MH1201 - PROBLEM SET 12
Date: 13 Apr 2016
Topics: The Gram-Schmidt process
Problem 1.
Let S be a subset in an inner product space V . Determine whether S is an orthogonal set,
an orthonormal set.
(1) V = C[0,
]
2
with the inner product hf, gi =
Z /2
f (x)
Answers to JC2 Preliminary Examination Paper 1 (H2 Physics)
Suggested Solutions:
1
2
3
4
5
6
7
8
9
10
No.
D
C
B
A
B
C
B
D
D
C
11
12
13
14
15
16
17
18
19
20
C
C
B
A
A
C
D
D
A
B
21
22
23
24
25
26
27
28
29
30
C
B
A
C
C
A
B
A
A
C
31
32
33
34
35
36
37
38
39
40
2010 NYJC JC2 Block Test 9740/1 Solutions
Qn
1
When x = 1, y = 7 :
a+bc+d = 7
When x = 2, y = 5 : 8a + 4b + 2c + d = 5
When x = 1, y = 1:
a + b + c + d = 1
(1)
(2)
(3)
dy
= 3ax 2 + 2bx + c
dx
dy
= 2 :
3a + 2b + c = 2 (4)
dx
Using GC, a = 1, b = 2, c =
NANYANG JUNIOR COLLEGE
JC2 BLOCK TEST
Higher 2
MATHEMATICS
9740/01
20th January 2010
Paper 1
3 Hours
Additional Materials:
Answer Paper
List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST
Write your name and class on all the work you hand in.
Write in d
1
2011 AJC Prelim JC2 H2 Phy Paper 1 suggested solutions
1
2
3
4
5
6
7
8
9
10
1
D
C
C
D
A
A
D
A
B
D
D
A
B
B
D
B
C
B
D
B
11
12
13
14
15
16
17
18
19
20
21
22
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30
B
A
B
A
D
B
A
D
B
B
31
32
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36
37
38
39
40
D
B
A
C
A
B
D
D
B
B
D
X =
Physics 9646
2011 C2 Preliminary Examination
Mark Scheme
Paper 1 Solutions
1
2
3
4
5
6
7
8
9
10
C
D
D
D
C
A
D
A
C
A
11
12
13
14
15
16
17
18
19
20
A
B
D
B
B
D
B
A
D
C
21
22
23
24
25
26
27
28
29
30
A
B
C
A
D
B
D
D
B
A
31
32
33
34
35
36
37
38
39
40
1
C
B
D
B
MH1201 - SOLUTIONS TO PROBLEM SET 10
Problem 1.
1) F
An eigenvector, by denition, should be nonzero, while the eigenspace E contains the zero
vector. Thats why the statement is false.
However, if we remove the zero vector from E , then the remaining vecto
MH1201 - SOLUTIONS TO PROBLEM SET 12
Problem 1.
Solution to part (1).
This set is not orthogonal, (and hence it is of course not orthonormal), because
2
, cos(x) =
2
/2
0
cos(x) dx = [sin(x)]/2 = 1.
0
Solution to part (2).
Put v1 = (cos , sin , 0), v2
MH1201 - SOLUTIONS TO PROBLEM SET 7
Problem 1.
As in Tutorial #5, you can proceed this problem in dierent ways, because for checking
one-to-one as well as onto properties, there are several criteria.
Below is presented one of those methods for both parts
MH1201 - SOLUTIONS TO PROBLEM SET 11
Problem 1.
(1) F
Hint: Take u = 0, then it works for any v, w.
(2) T
This is the last statement in Theorem 4.1.7.
The assumption u, v = u, w , u V , is equivalent to u, v w = 0, u V .
Put v w = s V . The last equation
MH1201 - SOLUTIONS TO PROBLEM SET 6
Problem 1.
Let = cfw_v1 , v2 , . . . , vn be an ordered basis for V . The only thing in this problem is that you
should remember well the denition of the coordinate vector of a vector x w.r.t. the ordered
basis : for e
MH1201 - SOLUTIONS TO PROBLEM SET 8
Problem 1.
No.
You know that the characteristic polynomial f () of a linear operator T is of degree n, which
has at most n roots, and moreover, not necessarily distinct.
For example:
- The polynomial f () = ( 1)(2 + 1)
MH1201 - SOLUTIONS TO PROBLEM SET 9
Problem 1.
(1) We have
T (1 )
T (2 )
1
2
= T = = (1)2
2
3
=
01 + (1)2
=
21 + 02 ,
2
2
= T = = 21
3
4
and hence
0 2
.
1 0
(2) It is clear, by 1), that the basis does not consist of eigenvectors of T .
[T ] =
Proble
MH1201 - SOLUTIONS TO PROBLEM SET 5
Problem 1.
Solution to part (1).
The linearity of T is easily checked (!).
Solution to part (2).
- Find a basis for N (T ):
We have
u = (a, b) N (T ) T (u) = 0
a+b
0
a+b=0
a = b = 0,
0 = 0
2a b = 0
2a b
0
whic
MH1201 - SOLUTIONS TO PROBLEM SET 1
Problem 1.
