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M341 Homework Solution 5
Section 2.2:
11(a)
: Let
X
1
= [
u
1
, u
2
,
· · ·
, u
n
]
and let
X
2
= [
v
1
, v
2
,
· · ·
, v
n
]
. Since
AX
=
O
we have for each
1
≤
i
≤
m
we have
a
i
1
u
1
+
a
i
1
u
2
+
· · ·
, a
in
u
n
= 0
and
a
i
1
v
1
+
a
i
1
v
2
+
· · ·
, a
in
v
n
= 0
. Therefore
a
i
1
(
u
1
+
v
1
)+
a
i
1
(
u
2
+
v
2
)+
· · ·
, a
in
(
u
n
+
v
n
) =
a
i
1
u
1
+
a
i
1
u
2
+
· · ·
, a
in
u
n
+
a
i
1
v
1
+
a
i
1
v
2
+
· · ·
, a
in
v
n
= 0
.
Therefore
X
1
+
X
2
is also a solution. Furthermore
a
i
1
cu
1
+
a
i
1
cu
2
+
· · ·
, a
in
cu
n
=
c
(
a
i
1
u
1
+
a
i
1
u
2
+
· · ·
, a
in
u
n
) =
c
0 = 0
. Therefore
c
X
1
is also a solution.
A simple proof is following:
A
(
X
1
+
X
2
) =
AX
1
+
AX
2
=
O
+
O
=
O
since
AX
1
=
AX
2
=
O
and
A
(
c
X
1
) =
c
AX
1
=
c
O
=
O
Hence
X
1
+
X
2
and
c
X
1
are also solutions.
11(b)
: Consider the linear system
x
1
+
x
2
+
x
3
=
1
x
1
+
x
2
−
x
3
=
1
has two solutions
X
1
= [1
,
0
,
0]
and
X
2
= [0
,
1
,
0]
. But
X
1
+
X
2
= [1
,
1
,
0]
is not a solution since
1 + 1 + 0
̸
= 1
.
11(c)
: Using the notation from part (a) we have
a
i
1
u
1
+
a
i
1
u
2
+
· · ·
, a
in
u
n
= 0
and
a
i
1
v
1
+
a
i
1
v
2
+
· · ·
, a
in
v
n
=
b
i
. Therefore
a
i
1
(
u
1
+
v
1
) +
a
i
1
(
u
2
+
v
2
) +
· · ·
, a
in
(
u
n
+
v
n
) =
a
i
1
u
1
+
a
i
1
u
2
+
· · ·
, a
in
u
n
+
a
i
1
v
1
+
a
i
1
v
2
+
· · ·
, a
in
v
n
= 0 +
b
i
=
b
i
. Therefore
X
1
+
X
2
is a solution of
AX
=
B
.
A simple proof is following:
A
(
X
1
+
X
2
) =
AX
1
+
AX
2
=
B
+
O
=
B
since
AX
1
=
B
and
AX
2
=
O
. Hence
X
1
+
X
2
is also a solution.
11(d)
: We will prove by contradiction. Suppose
AX
=
0
has a nontrivial solution
Y
and
AX
=
B
with
B
̸
=
O
has a unique solutions
X
1
by the assumption. Then by part(c),
X
2
=
X
1
+
Y
is also a
solution and
X
2
̸
=
X
1
since
Y
̸
=
0
. It contradicts to the fact
AX
=
B
has only one solution.
12
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 Spring '08
 HIETMANN

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