Math 340, Lecture 3
Solutions
to Exam # 1
Problem 1:
Let
A
=
1
5

3
0
2
1

3
0
2
and let
B
=
4

1
5
2
1
1
0
3
0
. Compute each of the following:
(a)
3
A

B
= 3
1
5

3
0
2
1

3
0
2

4

1
5
2
1
1
0
3
0
=
3
15

9
0
6
3

9
0
6

4

1
5
2
1
1
0
3
0
=

1
16

14

2
5
2

9

3
6
.
(b)
AB
=
1
5

3
0
2
1

3
0
2
4

1
5
2
1
1
0
3
0
=
14

5
10
4
5
2

12
9

15
.
(c)
Since (
AB
)
T
=
B
T
A
T
, we have
B
T
A
T
= (
AB
)
T
=
14

5
10
4
5
2

12
9

15
T
=
14
4

12

5
5
9
10
2

15
.
Problem 2:
(a)
What does it mean that an
n
×
n
matrix is
invertible
?
An
n
×
n
matrix
A
is invertible if and only if there is an
n
×
n
matrix
B
so that
AB
=
I
n
and
BA
=
I
n
where
I
n
is the identity
n
×
n
matrix.
(b)
Suppose that
A
and
B
are
n
×
n
matrices, and that both
A
and
B
are invertible with inverses
A

1
and
B

1
.
Prove
that the product
AB
is also invertible, and find its inverse.
We prove that
AB
is invertible by showing that the matrix
B

1
A

1
acts as an inverse. We have:
(
AB
)(
B

1
A

1
) =
A
(
BB

1
)
A

1
=
AI
n
A

1
=
AA

1
=
I
n
,
and
(
B

1
A

1
)(
AB
) =
B

1
(
A

1
A
)
B
=
B

1
I
n
B
=
B

1
B
=
I
n
.
Thus if we multiply
(
AB
)
on either side by the matrix
B

1
A

1
, we get the identity matrix, and so
(
AB
)
is
invertible with inverse
B

1
A

1
.
(c)
Suppose that
A
,
B
1
and
B
2
are
n
×
n
matrices, and that
B
1
A
=
I
n
and
AB
2
=
I
n
.
Prove
that
B
1
=
B
2
. (In this problem,
I
n
denotes the
n
×
n
identity matrix).
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 Spring '08
 Meyer
 Math, Algebra, Matrices, #, 2W, elementary row operations, 1 5 w, 3 3 w

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