MAT2611
ASSIGNMENT 1
Linear Algebra II
FULL SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 7
, SOLUTIONS:
Problem 1
(a) Let 𝐴 = {{1}} and 𝐵 = {1}. Clearly 𝐵 ∈ 𝐴 because {1} is the only element of 𝐴.
However, 𝐵 ⊈ 𝐴 since the element 1 in 𝐵 is not an element of 𝐴, which contains only {1}.
(b) Let 𝐴 = {1,2,3} and 𝐵 = {1}. Then 𝐵 ⊆ 𝐴 because the element 1 is indeed in 𝐴.
But 𝐵 ∉ 𝐴 since {1} itself is not listed among the elements of 𝐴.
SOLUTIONS:
Problem 2
1. 𝑃({∅}) = {∅, {∅}}. The set {∅} has one element, namely ∅, so its power set
contains the empty set and the set itself.
2. 𝑃({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}. The set has two distinct elements, so its
power set has 22 = 4 elements.
3. 𝑃({{∅}}) = {∅, {{∅}}}. The single element here is {∅}, so the power set contains
the empty set and the singleton containing that element.
4. We have 𝑃(∅) = {∅}, so 𝑃(𝑃(∅)) = 𝑃({∅}) = {∅, {∅}}.
5. First 𝑃({∅}) = {∅, {∅}}. Then 𝑃(𝑃({∅})) = 𝑃({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}.
Page 2 of 7
ASSIGNMENT 1
Linear Algebra II
FULL SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 7
, SOLUTIONS:
Problem 1
(a) Let 𝐴 = {{1}} and 𝐵 = {1}. Clearly 𝐵 ∈ 𝐴 because {1} is the only element of 𝐴.
However, 𝐵 ⊈ 𝐴 since the element 1 in 𝐵 is not an element of 𝐴, which contains only {1}.
(b) Let 𝐴 = {1,2,3} and 𝐵 = {1}. Then 𝐵 ⊆ 𝐴 because the element 1 is indeed in 𝐴.
But 𝐵 ∉ 𝐴 since {1} itself is not listed among the elements of 𝐴.
SOLUTIONS:
Problem 2
1. 𝑃({∅}) = {∅, {∅}}. The set {∅} has one element, namely ∅, so its power set
contains the empty set and the set itself.
2. 𝑃({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}. The set has two distinct elements, so its
power set has 22 = 4 elements.
3. 𝑃({{∅}}) = {∅, {{∅}}}. The single element here is {∅}, so the power set contains
the empty set and the singleton containing that element.
4. We have 𝑃(∅) = {∅}, so 𝑃(𝑃(∅)) = 𝑃({∅}) = {∅, {∅}}.
5. First 𝑃({∅}) = {∅, {∅}}. Then 𝑃(𝑃({∅})) = 𝑃({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}.
Page 2 of 7
