Math 215 HW #2 Solutions
1. Problem 1.4.6. Write down the 2 by 2 matrices
A
and
B
that have entries
a
ij
=
i
+
j
and
b
ij
= (

1)
i
+
j
. Multiply them to find
AB
and
BA
.
Solution:
Since
a
ij
indicates the entry in
A
which is in the
i
th row and in the
j
th column,
we see that
A
=
2
3
3
4
.
Likewise,
B
=
1

1

1
1
.
Therefore,
AB
=
2
3
3
4
1

1

1
1
=
2
·
1 + 3
·
(

1)
2
·
(

1) + 3
·
1
3
·
1 + 4
·
(

1)
3
·
(

1) + 4
·
1
=

1
1

1
1
.
Also,
BA
=
1

1

1
1
2
3
3
4
=
1
·
2 + (

1)
·
3
1
·
3 + (

1)
·
4
(

1)
·
2 + 1
·
3
(

1)
·
3 + 1
·
4
=

1

1
1
1
.
2. Problem 1.4.16. Let
x
be the column vector (1
,
0
, . . . ,
0). Show that the rule (
AB
)
x
=
A
(
Bx
)
forces the first column of
AB
to equal
A
times the first column of
B
.
Solution:
Suppose that
x
has
n
components. Then, in order for
Bx
to make sense,
B
must
be an
m
×
n
matrix for some
m
. In turn, for the matrix product
AB
to make sense,
A
must
be an
×
m
matrix for some
.
Now, suppose
A
=
a
11
· · ·
a
1
m
.
.
.
.
.
.
a
1
· · ·
a
m
and
B
=
b
11
· · ·
b
1
n
.
.
.
.
.
.
b
m
1
· · ·
b
mn
.
Then
AB
=
a
11
· · ·
a
1
m
.
.
.
.
.
.
a
1
· · ·
a
m
b
11
· · ·
b
1
n
.
.
.
.
.
.
b
m
1
· · ·
b
mn
=
∑
m
i
=1
a
1
i
b
i
1
]
· · ·
∑
m
i
=1
a
1
i
b
in
.
.
.
.
.
.
∑
m
i
=1
a
i
b
i
1
· · ·
∑
m
i
=1
a
i
b
in
.
Hence,
(
AB
)
x
=
∑
m
i
=1
a
1
i
b
i
1
]
· · ·
∑
m
i
=1
a
1
i
b
in
.
.
.
.
.
.
∑
m
i
=1
a
i
b
i
1
· · ·
∑
m
i
=1
a
i
b
in
1
0
.
.
.
0
=
∑
m
i
=1
a
1
i
b
i
1
∑
m
i
=1
a
2
i
b
i
1
.
.
.
∑
m
i
=1
a
i
b
i
1
,
1
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which is just a copy of the first column of
AB
.
On the other hand,
Bx
=
b
11
· · ·
b
1
n
.
.
.
.
.
.
b
m
1
· · ·
b
mn
1
0
.
.
.
0
=
b
11
b
21
.
.
.
b
m
1
,
which is the first column of
B
.
Therefore,
A
(
Bx
) is
A
times the first column of
B
; since
A
(
Bx
) = (
AB
)
x
and (
AB
)
x
is the first column of
AB
, we see that the first column of
AB
must be
A
times the first column of
B
.
Though it’s not part of the assigned problem, the same argument with different choices of
x
(e.g. (0
,
1
,
0
, . . . ,
0), etc.) will demonstrate that each column of
AB
must be equal to
A
times
the corresponding column in
B
.
3. Problem 1.4.20. The matrix that rotates the
xy
plane by an angle
θ
is
A
(
θ
) =
cos
θ

sin
θ
sin
θ
cos
θ
.
Verify that
A
(
θ
1
)
A
(
θ
2
) =
A
(
θ
1
+
θ
2
) from the identities for cos(
θ
1
+
θ
2
) and sin(
θ
1
+
θ
2
).
What is
A
(
θ
) times
A
(

θ
)?
Solution:
Using the definition of
A
(
θ
1
) and
A
(
θ
2
), we have that
A
(
θ
1
)
A
(
θ
2
) =
cos
θ
1

sin
θ
1
sin
θ
1
cos
θ
1
cos
θ
2

sin
θ
2
sin
θ
2
cos
θ
2
=
cos
θ
1
cos
θ
2

sin
θ
1
sin
θ
2

cos
θ
1
sin
θ
2

sin
θ
1
cos
θ
2
sin
θ
1
cos
θ
2
+ cos
θ
1
sin
θ
2

sin
θ
1
sin
θ
2
+ cos
θ
1
cos
θ
2
=
cos(
θ
1
+
θ
2
)

sin(
θ
1
+
θ
2
)
sin(
θ
1
+
θ
2
)
cos(
θ
1
+
θ
2
)
=
A
(
θ
1
+
θ
2
)
,
where I went from the second to the third lines using the identities for cos(
θ
1
+
θ
2
) and
sin(
θ
1
+
θ
2
). Geometrically, the fact that
A
(
θ
1
)
A
(
θ
2
) =
A
(
θ
1
+
θ
1
) corresponds to the fact
that rotating something by an angle of
θ
2
and then rotating the result by an angle of
θ
1
is
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 Spring '09
 Linear Algebra, Matrices, Row

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