Vector Worksheets With Answers - Math 125, Exam 3 Page 10
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19Name:
Section Registered In:
Math 125 – Exam 3 – Version 2
November 16, 2005
60 total points possible
1. Let v
= (1, 0, 1), v
= (2, 1, 3), v
= (4, 2, 6), and b = (3, 1, 2).
1
2
3
(a) (1 pt) Is b in v
, v
, v
? How many vectors are in v
, v
, v
?
1
2
3
1
2
3
Solution: b is not in the set. There are only three vectors in the set. Namely, v
, v
, and
1
2
v
.
3
(b) (1 pt) How many vectors are in the span of v
, v
, v
? What form do these vectors
1
2
3
take?
Solution: The vectors in the span of the set v
, v
, v
are any vector v that can be
1
2
3
represented as a linear combination of v
, v
, and v
. That is v has the form
1
2
3
v = c
v
+ c
v
+ c
v
1
1
2
2
3
3
for any real numbers c
, c
, and c
. Since the coefficients for the linear combination are
1
2
3
arbitrary, we see there are an infinite number of vectors in the span.
(c) (3 pts) Is b in the subspace spanned by v
, v
, v
? Explain your answer.
1
2
3
Solution: To determine if b is in the span of the set, we need to determine if there exists
specific coefficients c
, c
, and c
such that b = c
v
+ c
v
+ c
v
. To do this, we solve the
1
2
3
1
1
2
2
3
3
augmented system [ v
v
v
b]. Moving [ v
v
v
b] into reduced row echelon form, we get
1
2
3
1
2
3
1 0 0
1
.
0 1 2
1
0 0 0
0
Since this system is consistent – the rank of the coefficient matrix equals the rank of the
augmented matrix – the vector b lies in the span. (Although the question didn’t ask for it,
viewing the reduced row echelon form matrix as a dependency table, we see that b = v
+ v
.)
1
2
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education