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HOMEWORK ASSIGNMENT 1 SOLUTIONS
§
1.1
#
3.b
Let
A
= (3
,

6
,
7),
B
= (

2
,
0
,

4), and
C
= (5
,

9
,

2) be the three given points.
Set
u
=
B

A
= (

2
,
0
,

4)

(3
,

6
,
7) = (

5
,
6
,

11)
and
v
=
C

A
= (5
,

9
,

2)

(3
,

6
,
7) = (2
,

3
,

9)
.
Following example 2 in the book, the equation for the plane is
x
=
A
+
su
+
tv
= (3
,

6
,
7) +
s
(

5
,
6
,

11) +
t
(2
,

3
,

9)
.
§
1.1
#
3.d
Let
A
= (1
,
1
,
1),
B
= (5
,
5
,
5), and
C
= (

6
,
4
,
2) be the three given points. Set
u
=
B

A
= (5
,
5
,
5)

(1
,
1
,
1) = (4
,
4
,
4)
and
v
=
C

A
= (

6
,
4
,
2)

(1
,
1
,
1) = (

7
,
3
,
1)
.
Again, following example 2 in the book, the equation for the plane is
x
=
A
+
su
+
tv
= (1
,
1
,
1) +
s
(4
,
4
,
4) +
t
(

7
,
3
,
1)
.
§
1.1
#
7
This was done in class.
§
1.2
#
1
(a) True. This is axiom (VS 3) of a vector space.
(b) False. The zero vector is unique by Corollary 1, page 11.
(c) False. For example, let
a
= 0,
b
= 1, and
x
=
~
0. Then
ax
=
bx
=
~
0, but
a
6
=
b
.
(d) False. For example, consider the vector space
R
2
. If
a
= 0,
x
= (1
,
1) and
y
= (1
,
0)
then
ax
=
ay
= (0
,
0) but
x
6
=
y
.
(e) True. Just think of them as column vectors.
(f) False. An
m
×
n
matrix has
m
rows and
n
columns.
(g) False. For example, we can add
x
+ 1 and
x
2
to get
x
2
+
x
+ 1.
1
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HOMEWORK ASSIGNMENT 1 SOLUTIONS
(h) False. For example, let
f
=
x
+ 1 and
g
=

x
.
Then deg(
f
) = deg(
g
) = 1, but
f
+
g
= 1 has degree 0.
(i) True. To prove this, let
f
(
x
) =
a
n
x
n
+
a
n

1
x
n

1
+
···
+
a
1
x
+
a
0
, with
a
n
6
= 0 (since
deg(
f
) =
n
). Then
cf
(
x
) =
ca
n
x
n
+
···
ca
1
x
+
ca
0
, and if
c
6
= 0 then
ca
n
6
= 0. Therefore
deg(
cf
(
x
)) =
n
.
(j) True. Obvious.
(k) True. This is by de±nition of equality among functions. See example 3, page 9.
§
1.2
#
18
Let
V
=
{
(
a
1
, a
2
) :
a
1
, a
2
∈
R
}
. For (
a
1
, a
2
)
,
(
b
1
, b
2
)
∈
V
and
c
∈
R
, de±ne
(
a
1
, a
2
) + (
b
1
, b
2
) = (
a
1
+ 2
b
1
, a
2
+ 3
b
2
)
and
c
(
a
1
, a
2
) = (
ca
1
, ca
2
)
.
The trick to this problem is to notice that the multiplication looks normal, but the
addition looks funny.
So we should probably check the addition rules, to see if they
make sense.
Let’s check (VS 1).
From our addition rule, we have (
a
1
, a
2
) + (
b
1
, b
2
) =
(
a
1
+ 2
b
1
, a
2
+ 3
b
2
). Switching the summands, but using the same addition rule, we have
(
b
1
, b
2
) + (
a
1
, a
2
) = (
b
1
+ 2
a
1
, b
2
+ 3
a
2
). These two answers do not look like they are the
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 Fall '08
 GUREVITCH

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