Solution and Answer Guide for Linear Algebra: A Modern Introduction (5th Edition) by David Poole | Complete Step-by-Step Solutions

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 