Math 167 homework 3 solutions
October 20, 2010
2.1.22:
For which righthand sides (find a condition on
b
1
,
b
2
,
b
3
) are these
systems solvable?
(a)
1
4
2
2
8
4

1

4

2
x
1
x
2
x
3
=
b
1
b
2
b
3
r
2

2
r
1
and
r
3
+
r
1
give the conditions
b
2

2
b
1
= 0 and
b
3
+
b
1
= 0 for
solvability. These conditions are equivalent to
b
2
= 2
b
1
and
b
3
=

b
1
.
(b)
1
4
2
9

1

4
x
1
x
2
x
3
=
b
1
b
2
b
3
r
2

2
r
1
and
r
1
+
r
3
= 0 give the condition
b
3
+
b
1
= 0 for solvability.
This condition is equivalent to
b
3
=

b
1
.
2.1.26
The columns of
AB
are linear combinations of the columns of
A
.
This means:
The column space of
AB
is contained in
(possibly equal to)
the
column space of A
. Give an example where the column spaces of
A
and
AB
are not equal.
Let
A
=
1
0
0
0
1
0
0
0
1
and
AB
=
B
=
1
0
0
0
8
0
0
0
0
.
2.1.30
If the 9 by 12 system
Ax
=
b
is solvable for every
b
, then
C
(
A
) =
R
9
.
1
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2.2.24
Every column of
AB
is a combination of the columns of
A
. Then the
dimension of the column space gives rank(
AB
)
≤
rank(
A
). Prove also that
rank(
AB
)
≤
rank(
B
).
Note that
rank(
AB
) =
n

dim
N
(
AB
)
,
where
n
is the number of columns of
AB
and dim
N
(
AB
) is dimension of
the nullspace of
AB
. It is true that
dim
N
(
AB
)
≥
dim
N
(
B
)
,
since
x
∈ N
(
B
) implies
x
∈ N
(
AB
). Also, note that
rank(
B
) =
n

dim
N
(
B
)
.
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