2
b)
If (6
,
0
,
0
,
3) can be expressed as a linear combination of vectors of
V
then there exists
scalars
x
1
,
x
2
,
and
x
3
such that
x
1
(1
,
1
,
0
,
1) +
x
2
(2
,
0
,
0
,
2) +
x
3
(0
,
2
,

1
,
1) = (6
,
0
,
0
,
3).
Thus we are trying to solve the system with augmented matrix
1
2
0
6
1
0
2
0
0
0

1
0
1
2
1
3
.
Rowreducing gives
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
.
Hence the system is not consistent so (6
,
0
,
0
,
3) is not in Sp(
V
).
B6 a)
The nullspace of the linear mapping
L
with matrix
A
is the same as the solution
space of the homogeneous system
A
x
=
0
. Since there is no leading 1 in the third or fifth
column of the matrix,
x
3
and
x
5
are arbitrary. let
x
3
=
s
and
x
5
=
t
. By backsubstitution
we get
x
4
=

4
t
,
x
2
=

4
t
and
x
1
=
t
. Thus, the solution space is all vectors of the form
(
x
1
, x
2
, x
3
, x
4
, x
5
) = (
t,

4
t, s,

4
t, t
) =
s
(0
,
0
,
1
,
0
,
0) +
t
(1
,

4
,
0
,

4
,
1)
so the nullspace of
L
is Sp
{
(0
,
0
,
1
,
0
,
0)
,
(1
,

4
,
0
,

4
,
1)
}
.
b)
The nullspace of the linear mapping
L
with matrix
A
is the same as the solution space
of the homogeneous system
A
x
=
0
. Since there is no leading 1 in the second, fourth or
sixth column we have
x
2
,
x
4
, and
x
6
are arbitrary. Let
x
2
=
r
,
x
4
=
s
and
x
6
=
t
. Then by
back substitution we have
x
1
= 2
r
+ 4
s
+ 6
t
,
x
3
= 3
s
+ 6
t
and
x
5
= 5
t
. Hence the solution
space is all vectors of the form
(
x
1
, x
2
, x
3
, x
4
, x
5
, x
6
) =
r
(2
,
1
,
0
,
0
,
0
,
0) +
s
(4
,
0
,
3
,
1
,
0
,
0) +
t
(6
,
0
,
6
,
0
,
5
,
1)
hence the nullspace of
L
is Sp
{
(2
,
1
,
0
,
0
,
0
,
0)
,
(4
,
0
,
3
,
1
,
0
,
0)
,
(6
,
0
,
6
,
0
,
5
,
1)
}
.
B10.
Since the range of
L
consists of all vectors that are multiples of (2
,
1), the columns
of the matrix must be multiples of (2
,
1). Since the nullspace of
L
consists of all vectors
that are multiplies of (2
,
1) we must have the dot product of this vector with any row equal
to 0. Hence the rows of the matrix must be multiples of (1