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