and obtain:
x1 + 5 x2 = 7
7 x2 = 9
x1 + 5 x2 = 7
Scale R2 by 1/7:
x2
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 com