Assignment 5
Math 417 — Winter 2009
Due February 20
§
3.1#2.
A vector
x
=
x
y
is in the kernel of the given
A
if and only if
0 =
Ax
=
2
3
6
9
x
y
=
2
x
+ 3
y
6
x
+ 9
y
,
i.e, if and only if
x
and
y
satisfy the simultaneous linear equations 2
x
+3
y
= 0 and 6
x
+9
y
= 0.
The second equation here is redundant, being 3 times the first. (If you didn’t notice this
right away, you’d see it when you start rowreducing the augmented matrix of this system
of equations.) So the kernel consists of vectors satisfying 2
x
+ 3
y
= 0. Such vectors have the
form

3
2
y
y
(obtained by solving the linear equation for
x
in terms of
y
). In other words,
they are obtained by multiplying the vector
v
=

3
2
1
by an arbitrary number.
So this
single vector
v
spans the kernel of
A
.
The kernel is also spanned by any vector obtained by multiplying
v
by a nonzero number.
So people who dislike fractions could use

3
2
or
3

2
just as well as
v
.
§
3.1#6.
A calculation like that in the previous problem shows that
x
=
x
y
z
is in the
kernel of
A
if and only if
x
+
y
+
z
= 0. (There are three equations, but the second and
third are not only redundant as in the previous problem but actually identical to the first
equation.) The general solution is
x
=

y

z
y
z
=
y

1
1
0
+
z

1
0
1
.
So the kernel is spanned by the two vectors

1
1
0
and

1
0
1
.
1
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§
3.1#10.
By chance, the matrix
A
is already in echelon form. The system
Ax
= 0 has
only one free variable, namely
x
3
. Setting
x
3
= 1, we find the solution
x
=
1

2
1
0
and so
ker(
A
) is the span of this vector.
§
3.1#16.
The image of
A
consists of the vectors of the form
Ax
=
1
2
3
1
2
3
1
2
3
x
y
z
=
x
+ 2
y
+ 3
z
x
+ 2
y
+ 3
z
x
+ 2
y
+ 3
z
= (
x
+ 2
y
+ 3
z
)
1
1
1
.
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