An Introduction to Optimization 4th Edition Solution Manual PDF
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Course
MATH101
Institution
University Of San Andrés (UdeSA
)
Book
An Introduction to Optimization
Complete Solutions Manual PDF for An Introduction to Optimization Fourth Edition by Edwin K. P. Chong and Stanislaw H. Zak
An Introduction to Optimization Solution Manual
Exam (elaborations) TEST BANK FOR An Introduction to Optimization 4th edition By Edwin K. P. Chong, Stanislaw H. Zak (solution manual)
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AN INTRODUCTION TO
OPTIMIZATION
SOLUTIONS MANUAL
Fourth Edition
Edwin K. P. Chong and Stanislaw H. Żak
A JOHN WILEY & SONS, INC., PUBLICATION
,1. Methods of Proof and Some Notation
1.1
A B not A not B A⇒B (not B)⇒(not A)
F F T T T T
F T T F T T
T F F T F F
T T F F T T
1.2
A B not A not B A⇒B not (A and (not B))
F F T T T T
F T T F T T
T F F T F F
T T F F T T
1.3
A B not (A and B) not A not B (not A) or (not B))
F F T T T T
F T T T F T
T F T F T T
T T F F F F
1.4
A B A and B A and (not B) (A and B) or (A and (not B))
F F F F F
F T F F F
T F F T T
T T T F T
1.5
The cards that you should turn over are 3 and A. The remaining cards are irrelevant to ascertaining the
truth or falsity of the rule. The card with S is irrelevant because S is not a vowel. The card with 8 is not
relevant because the rule does not say that if a card has an even number on one side, then it has a vowel on
the other side.
Turning over the A card directly verifies the rule, while turning over the 3 card verifies the contraposition.
2. Vector Spaces and Matrices
2.1
We show this by contradiction. Suppose n < m. Then, the number of columns of A is n. Since rank A is
the maximum number of linearly independent columns of A, then rank A cannot be greater than n < m,
which contradicts the assumption that rank A = m.
2.2
.
⇒: Since there exists a solution, then by Theorem 2.1, rank A = rank[A ..b]. So, it remains to prove that
rank A = n. For this, suppose that rank A < n (note that it is impossible for rank A > n since A has
only n columns). Hence, there exists y ∈ Rn , y 6= 0, such that Ay = 0 (this is because the columns of
1
,A are linearly dependent, and Ay is a linear combination of the columns of A). Let x be a solution to
Ax = b. Then clearly x + y 6= x is also a solution. This contradicts the uniqueness of the solution. Hence,
rank A = n.
⇐: By Theorem 2.1, a solution exists. It remains to prove that it is unique. For this, let x and y be
solutions, i.e., Ax = b and Ay = b. Subtracting, we get A(x − y) = 0. Since rank A = n and A has n
columns, then x − y = 0 and hence x = y, which shows that the solution is unique.
2.3
⊤
Consider the vectors āi = [1, a⊤
i ] ∈ R
n+1
, i = 1, . . . , k. Since k ≥ n + 2, then the vectors ā1 , . . . , āk must
be linearly independent in Rn+1. Hence, there exist α1 , . . . αk , not all zero, such that
k
X
α i ai = 0.
i=1
Pk
The first component of the above vector equation is i=1 αi = 0, while the last n components have the form
Pk
i=1 αi ai = 0, completing the proof.
2.4
a. We first postmultiply M by the matrix
" #
Ik O
−M m−k,k I m−k
to obtain " #" # " #
M m−k,k I m−k Ik O O I m−k
= .
M k,k O −M m−k,k I m−k M k,k O
Note that the determinant of the postmultiplying matrix is 1. Next we postmultiply the resulting product
by " #
O Ik
I m−k O
to obtain " #" # " #
O I m−k O Ik Ik O
= .
M k,k O I m−k O O M k,k
Notice that " #! " #!
Ik O O Ik
det M = det det ,
O M k,k I m−k O
where " #!
O Ik
det = ±1.
I m−k O
The above easily follows from the fact that the determinant changes its sign if we interchange columns, as
discussed in Section 2.2. Moreover,
" #!
Ik O
det = det(I k ) det(M k,k ) = det(M k,k ).
O M k,k
Hence,
det M = ± det M k,k .
b. We can see this on the following examples. We assume, without loss of generality that M m−k,k = O and
let M k,k = 2. Thus k = 1. First consider the case when m = 2. Then we have
" # " #
O I m−k 0 1
M= = .
M k,k O 2 0
2
, Thus,
det M = −2 = det (−M k,k ) .
Next consider the case when m = 3. Then
...
0 1 0
...
" #
O I m−k 0 0 1 = 2 6= det (−M k,k ) .
det = det
M k,k O · · ·
··· ··· · · ·
...
2 0 0
Therefore, in general,
det M 6= det (−M k,k )
However, when k = m/2, that is, when all sub-matrices are square and of the same dimension, then it is
true that
det M = det (−M k,k ) .
See [121].
2.5
Let " #
A B
M=
C D
and suppose that each block is k × k. John R. Silvester [121] showed that if at least one of the blocks is
equal to O (zero matrix), then the desired formula holds. Indeed, if a row or column block is zero, then the
determinant is equal to zero as follows from the determinant’s properties discussed Section 2.2. That is, if
A = B = O, or A = C = O, and so on, then obviously det M = 0. This includes the case when any three
or all four block matrices are zero matrices.
If B = O or C = O then " #
A B
det M = det = det (AD) .
C D
The only case left to analyze is when A = O or D = O. We will show that in either case,
det M = det (−BC) .
Without loss of generality suppose that D = O. Following arguments of John R. Silvester [121], we premul-
tiply M by the product of three matrices whose determinants are unity:
" #" #" #" # " #
I k −I k I k O I k −I k A B −C O
= .
O Ik Ik Ik O Ik C O A B
Hence,
" # " #
A B −C O
det =
C O A B
= det (−C) det B
= det (−I k ) det C det B.
Thus we have " #
A B
det = det (−BC) = det (−CB) .
C O
3
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