Solutions of Selected Problems in HW6
3.4.8
(a)
For arbitrary (
a, b, c
)
T
∈
R
3
, set
C
1
1
1
1
+
C
2
3

1
4
=
a
b
c
.
The augmented matrix is
1
3
a
1

1
b
1
4
c
of which RREF is
1
0
4
a

3
c
0
1

a
+
c
0
0

5
a
+
b
+ 4
c
.
There will be no solutions for
C
1
, C
2
if

5
a
+
b
+4
c
6
= 0
.
For example, (
a, b, c
)
T
= (0
,
1
,
1)
T
cannot be expressed as a linear combination of (1
,
1
,
1)
T
,
(3
,

1
,
4)
T
.
Thus
R
3
cannot be
spanned by (1
,
1
,
1)
T
,
(3
,

1
,
4)
T
.
(b)
RREF(
X
) should have no free variables so that the corresponding system for
C
1
, C
2
, C
3
is consistent. Equivalently, you can also show det(
X
)
6
= 0
.
(Recall the theorem that for
v
1
,
· · ·
,
v
n
∈
R
n
, linear independence of the vectors are
equivalent to their spanning
R
n
.
Thus it suffices to show either linear independence or
spanning ability to show the vectors are a basis.)
(c)
From the part (a), you can set
x
3
= (
a, b, c
)
T
with

5
a
+
b
+ 4
c
6
= 0
.
For example,
x
3
= (0
,
1
,
1)
T
.
Confirm det(
X
)
6
= 0
.
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 Spring '08
 KIM
 Linear Algebra, Algebra, Vector Space, basis, Linear combination, linearly independent vectors

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