M341 Homework Solution 9
Section 4.3:
1(b)
: We form the matrix whose rows are the vectors in
S
and row reduce.
A
=
3
1
−
2
−
3
−
1
2
6
2
−
4
⇒
3
1
−
2
0
0
0
0
0
0
=
C
Since Row
(
A
)
=Row
(
C
)
=Span
S
. This tells us that Span
S
=
{
a
[3
,
1
,
−
2] :
a
∈
R
}
2(b)
:
We use the correspondence between
R
4
and
P
3
. We let
ˆ
S
=
{
[1
,
2
,
0
,
0]
,
[0
,
−
4
,
0
,
1]
,
[
−
5
,
0
,
0
,
12]
,
[1
,
−
1
,
0
,
0]
}
We form the matrix whose rows are the vectors in
ˆ
S
. We form the matrix whose rows are the vectors
in
ˆ
S
and row reduce.
A
=
1
2
0
0
0
−
4
0
1
−
5
0
0
12
1
−
1
0
0
⇒
1
1
0
0
0
1
0
0
0
0
0
1
0
0
0
0
=
C
This tells us that Span
ˆ
S
=
{
a
[1
,
0
,
0
,
0]+
b
[0
,
1
,
0
,
0]+
c
[0
,
0
,
0
,
1] :
a,b,c
∈
R
}
=
{
[
a,b,
0
,c
] :
a,b,c
∈
R
}
. Correspondingly Span
S
=
{
ax
3
+
bx
2
+
c
:
a,b,c
∈
R
}
.
3(b)
:
We use the correspondence between
M
22
and
R
4
. We let
ˆ
S
=
{
[1
,
3
,
−
2
,
1]
,
[
−
2
,
−
5
,
3
,
1]
,
[1
,
4
,
−
3
,
4]
}
We form the matrix whose rows are the vectors in
ˆ
S
. We form the matrix whose rows are the vectors
in
ˆ
S
and row reduce.
A
=
1
3
−
2
1
−
2
−
5
3
1
1
4
−
3
4
⇒
1
0
1
−
8
0
1
−
1
3
0
0
0
0
=
C
This tells us that Span
ˆ
S
=
{
a
[1
,
0
,
1
,
−
8] +
b
[0
,
1
,
−
1
,
3] :
a,b
∈
R
}
. Correspondingly Span
S
=
{
a
[
1
0
1
−
8
]
+
b
[
0
1
−
1
3
]
:
a,b
∈
R
}
=
{
[
a
b
a
−
b
−
8
a
+ 3
b
]
:
a,b
∈
R
}
.
6
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 HIETMANN
 Linear Algebra, Vectors, Row

Click to edit the document details