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with
x
5
arbitrary. Then
X
=
x
5
0

x
5

3
x
5
x
5
=
x
5
1
0

1

3
1
,
so [1
,
0
,

1
,

3
,
1]
t
is a basis for
N
(
A
).
6. In Section 1.6, problem 12, we found that the matrix
A
=
1
0
1
0
1
0
1
0
1
1
1
1
1
1
0
0
0
1
1
0
has reduced row–echelon form
B
=
1
0
0
1
1
0
1
0
1
1
0
0
1
1
0
0
0
0
0
0
.
From
B
we read o± the following:
(a) The three non–zero rows of
B
form a basis for
R
(
A
).
(b) The ²rst three columns of
A
form a basis for
C
(
A
).
(c) To ²nd a basis for
N
(
A
), we solve
AX
= 0 and equivalently
BX
= 0.
From
B
we see that the solution is
x
1
=

x
4

x
5
=
x
4
+
x
5
x
2
=

x
4

x
5
=
x
4
+
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 Fall '10
 Dreibelbis

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