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ma253
,Fa
l
l
2007
— Problem Set
7
Solutions
1.
Problems from the textbook:
a. Section
3
.
3
, problems *
18
,*
22
,*
26
,*
28
,*
30
,*
36
,*
38
.
*18
Well, if el professor is going to ask you to do things involving nasty
matrices, it’s nice when they come prerowreduced, eh? Reading off from
the matrix, the image is
R
3
,soabas
isistheusua
l
e
1
,
e
2
and
e
3
.Thekerne
l
is
ker
(
A
)=
Span
⎛
⎜
⎜
⎜
⎜
⎝
⎡
⎢
⎢
⎢
⎢
⎣
1
0

5
1
0
⎤
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎣
2
1
0
0
0
⎤
⎥
⎥
⎥
⎥
⎦
⎞
⎟
⎟
⎟
⎟
⎠
.
*22
Rats
,th
ist
imewehavetorowreduce.
rref
(
A
)=
⎡
⎣
10

6
01 5
00 0
⎤
⎦
,
so a basis of the kernel is the single vector
⎡
⎣
6

5
1
⎤
⎦
and a basis of the image
consists of the two vectors
⎡
⎣
2
4
7
⎤
⎦
and
⎡
⎣
4
5
9
⎤
⎦
.
*26
For part (a), start by ±nding the kernel of
C
:i
ti
sthespano
f
⎡
⎣
0
1

1
⎤
⎦
.
Now just ±nd which of the others have this vector in the kernel:
L
is the
only one. (Notice that this vector is in the kernel if and only if the last two
columns are equal, so this is easy to check!). Now rowreduce
L
to make
sureitskerne
lisa
lsojustthespanof
⎡
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View Full Document For (b), start by noting that the image of
C
is the span of
⎡
⎣
1
1
1
⎤
⎦
and
⎡
⎣
1
0
1
⎤
⎦
.I
t
is not hard to see that this span is just the set of all vectors in
R
3
such that
the ±rst and the third coordinates are equal. Now it’s easy to see that
H
and
X
have the same image as
C
.
For (c), we can eliminate
C
,
H
,and
X
at once, by part (b). Now notice that
T
and
Y
have the same image: both images contain
⎡
⎣
1
0
0
⎤
⎦
,andthefactthat
⎡
⎣
1
0
0
⎤
⎦
+
⎡
⎣
0
1
1
⎤
⎦
=
⎡
⎣
1
1
1
⎤
⎦
shows that they are the same. So
L
is the matrix whose
image is different from that of all the others.
*28
Since there are
4
vectors, they will be a basis if and only if they are
independent. Now notice that if
α
⎡
⎢
⎢
⎣
1
0
0
2
⎤
⎥
⎥
⎦
+
β
⎡
⎢
⎢
⎣
0
1
0
3
⎤
⎥
⎥
⎦
+
γ
⎡
⎢
⎢
⎣
0
0
1
4
⎤
⎥
⎥
⎦
+
δ
⎡
⎢
⎢
⎣
2
3
4
k
⎤
⎥
⎥
⎦
=
⎡
⎢
⎢
⎣
0
0
0
0
⎤
⎥
⎥
⎦
,
we can read off from the ±rst three lines that
α
=
2δ
,
β
=
3δ
,
γ
=

4δ
. Plugging those into the fourth line, we get
0
=
4δ

9δ

16δ
+
kδ
=(
k

29
)
δ.
So the vectors are dependent (i.e., there is a nonzero choice for
δ
that
makes the equation true) if and only if
k
=
29
. For all other values of
k
,the
vectors are independent, hence are a basis of
R
4
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This homework help was uploaded on 01/23/2008 for the course MATH 253 taught by Professor Ghitza during the Spring '07 term at Colby.
 Spring '07
 GHITZA
 Linear Algebra, Algebra, Matrices

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