2019 MATH1030
Classwork 3
Last updated: September 27, 2019
•
online exercise
The following webpage will be set up on Thursday Sep 26
.
Link:
Login: your student ID
Password: your student ID (please change it right after you login)
0.2 pt for each question set
.
•
Midterm: Oct 23 Wed 7:309:30.
Important, Venue changed
: LSB LT2, 3, 5, 6
•
Hand in
2a, 4b, 5b, 6, 7, 11, 12b, 13, 16. For
section A
try also 14, 15, 17.
Then do as many questions as possible. You can get full credit as long as you show
you effort.
•
You can discuss the questions with your classmates or TA. But don’t copy answers
directly. Answer will be posted online.
•
If you cannot finish the class work,
take the question paper home and try to
finish the rest of them
.
•
CW2 Solution has been posted on math1030 class webpage:
For password protected area (lecture notes or exercises):
login
: math1030
password
: like1030
Questions
Notation
: Given a
n
×
n
matrix
A
, we use the notation
A
0
=
I
n
,
A
1
=
A
,
A
2
=
AA
,
A
3
=
AAA
and so on. Obviously
A
k
+1
=
A
k
A
=
AA
k
for each integer
k
≥
0.
1. Let
A
=
1
3
5
7
2
4
6
8
1
1
1
1
,
u
=
1
10
100
1000
(a) Write down [
A
]
23
, [
A
]
14
, [
u
]
2
.
(b) Compute
A
u
.
Answer.
(a) [
A
]
23
= 6, [
A
]
14
= 7, [
u
]
2
= 10.
(b)
A
u
=
7531
8642
1111
2. Solve each of the following equations for
x
, or explain why a solution does not exist.
1
Subscribe to view the full document.
(a)
2
1
2
3
0
4
2

3
1
1
2
0
1
x
=

1
1
0
0
5

2
(b)
x
1
3
4
2
1

1
+
4
3
6
0
1
1
=
7
12
6
6
4
2
Answer.
(a) The left hand side of the equation is equal to

1
1
0
0
5
4

3
x
.
Hence
x
must satisfy the equation 4

3
x
=

2, implying that
x
= 2.
(b) The left hand side of the equation is equal to
x
+ 4
3
x
+ 3
4
x
+ 6
2
x
x
+ 1

x
+ 1
.
Equating the (1
,
1) entries of both sides gives the equation
x
+ 4 = 7, so that
x
= 3. Equating the (2
,
3) entries of both sides gives the equation

x
+ 1 = 2,
which is not satisfied by
x
= 3. Hence, the original equation has no solution.
3. Compute the following.
(a)
1
2

1
1
1
1
1
1

1

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

1
1
0

2
1
1
1

1
1
1
0
1
1
2
0
0
(c)
2
1
2
3
4
0
1
0
1

1

1
1
0

3
0
1
1
2
1
1
1
1
0

1

1

1
1
1
2
1

1
0
0

1
Answer.
(a)
1
1
3
5

1

1
3
4
1
1
1
2
2
(b)
4
8

1
0
2
4
(c)
1
1

2

3

5

4
4. (a) Let
A
=
1
0
1
0
1
1
0
0
2
,
B
=
1
1
2
1
0
0
0
1
1
.
Find
AB
and
BA
.
(b) Let
a, b, c, d
be real numbers and let
A
=
1
0
a
b
0
1
c
d
,
B
=

a

b

c

d
1
0
0
1
.
Find
AB
and
BA
.
Answer.
(a)
AB
=
1
2
3
1
1
1
0
2
2
BA
=
1
1
6
1
0
1
0
1
3
(b)
AB
=
0
0
0
0
BA
=

a

b

a
2

bc

ab

bd

c

d

ac

cd

bc

d
2
1
0
a
b
0
1
c
d
5. (a) Let
A
=
0
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
.
Find
A
2
.
(b) Let
B
=
1
0
1
5
1
3
1

1
2
,
C
=

5
1
1
7

1

2
6

1

1
.
Subscribe to view the full document.
 Summer '14
 LecturerFuXiaoye