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THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
MAT2310 Linear Algebra and Applications (Fall 2007)
Suggested Solutions For Homework 4
1. Let
B
be the space of all the solutions
x
such that
A
x
=
0
. Since the rank of the matrix
A
is
n

r
, the dimension of
B
must be
n

(
n

r
) =
r
. By assumption,
x
1
,
x
2
,
···
,
x
r
be linearly
independent vectors satisfying
A
x
=
0
, it is a base for
B
. Hence any
x
∈
B
can be expressed as a
linear combination of
x
1
,
x
2
,
···
,
x
r
.
2. (a)
A
=
1

1
3
5

4

4
7

6
2
•
(5) times the ﬁrst row was added to the second row,
•
(7) times the ﬁrst row was added to the third row,
we obtain
1

1
3
0
1

19
0
1

19
•
(1) times the second row was added to the ﬁrst row,
•
(1) times the second row was added to the third row,
we obtain
1 0

16
0 1

19
0 0
0
A basis for the row space is
{
(1
,
0
,

16)
,
(0
,
1
,

19)
}
, a basis for the null space is
16
19
1
.
The transpose of
A
is
1
5
7

1

4

6
3

4
2
Using elementary row operations, we obtain
1 0 2
0 1 1
0 0 0
Hence a basis for the column space is
1
0
2
,
0
1
1
.
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(b) Similar to (a), performing the elementary row operations, we obtain
1 0 1

2
/
7
0 1 1
4
/
7
0 0 0
0
Hence a basis for the row space is
{
(1
,
0
,
1
,

2
/
7)
,
(0
,
1
,
1
,
4
/
7)
}
, a basis for the null space is
{
(

1
,

1
,
1
,
0)
T
,
(2
/
7
,

4
/
7
,
0
,
1)
T
}
. The transpose of
A
is
1 2

1
4 1
3
5 3
2
2 0
2
Using the elementary row operations, we obtain
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 Spring '06
 Dr.JeffChakFuWONG
 Linear Algebra, Algebra

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