Homework solutions: Section 2.1 #18,20-24, Section 2.2 #14,
16, 18, 20 and Section 2.3 #16, 18, 22
Section 2.1 #18,20-24.
18. Suppose the ﬁrst two columns,
b
1
and
b
2
, of
B
are equal. What
can you say about the columns of
AB
(if
AB
is deﬁned)? Why?
We start with the assumption (the ”Suppose” part) that
b
1
=
b
2
. So
A
b
1
=
A
b
2
. Now we know
AB
= [
A
b
1
,A
b
2
,...,A
b
n
]
, so we see that
the ﬁrst two columns of
AB
are also equal.
20. Suppose the second column of
B
is all zeros. What can you say
about the second column of
AB
?
We start with the assumption that
b
2
=
0
. So
A
b
2
=
A
0
=
0
. Now
we know
AB
= [
A
b
1
,A
b
2
,...,A
b
n
] = [
A
b
1
,
0
,...,A
b
n
]
, so we see
that the second column of
AB
is all zeros as well.
21. Suppose the last column of
AB
is entirely zero but
B
itself has no
column of zeros. What can you say about the columns of
A
?
Let
B
= [
b
1
,...,
b
n
]
. Then
AB
= [
A
b
1
,A
b
2
,...,A
b
n
]
. We start
with the assumption that the last column of
AB
is the zero vector. So we
get that
A
b
n
=
0
. Thus
x
=
b
n
is a solution to the equation
A
x
=
0
.
Since
B
has no columns that are entirely zero, we know
b
n
6
=
0
, so
x
=
b
n
is a non-trivial solution to the equation
A
x
=
0
. Since the
homogeneous equation
A
x
=
0
has non-trivial solutions, the columns
of
A
are linearly dependent.
22. Show that if the columns of
B
are linearly dependent, then so are
the columns of
AB
.
You might want to reword this problem into the “Suppose.
.., Explain
why.
..” form that you are used to. In that form the statement would
be: Suppose that the columns of
B
are linearly dependent. Explain why
the columns of
AB
are also independent.
Now we begin with the assumption (the ”Suppose” part) that the
columns of
B
are linearly dependent. One way for the columns of
B
to be linearly dependent is for one column to be a multiple of another
column. However we cannot assume that that is the case, because there
are other relationships between the columns of
B
that would cause the