Instructor
Time
Duration
Signature
:
:
:
:
:
:
Last Name
Name
ID Number
M E T U Department of Mathematics
Math 260
Basic Linear Algebra
Exam 2
01.08.2008
13: 00
100
minutes
4 QUESTIONS ON 4 PAGES
TOTAL 30 POINTS
ANSWER KEY
1
2
3
4
(1)
(3 + 3 + 1 = 7 pts.)
Let
W
=
{
(
x
1
,x
2
,x
3
,x
4
,x
5
)
∈
R
5
:
x
1

2
x
2
= 0
, x
3
+ 4
x
4
= 0
}
.
(a) Show that
W
is a subspace of
R
5
.
The zero vector (0
,
0
,
0
,
0
,
0) is contained in
W
since its components satisfy the relations
x
1

2
x
2
= 0
, x
3
+ 4
x
4
= 0
.
So,
W
6
=
∅
.
Let
u
= (
x
1
,
···
,x
5
)
,v
= (
y
1
,
···
,y
5
)
∈
W,
that is,
x
1

2
x
2
= 0
, x
3
+ 4
x
4
= 0 and
y
1

2
y
2
= 0
, y
3
+ 4
y
4
= 0
,
and let
c
∈
R
.
Now,
u
+
v
= (
x
1
+
y
1
,
···
,x
5
+
y
5
)
,
and
cu
= (
cx
1
,
···
,cx
5
)
.
Since (
x
1
+
y
1
)

2(
x
2
+
y
2
) =
x
1

2
x
2
+
y
1

2
y
2
= 0 and (
x
3
+
y
3
) + 4(
x
4
+
y
4
) =
x
3
+4
x
4
+
y
3
+4
y
4
= 0 we have
u
+
v
∈
W
which means that
W
is closed under addition.
Moreover, (
cx
1
)

2(
cx
2
) =
c
(
x
1

2
x
2
) = 0 and (
cx
3
)+4(
cx
4
) =
c
(
x
3
+4
x
2
) = 0 implies
that
cv
∈
W,
that is,
W
is closed under scalar multiplication.
Thus,
W
is a subspace of
R
5
.
(b) Find a basis for
W.
We should solve the homogeneous system
x
1

2
x
2
= 0
, x
3
+ 4
x
4
= 0
,
which has the
coeﬃcient matrix
"
1

2 0 0 0
0
0 1 4 0
#
,
from which it can be seen that
x
2
,x
4
and
x
5
are free variables.
Letting
x
2
= 1
,x
4
=
x
5
= 0 we obtain
x
1
= 2
,x
3
= 0 and hence
u
1
= (2
,
1
,
0
,
0
,
0) is a
fundamental solution.
For