INTRODUCTION
TO
LINEAR
ALGEBRA
Fourth Edition
MANUAL FOR INSTRUCTORS
Gilbert Strang
Massachusetts Institute of Technology
math.mit.edu/linearalgebra
web.mit.edu/18.06
video lectures: ocw.mit.edu
math.mit.edu/
±
gs
www.wellesleycambridge.com
email: [email protected]
Wellesley  Cambridge Press
Box 812060
Wellesley, Massachusetts 02482
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Solutions to Exercises
Problem Set 1.1, page 8
1
The combinations give (a) a line in
R
3
(b) a plane in
R
3
(c) all of
R
3
.
2
v
C
w
D
.2;3/
and
v
N
w
D
.6;
N
1/
will be the diagonals of the parallelogram with
v
and
w
as two sides going out from
.0;0/
.
3
This problem gives the diagonals
v
C
w
and
v
N
w
of the parallelogram and asks for
the sides: The opposite of Problem 2. In this example
v
D
.3;3/
and
w
D
.2;
N
2/
.
4
3
v
C
w
D
.7;5/
and
c
v
C
d
w
D
.2c
C
d;c
C
2d/
.
5
u
C
v
D
.
N
2;3;1/
and
u
C
v
C
w
D
.0;0;0/
and
2
u
C
2
v
C
w
D
.
add first answers
/
D
.
N
. The vectors
u
;
v
;
w
are in the same plane because a combination gives
. Stated another way:
u
D N
v
N
w
is in the plane of
v
and
w
.
6
The components of every
c
v
C
d
w
add to zero.
c
D
3
and
d
D
9
give
.3;3;
N
6/
.
7
The nine combinations
c.2;1/
C
d.0;1/
with
c
D
0;1;2
and
d
D
.0;1;2/
will lie on
a lattice. If we took all whole numbers
c
and
d
, the lattice would lie over the whole
plane.
8
The other diagonal is
v
N
w
(or else
w
N
v
). Adding diagonals gives
2
v
(or
2
w
).
9
The fourth corner can be
.4;4/
or
.4;0/
or
.
N
2;2/
. Three possible parallelograms!
10
i
N
j
D
.1;1;0/
is in the base (
x

y
plane).
i
C
j
C
k
D
.1;1;1/
is the opposite corner
from
. Points in the cube have
0
±
x
±
1
,
0
±
y
±
1
,
0
±
z
±
1
.
11
Four more corners
.1;1;0/;.1;0;1/;.0;1;1/;.1;1;1/
. The center point is
.
1
2
;
1
2
;
1
2
/
.
Centers of faces are
.
1
2
;
1
2
;0/;.
1
2
;
1
2
;1/
and
.0;
1
2
;
1
2
/;.1;
1
2
;
1
2
/
and
.
1
2
;0;
1
2
/;.
1
2
;1;
1
2
/
.
12
A fourdimensional cube has
2
4
D
16
corners and
2
±
4
D
8
threedimensional faces
and
24
twodimensional faces and
32
edges in Worked Example
2.4 A
.
13
Sum
D
zero vector. Sum
D N
2
:
00
vector
D
8
:
00
vector.
2
:
00
is
30
ı
from horizontal
D
.
cos
±
6
;
sin
±
6
/
D
.
p
3=2;1=2/
.
14
Moving the origin to
6
:
00
adds
j
D
.0;1/
to every vector. So the sum of twelve vectors
changes from
0
to
12
j
D
.0;12/
.
15
The point
3
4
v
C
1
4
w
is threefourths of the way to
v
starting from
w
. The vector
1
4
v
C
1
4
w
is halfway to
u
D
1
2
v
C
1
2
w
. The vector
v
C
w
is
2
u
(the far corner of the
parallelogram).
16
All combinations with
c
C
d
D
1
are on the line that passes through
v
and
w
.
The point
V
D N
v
C
2
w
is on that line but it is beyond
w
.
17
All vectors
c
v
C
c
w
are on the line passing through
and
u
D
1
2
v
C
1
2
w
. That
line continues out beyond
v
C
w
and back beyond
. With
c
²
0
, half of this line
is removed, leaving a
ray
that starts at
.
18
The combinations
c
v
C
d
w
with
0
±
c
±
1
and
0
±
d
±
1
fill the parallelogram
with
sides
v
and
w
. For example, if
v
D
.1;0/
and
w
D
then
c
v
C
d
w
fills the unit
square.
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