Math 330 Exam 2, March 2, 2009
No calculators are allowed on the following.
Be sure to show all your work, especially any row
operations.
Matrix Operations  20 points
Consider the following matrices:
A
=
bracketleftBigg
1
2
3
0

1

1
bracketrightBigg
, B
=
1
5
1
0
0
10
Compute each of the following and answer the indicated questions (5 points each part)
(a)
What is
A
T
? What is
A
T
+
B
?
SOLUTION:
bracketleftBigg
1
2
3
0

1

1
bracketrightBigg
T
+
1
5
1
0
0
10
=
1
0
2

1
3

1
+
1
5
1
0
0
10
=
2
5
3

1
3
9
(b)
Find a matrix
C
so that
C
+ 3
A
is the appropriate sized matrix with all entries zero.
SOLUTION:
C
+ 3
A
= 0 means that
C
=

3
A
. So
C
=

3
A
=

3
bracketleftBigg
1
2
3
0

1

1
bracketrightBigg
=
bracketleftBigg

3

6

9
0
3
3
bracketrightBigg
(c)
Calculate
BA
. Does
AB
=
BA
? Explain and/or calculate.
SOLUTION:
1
5
1
0
0
10
bracketleftBigg
1
2
3
0

1

1
bracketrightBigg
=
1

3

2
1
2
3
0

10

10
Since
AB
is a 2
×
2 matrix and
BA
is a 3
×
3 matrix they cannot be equal.
(d)
Specify any matrix
D
(with all entries nonzero), so that
DB
=
bracketleftBigg
0
0
0
0
bracketrightBigg
.
We should make
D
a 2
×
3 matrix so that
DB
is 2
×
2. The two columns of
B
give weights
for linear combinations of the columns of
D
which produce the zero vector. Therefore, with
D
= [
vector
d
1
vector
d
2
vector
d
3
] then we need (from the first column of
B
) that
vector
d
1
+
vector
d
2
=
vector
0. From the second
column of
B
we need that 5
vector
d
1
+ 10
vector
d
3
=
vector
0. There are many possibilities, one of the simpler
ones
D
=
bracketleftBigg
2

2
1
2

2
1
bracketrightBigg
1
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Inverse Matrices  20 points
Find the inverses (or explain why these do not exist) of each of the following matrices: (10 pts each)
(a)
A
=

1
0
0
0
0
1
0
0
0
0
0
5
0
0
1
0
SOLUTION:

1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
0
0
5
0
0
1
0
0
0
1
0
0
0
0
1
R
1
←
R
1 &
R
3
←
R
3
5
=
⇒
1
0
0
0

1
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
1
5
0
0
0
1
0
0
0
0
1
R
3
↔
R
4
=
⇒
1
0
0
0

1
0
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
1
5
0
So the inverse matrix is
A

1
=

1
0
0
0
0
1
0
0
0
0
0
1
0
0
1
5
0
(b)
B
=
1
1
0
2
3
2

1
0
1
SOLUTION:
1
1
0
1
0
0
2
3
2
0
1
0

1
0
1
0
0
1
R
2
←
R
2

2
R
1 &
R
3
←
R
3+
R
1
=
⇒
1
1
0
1
0
0
0
1
2

2
1
0
0
1
1
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 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra, Matrices, Matrix Operations

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