SOLUTION AND ANSWER GUIDE
LINEAR ALGEBRA: A MODERN INTRODUCTION
5TH EDITION
CHAPTER NO. 01: VECTORS
1.1 THE GEOMETRY AND ALGEBRA OF VECTORS
1.
Solution:
2.
Solution:
Since
1 3 4 1 2 3 1 −2 −1 1 3 4
= + , = + , = + , = + ,
−3 0 −3 −3 3 0 −3 3 0 −3 −2 −5
plotting those vectors gives
3.
Solution:
,4.
Solution:
( )
Since the heads are all at 1, 2, 3 , the tails are at
1 0 1 1 3 −2 1 1 0 1 −1 2
2 − 2
= 0 , 2 − 2
= 0 , 2= − −2 4 , 2= − −1 3 .
3 0 3 3 1 2 3 1 2 3 −2 5
5.
Solutions:
The four vectors AB are
In standard position, the vectors are
( )
(a) AB= 4 − 1, 2 − −1 = 3, 3 .
( )
(b) AB= 2 − 0, −1− −2 = 2, 1
1 3 3 3
(c) AB = − 2, 3 − = − ,
2 2 2 2
1 1 1 1 1 1
(d) AB = − , − = − , .
6 3 2 3 6 6
,6.
Solution:
Recall the notation that a , b denotes a move of a units horizontally and b units vertically. Then during
the first part, the hiker travels 4 km north, so a = 0, 4 . During the second part, the hiker travels 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
7.
Solution:
3 2 3 + 2 5
a + b= + = = .
0 3 0 + 3 3
, 8.
Solution:
b + c= + =
( )
2 −2 2 + −2 0
= .
3 3 3 + 3 6
9.
Solution:
3 −2 5
d−c
= − = .
−2 3 −5
LINEAR ALGEBRA: A MODERN INTRODUCTION
5TH EDITION
CHAPTER NO. 01: VECTORS
1.1 THE GEOMETRY AND ALGEBRA OF VECTORS
1.
Solution:
2.
Solution:
Since
1 3 4 1 2 3 1 −2 −1 1 3 4
= + , = + , = + , = + ,
−3 0 −3 −3 3 0 −3 3 0 −3 −2 −5
plotting those vectors gives
3.
Solution:
,4.
Solution:
( )
Since the heads are all at 1, 2, 3 , the tails are at
1 0 1 1 3 −2 1 1 0 1 −1 2
2 − 2
= 0 , 2 − 2
= 0 , 2= − −2 4 , 2= − −1 3 .
3 0 3 3 1 2 3 1 2 3 −2 5
5.
Solutions:
The four vectors AB are
In standard position, the vectors are
( )
(a) AB= 4 − 1, 2 − −1 = 3, 3 .
( )
(b) AB= 2 − 0, −1− −2 = 2, 1
1 3 3 3
(c) AB = − 2, 3 − = − ,
2 2 2 2
1 1 1 1 1 1
(d) AB = − , − = − , .
6 3 2 3 6 6
,6.
Solution:
Recall the notation that a , b denotes a move of a units horizontally and b units vertically. Then during
the first part, the hiker travels 4 km north, so a = 0, 4 . During the second part, the hiker travels 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
7.
Solution:
3 2 3 + 2 5
a + b= + = = .
0 3 0 + 3 3
, 8.
Solution:
b + c= + =
( )
2 −2 2 + −2 0
= .
3 3 3 + 3 6
9.
Solution:
3 −2 5
d−c
= − = .
−2 3 −5
