36.
Ex. 1.5.30:
A
is a
2
×
5
matrix with two pivot positions.
Solution.
(a)
A
has five columns, so
A
does not have a pivot in every column, which means there are free variables in
the matrix equation
A
x
=
0
, so
A
x
=
0
has nontrivial solutions
.
(b)
A
has a pivot in every row, so
A
x
=
b
has a solution for every
b
in
R
2
.
37.
Ex. 1.5.40:
Suppose
A
is a
3
×
3
matrix and
b
is a vector in
R
3
such that the equation
A
x
=
b
does
not
have a
solution. Does there exist a vector
y
in
R
3
such that the equation
A
x
=
y
has a unique solution? Discuss.
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Solution.
Since there is a vector
b
in
R
3
for which
A
x
=
b
has no solution, we know that
A
does not have a pivot
in every row. Since
A
is square, this means
A
does not have a pivot in every column. This means that
A
x
=
0
has
infinitely many solutions, which means that for every
y
,
A
x
=
y
has either infinitely many solutions or no solutions.
Therefore,
there is no vector
y
in
R
3
for which
A
x
=
y
has a unique solution
.
Section 1.7
38.
Ex. 1.7.8:
In Exercises 5–8, determine if the columns of the matrix form a linearly independent set. Justify each
answer.
1

2
3
2

2
4

6
2
0
1

1
3
Solution.
There are more columns than entries in each column, so the columns are
not linearly independent
.
11
39.
Ex. 1.7.10:
In Exercises 9 and 10,
(a)
for what values of
h
is
v
3
in
Span
{
v
1
,
v
2
}
, and
(b)
for what values of
h
is
{
v
1
,
v
2
,
v
3
}
linearly
dependent
? Justify each answer.
v
1
=
1

3

5
,
v
2
=

3
9
15
,
v
3
=
2

5
h
.
Solution 1.
(a) Notice that
v
2
=

3
v
1
. Then if
v
3
were in Span
{
v
1
,
v
2
}
, it would be in Span
{
v
1
}
as well, i.e.
v
3
would
be a constant multiple of
v
1
. But the second entry of
v
1
is

3 times its first entry, so the same will be true of any
constant multiple of
v
1
. The second entry of
v
3
is not

3 times its first entry, so
v
3
is not in Span
{
v
1
,
v
2
}
for any
h
.
(b) Since
v
2
=

3
v
1
,
{
v
1
,
v
2
,
v
3
}
is linearly dependent for every
h
.
Solution 2.
We rowreduce the matrix
v
1
v
2
v
3
:
1

3
2

3
9

5

5
15
h
R
2
→
R
2
+3
R
1
→
R
3
→
R
3
+5
R
1
1

3
2
0
0
1
0
0
h
+ 10
R
3
→
R
3

(
h
+10)
R
2
→
1

3
2
0
0
1
0
0
0
(a) Since the augmented matrix
v
1
v
2
v
3
has a pivot in its rightmost column regardless of
h
, there are no
solutions
x
1
, x
2
to
x
1
v
1
+
x
2
v
2
=
v
3
, so
v
3
is not in Span
{
v
1
,
v
2
}
for any
h
.
(b) Since the coefficient matrix
v
1
v
2
v
3
never has a pivot in its second column regardless of
h
, there is no
h
which makes
v
1
v
2
v
3
have a pivot in every column, so
{
v
1
,
v
2
,
v
3
}
is linearly dependent for every
h
.
40.
Ex. 1.7.12:
In Exercises 11–14, find the value(s) of
h
for which the vectors are linearly
dependent
. Justify each
answer.
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 Spring '08
 Chorin
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Linear combination