Solutions to Second Homework
Problem
3(
b
) (Section 1
.
4)Assume that the ﬁrst vector is a linear combi
nation of the other two. Then,
(1
,
2
,

3) =
a
(

3
,
2
,
1) +
b
(2
,

1
,

1)
.
Doing some algebra we get that
(1
,
2
,

3) = (

3
a
+ 2
b,
2
a

b,
1
a

b
)
.
So, we have the system

3
a
+ 2
b
=
1
2
a

b
=
2
1
a

b
=

3
,
with corresponding matrix:

3
2
1
2

1
2
1

1

3
.
Some computations here show that:

3
2
1
2

1
2
1

1

3
∼
1

1

3
2

1
2

3
2
1
∼
1

1

3
0
1
8
0

1

8
∼
1

1

3
0
1
8
0
0
0
,
i.e. the system of equations has a solution (can you ﬁnd it?), which implies that
(1
,
2
,

3) is a linear combination of (

3
,
2
,
1)
,
(2
,

1
,

1)
.
±
Problem
4(
f
) (Section 1
.
4)We work as above. Say that
6
x
3

3
x
2
+
x
+ 2 =
a
(
x
3

x
2
+ 2
x
+ 3) +
b
(2
x
3

3
x
+ 1)
.
Doing, again, some algebra, (and after some work) we get the system
a
+ 2
b
=
6

a
=

3
2
a

3
b
=
1
3
a
+
b
=
2
1