Solutions Manual
Linear Algebra A Modern Introduction 4th Edition
By
David Poole
( All Chapters Included - 100% Verified Solutions )
1
,Chapter 1
Vectors
1.1 The Geometry and Algebra of Vectors
H-2, 3L 3 H2, 3L
1.
2
1
H3, 0L
-2 -1 1 2 3
-1
H3, -2L
-2
2. Since
2 3 5 2 2 4 2 −2 0 2 3 5
+ = , + = , + = , + = ,
−3 0 −3 −3 3 0 −3 3 0 −3 −2 −5
plotting those vectors gives
1 2 3 4 5
-1
c b
-2
a
-3
d
-4
-5
3
2
,4 CHAPTER 1. VECTORS
3. 2 z
c 1 b
-2 -1 0
0
1 y2
0 -1
2
1 a
3
x
-1
d
-2
4. Since the heads are all at (3, 2, 1), the tails are at
3 0 3 3 3 0 3 1 2 3 −1 4
2 − 2 = 0 , 2 − 2 = 0 , 2 − −2 = 4 , 2 − −1 = 3 .
1 0 1 1 1 0 1 1 0 1 −2 3
# »
5. The four vectors AB are
3
c
2
1
d
a
1 2 3 4
-1
b
-2
In standard position, the vectors are
# »
(a) AB = [4 − 1, 2 − (−1)] = [3, 3].
# »
(b) AB = [2 − 0, −1 − (−2)] = [2, 1]
# »
(c) AB = 21 − 2, 3 − 32 = − 23 , 32
# »
(d) AB = 61 − 13 , 12 − 31 = − 61 , 16 .
3
2
a
1
c
b
d
-1 1 2 3
3
, 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 5
6. Recall the notation that [a, b] denotes a move of a units horizontally and b units vertically. Then during
the first part of the walk, the hiker walks 4 km north, so a = [0, 4]. During the second part of the
walk, the hiker walks a distance of 5 km northeast. From the components, we get
" √ √ #
◦ ◦ 5 2 5 2
b = [5 cos 45 , 5 sin 45 ] = , .
2 2
Thus the net displacement vector is
" √ √ #
5 2 5 2
c=a+b= , 4+ .
2 2
3 2 3+2 5 3
7. a + b = + = = .
0 3 0+3 3
2
a+b
1 b
a
1 2 3 4 5
2 −2 2 − (−2) 4 3
8. b−c = − = = .
3 3 3−3 0
2
b -c
1
b-c
1 2 3 4
3 −2 5
9. d − c = − = . 1 2 3 4 5
−2 3 −5
-1 d
-2
-3 d-c -c
-4
-5
3 3 3+3 a
10. a + d = + = = 1 2 3 4 5 6
0 −2 0 + (−2)
6 -1 d
. a+d
−2
-2
11. 2a + 3c = 2[0, 2, 0] + 3[1, −2, 1] = [2 · 0, 2 · 2, 2 · 0] + [3 · 1, 3 · (−2), 3 · 1] = [3, −2, 3].
12.
3b − 2c + d = 3[3, 2, 1] − 2[1, −2, 1] + [−1, −1, −2]
= [3 · 3, 3 · 2, 3 · 1] + [−2 · 1, −2 · (−2), −2 · 1] + [−1, −1, −2]
= [6, 9, −1].
4
Linear Algebra A Modern Introduction 4th Edition
By
David Poole
( All Chapters Included - 100% Verified Solutions )
1
,Chapter 1
Vectors
1.1 The Geometry and Algebra of Vectors
H-2, 3L 3 H2, 3L
1.
2
1
H3, 0L
-2 -1 1 2 3
-1
H3, -2L
-2
2. Since
2 3 5 2 2 4 2 −2 0 2 3 5
+ = , + = , + = , + = ,
−3 0 −3 −3 3 0 −3 3 0 −3 −2 −5
plotting those vectors gives
1 2 3 4 5
-1
c b
-2
a
-3
d
-4
-5
3
2
,4 CHAPTER 1. VECTORS
3. 2 z
c 1 b
-2 -1 0
0
1 y2
0 -1
2
1 a
3
x
-1
d
-2
4. Since the heads are all at (3, 2, 1), the tails are at
3 0 3 3 3 0 3 1 2 3 −1 4
2 − 2 = 0 , 2 − 2 = 0 , 2 − −2 = 4 , 2 − −1 = 3 .
1 0 1 1 1 0 1 1 0 1 −2 3
# »
5. The four vectors AB are
3
c
2
1
d
a
1 2 3 4
-1
b
-2
In standard position, the vectors are
# »
(a) AB = [4 − 1, 2 − (−1)] = [3, 3].
# »
(b) AB = [2 − 0, −1 − (−2)] = [2, 1]
# »
(c) AB = 21 − 2, 3 − 32 = − 23 , 32
# »
(d) AB = 61 − 13 , 12 − 31 = − 61 , 16 .
3
2
a
1
c
b
d
-1 1 2 3
3
, 1.1. THE GEOMETRY AND ALGEBRA OF VECTORS 5
6. Recall the notation that [a, b] denotes a move of a units horizontally and b units vertically. Then during
the first part of the walk, the hiker walks 4 km north, so a = [0, 4]. During the second part of the
walk, the hiker walks a distance of 5 km northeast. From the components, we get
" √ √ #
◦ ◦ 5 2 5 2
b = [5 cos 45 , 5 sin 45 ] = , .
2 2
Thus the net displacement vector is
" √ √ #
5 2 5 2
c=a+b= , 4+ .
2 2
3 2 3+2 5 3
7. a + b = + = = .
0 3 0+3 3
2
a+b
1 b
a
1 2 3 4 5
2 −2 2 − (−2) 4 3
8. b−c = − = = .
3 3 3−3 0
2
b -c
1
b-c
1 2 3 4
3 −2 5
9. d − c = − = . 1 2 3 4 5
−2 3 −5
-1 d
-2
-3 d-c -c
-4
-5
3 3 3+3 a
10. a + d = + = = 1 2 3 4 5 6
0 −2 0 + (−2)
6 -1 d
. a+d
−2
-2
11. 2a + 3c = 2[0, 2, 0] + 3[1, −2, 1] = [2 · 0, 2 · 2, 2 · 0] + [3 · 1, 3 · (−2), 3 · 1] = [3, −2, 3].
12.
3b − 2c + d = 3[3, 2, 1] − 2[1, −2, 1] + [−1, −1, −2]
= [3 · 3, 3 · 2, 3 · 1] + [−2 · 1, −2 · (−2), −2 · 1] + [−1, −1, −2]
= [6, 9, −1].
4