(1) Answer: NO.
We have
(a
1 , a2 )
+ (b1 , b2 ) = (a1 + b1 , a2 + b2 )
(a1 , a2 ) = (a1 , 0).
Its clear that axioms A1A2 are satised.
Furthermore, axioms A3A6 are also satised, because these axioms just refe
MH1201 - SOLUTIONS TO PROBLEM SET 3
Problem 1.
Solution to part (1).
Since dim(R3 ) = 3, any basis of R3 consists of exactly 3 vectors. So it doesnt matter what
choice you make for x, the given set cannot be a basis for R3 , as it consists of at most 2 v
MH1201 - SHORT ANSWERS TO CHALLENGING PROBLEMS
Notation: #n.x stand for Tutorial #n, Challenging Problem x.
Answer to #1.6.
In this problem V = F(R, R).
For the set A of even functions, f (x) = f (x) for all x.
f 0 belongs to A, because f (x) = 0 = f (
MH1201 - SOLUTIONS TO PROBLEM SET 4
Problem 1.
The problem is essentially asking us to determine the coordinates of a vector (a1 , a2 , a3 , a4 )
in terms of this basis.
Because the given set is a basis, we know that there are unique scalars 1 , 2 , 3 , a
Preliminary Examination
JC2 H2 Physics 2011
Answers to JC2 Preliminary Examination Paper 1 (H2 Physics)
Suggested Solutions:
1
2
3
4
5
6
7
8
9
10
A
C
B
B
C
D
A or B
B
B
C
11
12
13
14
15
16
17
18
19
20
C
C
B
A
A
C
D
C
B
B
21
22
23
24
25
26
27
28
29
30
A
C
MERIDIAN JUNIOR COLLEGE
Preliminary Examination
Higher 2
_
H2 Physics
9646/1
Paper 1
23 September 2011
1 hour 15 min
_
Class
Reg Number
Candidate Name _
READ THESE INSTRUCTIONS FIRST
Do not open this booklet until you are told to do so.
There are forty qu
VICTORIA JUNIOR COLLEGE
2011 JC2 PRELIMINARY EXAMINATIONS
9646/01
PHYSICS
Higher 2
21 Sep 2011
WEDNESDAY
Paper 1 Multiple Choice
8 am 9.15 am
1 Hour 15 min
Additional Materials: Multiple Choice Answer Sheet
READ THESE INSTRUCTIONS FIRST
Write in soft penc
2012 JC2 H1 & H2 ECONOMICS
16. Market Failure & Government Intervention
NYJC
Review on: Market System And Market Failure
Below are general issues you need to know before attempting the tutorial questions.
Please put a tick in the column, to indicate that
2012 JC2 H1 & H2 ECONOMICS
NYJC
International Economics:
13. International Trade, Finance & Globalization
Review on: International Economics:
International Trade, Finance & Globalization
Below are general issues you need to know before attempting the tuto
2010 NYJC JC2 Block Test 9740/1 Solutions
Qn
1
When x = 1, y = 7 :
a+bc+d = 7
When x = 2, y = 5 : 8a + 4b + 2c + d = 5
When x = 1, y = 1:
a + b + c + d = 1
(1)
(2)
(3)
dy
= 3ax 2 + 2bx + c
dx
dy
= 2 :
3a + 2b + c = 2 (4)
dx
Using GC, a = 1, b = 2, c =
NAME Mm MU 9r
MATH 15A QUIZ 1 [QUIZ A]
(1) (2 pts) Is the following matrix in Reduced Echelon Form (just circle one answer,
no explanation needed):
@ No
1
(2) (3 pts) Are the vectors »1 ,
—1
yourwork)
) I / O
“I 1 ‘l a -
“I ~J r a
I
No, 4F8‘5‘5 0656:559
SELF QUIZ 1 SOLUTION
a. False. (The word reduced is missing.) Counterexample:
1
A=
3
2
1
, B = 0
4
2
1
, C = 0
2
2
1
The matrix A is row equivalent to matrices B and C, both in echelon form.
b. False. Counterexample: Let A be any nn matrix with fewer t
Solutions for Linear Algebra Midterm I
1. True or False?
(a) If an n x n matrix A is not symmetric, then ATA is not symmetric.
(b) In general, the determinant of the sum of two matrices equals the sum of the
determinants of the matrices.
(c) If A and B ar
MH1200 Quiz 2 Solutions
September 17, 2015
Problem 1. Let A = [ai,j ]nn be an n-by-n matrix. Prove the following statement, or give a
counterexample: If A is invertible then
a1,1
a1,2 a1,n1
a2,1
a2,2 a2,n1
B= .
.
.
.
.
.
.
an1,1 an1,2 an1,n1
is also in
NAME: SOLUTIONS
LINEAR ALGEBRA MIDTERM [EXAM A]
HAROLD SULTAN
I NSTRUCTIONS
(1) Timing: You have 80 minutes for this exam. Some of the questions are harder than
others. Please use your time carefully and do not dwell on any single question for
too long be