Solutions of Selected Problems in Chapter Test B
3.
(a)
Note that
N
(
A
) =
{
x
∈
R
5

A
x
=
0
}
.
We can derive RREF of the corresponding augmented matrix (
A

0
) as
1 3 0 2 3
0
0 0 1 1 1
0
0 0 0 0 0
0
0 0 0 0 0
0
.
For
x
= (
x
1
, x
2
, x
3
, x
4
, x
5
)
T
, we have a general solution as
x
2
=
α, x
4
=
β, x
5
=
γ, x
1
=

3
α

2
β

3
γ, x
3
=

β

γ.
Now observe
x
1
x
2
x
3
x
4
x
5
=
α

3
1
0
0
0
+
β

2
0

1
1
0
+
γ

3
0

1
0
1
.
This means
N
(
A
) is spanned by
{
(

3
,
1
,
0
,
0
,
0)
T
,
(

2
,
0
,

1
,
1
,
0)
T
,
(

3
,
0
,

1
,
0
,
1)
T
}
.
Since these vectors are also linearly independent, they form a basis and nullity(
A
)=dim(
N
(
A
)) =
3
.
(b)