MAT2611 EXAM
PACK 2023
LATEST QUESTIONS
AND ANSWERS
,MAT2611
Assignment PACK
2022
Assigment Feedback
Questions. Answers
, MAT2611/201/1/2020
LINEAR ALGEBRA
MAT2611
Department of Mathematical Sciences
This tutorial letter contains solutions for assignment 01.
BARCODE
university
Define tomorrow. of south africa
, Problem 1. Answer Exercise 2.22 from Addendum C. [10 marks]
Solution 1. The power set P(A) of a set A is the set of all subsets of A, i.e. we have
T ∈ P(A) ⇔ T ⊆ A.
Since ∅ ⊆ A (always, Why?)
and A ⊆ A always, the set P(A) always has at least the members ∅
and A, always.
As for example, since T ⊆ ∅ ⇔ T = ∅, we have:
P(∅) = {∅}.
Similarly, since T ⊆ {∅} if and only if either T = ∅ or else T = {∅}, we have:
P({∅}) =∅, {∅}
.
Similarly:
P({{∅}}) = ∅, {{∅}}
P({∅, {∅}}) = ∅, {∅}, {{∅}}, {∅, {∅}}
The rest can be figured out now.
Guess the number of elements of P(A) if A has exactly n elements and prove your guess.
Problem 2. Answer Exercise 3.12 & Exercise 3.13 from Addendum C. [5 + 5 = 10 marks]
f
Solution 2. Recall that X→
− Y if the following three conditions are satisfied:
(a) f ⊆ X × Y
(b) For each p ∈ X there exists a q ∈ Y such that (p, q) ∈ f .
0
(c) If (p, q) ∈ and
f (p, q) ∈ fthen q = 0q
.
f
Given X−
→ Y to be a one-to-one correspondence there is the additional property:
(d) For each q ∈ Y there exists a unique p ∈ X such that (p, q) ∈ f .
Hence the set:
(?) f −1 = (y , x ) : (x , y ) ∈ f
satisfies all the conditions (a)-(d) with X and Y interchanged.
−1
Verification for (a) From the definition in (?): (x , y ) ⇔
∈ (y
f , x) ∈ ⊆
f X × Y ⇒ (y, x) ∈ Y × X .
−1
Hence f ⊆ Y × X.
Verification for (b)&(c) Choose and fix any q ∈ Y . Using (d), for each q ∈ Y there exists a uniq
p ∈ X such that (p, q) ∈⇔ f(q, p) ∈ −1
f .
Hence for each q ∈ Y there exists a unique p ∈ X such that(q, p)∈ f −1 verifying the
conditions (b) & (c).
2
PACK 2023
LATEST QUESTIONS
AND ANSWERS
,MAT2611
Assignment PACK
2022
Assigment Feedback
Questions. Answers
, MAT2611/201/1/2020
LINEAR ALGEBRA
MAT2611
Department of Mathematical Sciences
This tutorial letter contains solutions for assignment 01.
BARCODE
university
Define tomorrow. of south africa
, Problem 1. Answer Exercise 2.22 from Addendum C. [10 marks]
Solution 1. The power set P(A) of a set A is the set of all subsets of A, i.e. we have
T ∈ P(A) ⇔ T ⊆ A.
Since ∅ ⊆ A (always, Why?)
and A ⊆ A always, the set P(A) always has at least the members ∅
and A, always.
As for example, since T ⊆ ∅ ⇔ T = ∅, we have:
P(∅) = {∅}.
Similarly, since T ⊆ {∅} if and only if either T = ∅ or else T = {∅}, we have:
P({∅}) =∅, {∅}
.
Similarly:
P({{∅}}) = ∅, {{∅}}
P({∅, {∅}}) = ∅, {∅}, {{∅}}, {∅, {∅}}
The rest can be figured out now.
Guess the number of elements of P(A) if A has exactly n elements and prove your guess.
Problem 2. Answer Exercise 3.12 & Exercise 3.13 from Addendum C. [5 + 5 = 10 marks]
f
Solution 2. Recall that X→
− Y if the following three conditions are satisfied:
(a) f ⊆ X × Y
(b) For each p ∈ X there exists a q ∈ Y such that (p, q) ∈ f .
0
(c) If (p, q) ∈ and
f (p, q) ∈ fthen q = 0q
.
f
Given X−
→ Y to be a one-to-one correspondence there is the additional property:
(d) For each q ∈ Y there exists a unique p ∈ X such that (p, q) ∈ f .
Hence the set:
(?) f −1 = (y , x ) : (x , y ) ∈ f
satisfies all the conditions (a)-(d) with X and Y interchanged.
−1
Verification for (a) From the definition in (?): (x , y ) ⇔
∈ (y
f , x) ∈ ⊆
f X × Y ⇒ (y, x) ∈ Y × X .
−1
Hence f ⊆ Y × X.
Verification for (b)&(c) Choose and fix any q ∈ Y . Using (d), for each q ∈ Y there exists a uniq
p ∈ X such that (p, q) ∈⇔ f(q, p) ∈ −1
f .
Hence for each q ∈ Y there exists a unique p ∈ X such that(q, p)∈ f −1 verifying the
conditions (b) & (c).
2
