SOLUTIONS FOR HOMEWORK 2
Section 1.4
3. (b) Yes
, the first vector is a linear combination of the other two. To verify this, we
need to find
a,b
∈
R
s.t. (1
,
2
,
−
3) =
a
(
−
3
,
2
,
1) +
b
(2
,
−
1
,
−
1). This gives rise to a system
of three equations with two unknowns:
−
3
a
+
2
b
=
1
2
a
−
b
=
2
a
−
b
=
−
3
To solve this system of equations, we eliminate the unknown
a
from Equations 1 and 2.
More precisely, we keep
Eq
3, and use
Eq
1 + 3
Eq
3 and
Eq
2
−
Eq
3 instead of
Eq
1 and
Eq
2, respectively. This results in a new system of equations, with the same solution(s) as
the old one.
−
b
=
−
8
b
=
8
a
−
b
=
−
3
This leads to
b
= 8, and
a
=
b
−
3 = 5. Doublechecking, we verify the equality (1
,
2
,
−
3) =
5(
−
3
,
2
,
1) + 8(2
,
−
1
,
−
1).
(d) No
, the first vector is not a linear combination of the other two. To verify this, we
need to prove that no
a,b
∈
R
satisfy (2
,
−
1
,
0) =
a
(1
,
2
−
3)+
b
(1
,
−
3
,
2). The last equality
gives rise to a system of three equations with two unknowns:
a
+
b
=
2
2
a
−
3
b
=
−
1
−
3
a
+
2
b
=
0
To solve this system of equations, we eliminate the unknown
a
from Equations 2 and 3.
More precisely, we keep
Eq
1, and use
Eq
2
−
2
Eq
1 and
Eq
3 + 3
Eq
1 instead of
Eq
2 and
Eq
3, respectively. This yields a new system of equations, with the same solution(s) as the
original one.
a
+
b
=
2
−
5
b
=
−
5
+
7
b
=
6
The last two equalities are mutually contradictory, hence the system has no solution.
4. (b) No
, 4
x
3
+2
x
2
−
6 is not a linear combination of
x
3
−
2
x
2
+4
x
+1 and 3
x
3
−
6
x
2
+
x
+4.
Indeed, the first polynomial is a linear combination of the other two iff there exist
a,b
∈
R
for which
4
x
3
+ 2
x
2
+ 0
·
x
−
6 =
a
(
x
3
−
2
x
2
+ 4
x
+ 1) +
b
(3
x
3
−
6
x
2
+
x
+ 4)
.
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 Spring '10
 FUCKHEAD
 Linear Algebra, Equations, Vector Space, Linear combination

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