Exam (elaborations)
MAT2611 EXAM PACK 2023
- Institution
- University Of South Africa (Unisa)
MAT2611 EXAM PACK 2023
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MAT2611EXAM PACK
2023
QUESTIONS
AND
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
, MAT2611/201/1/2020
Verification for (d) For each p ∈ X there exists by (c) for
a unique
f q ∈ Y such that (p, q) ⇔
∈f
(q, p) ∈ −1 −1
f , verifying (d) for f.
The proof of f −1
◦f = 1Y and f−1◦f = 1X should now be clear from (a)-(d) for both
and ff−1.
[Total: 20 marks]
3
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