UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Semester 1 — 2026
⋄
Module Code: MAT2611
Module Name: Linear Algebra
Assignment No.: 01
Due Date: 30 April 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for MAT2611 Linear Algebra
at the University of South Africa.
,UNISA | MAT2611 Assignment 01 — 2026
Problem 1: Elements and Subsets
Question (a): Give an example of a set A such that there is a set B with B ∈ A but B ̸⊆ A.
Question (b): Give an example of a set A such that there is a set B with B ⊆ A but B ∈
/ A.
1(a) Solution
Let A = {{1, 2}, 3} and B = {1, 2}.
Step 1: Verify B ∈ A.
The elements of A are {1, 2} and 3. Since {1, 2} is listed as a member of A, we have B =
{1, 2} ∈ A. ✓
Step 2: Verify B ̸⊆ A.
For B ⊆ A, every element of B must also be an element of A. The elements of B are 1 and 2.
Checking whether 1 ∈ A: the elements of A are {1, 2} and 3; the number 1 is not among them.
Therefore 1 ∈
/ A.
Since at least one element of B (namely 1) is not in A, it follows that B ̸⊆ A. ✓
Implementation Insight
The key idea: B ∈ A means B is itself an element (a member) of A. The condition
B ⊆ A means every element of B belongs to A. These two conditions are independent,
so one can hold without the other.
1(b) Solution
Let A = {1, 2, 3} and B = {1, 2}.
Step 1: Verify B ⊆ A.
The elements of B are 1 and 2. Since 1 ∈ A and 2 ∈ A, every element of B is in A. Therefore
B ⊆ A. ✓
Step 2: Verify B ∈
/ A.
The elements of A are 1, 2, and 3; none of these is the set {1, 2}. Therefore B = {1, 2} ∈
/ A. ✓
Page 1 of 15
, UNISA | MAT2611 Assignment 01 — 2026
Key Distinction
Subset vs. element: B ⊆ A is a relationship between two sets (containment). B ∈ A
treats B as an object that lives inside A. A set can be a subset of A without being a
member of A, and vice versa.
Page 2 of 15
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Semester 1 — 2026
⋄
Module Code: MAT2611
Module Name: Linear Algebra
Assignment No.: 01
Due Date: 30 April 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for MAT2611 Linear Algebra
at the University of South Africa.
,UNISA | MAT2611 Assignment 01 — 2026
Problem 1: Elements and Subsets
Question (a): Give an example of a set A such that there is a set B with B ∈ A but B ̸⊆ A.
Question (b): Give an example of a set A such that there is a set B with B ⊆ A but B ∈
/ A.
1(a) Solution
Let A = {{1, 2}, 3} and B = {1, 2}.
Step 1: Verify B ∈ A.
The elements of A are {1, 2} and 3. Since {1, 2} is listed as a member of A, we have B =
{1, 2} ∈ A. ✓
Step 2: Verify B ̸⊆ A.
For B ⊆ A, every element of B must also be an element of A. The elements of B are 1 and 2.
Checking whether 1 ∈ A: the elements of A are {1, 2} and 3; the number 1 is not among them.
Therefore 1 ∈
/ A.
Since at least one element of B (namely 1) is not in A, it follows that B ̸⊆ A. ✓
Implementation Insight
The key idea: B ∈ A means B is itself an element (a member) of A. The condition
B ⊆ A means every element of B belongs to A. These two conditions are independent,
so one can hold without the other.
1(b) Solution
Let A = {1, 2, 3} and B = {1, 2}.
Step 1: Verify B ⊆ A.
The elements of B are 1 and 2. Since 1 ∈ A and 2 ∈ A, every element of B is in A. Therefore
B ⊆ A. ✓
Step 2: Verify B ∈
/ A.
The elements of A are 1, 2, and 3; none of these is the set {1, 2}. Therefore B = {1, 2} ∈
/ A. ✓
Page 1 of 15
, UNISA | MAT2611 Assignment 01 — 2026
Key Distinction
Subset vs. element: B ⊆ A is a relationship between two sets (containment). B ∈ A
treats B as an object that lives inside A. A set can be a subset of A without being a
member of A, and vice versa.
Page 2 of 15
