⋆
]]
⋆
MAT2611: Linear Algebra II
OCT/NOV Examination 2026 Preparation
Covers Past Papers: Oct/Nov 2023 • Oct/Nov 2024 • Oct/Nov 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Mathematics – Linear Algebra [
_ Exam Revision Guide
MAT2611
Module Code:
Linear Algebra II
Module Name:
Oct/Nov 2023, 2024, 2025
Exam Papers:
Oct/Nov 2026 Examination
Preparation For:
20 MCQ Questions
Format:
100 Marks (5 per question)
Total Marks:
2 Hours
Duration:
Study for understanding. Each question is followed by a fully worked answer with
key concepts and exam tips.
Exam Revision Notes | MAT2611 | 2023–2026
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
UNIVERSITY EXAMINATIONS
OCT/NOV 2025 — MAT2611: Linear Algebra II
100 Marks | 2 Hours | 20 Questions (5 marks each)
All questions are multiple choice. Select the single best answer.
Question 1 [5] marks
Ù Oct/Nov 2025
Question: Consider the sets A = {x ∈ Z : x = 2(y − 3) for some y ∈ Z} and B = {x ∈ Z :
x = 2z for some z ∈ Z}. Which statement is true?
(1) A and B are equal (2) A ⊊ B (3) B ⊊ A (4) None of the above
Answer:
Correct answer: (1) A and B are equal.
Key Concept
Two sets are equal if and only if every element of one belongs to the other and vice
versa.
Proof:
• Any x ∈ A has the form x = 2(y − 3) = 2y − 6 = 2(y − 3). Since y ∈ Z, set z = y − 3 ∈ Z,
so x = 2z ∈ B. Thus A ⊆ B.
• Any x ∈ B has the form x = 2z for some z ∈ Z. Set y = z + 3 ∈ Z, then 2(y − 3) = 2z =
x ∈ A. Thus B ⊆ A.
Therefore A = B.
Exam Tip
When comparing sets defined by different expressions, always test both A ⊆ B and B ⊆
A by direct substitution.
Page 2 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Question 2 [5] marks
Ù Oct/Nov 2025
Question: Let f : R → R+ be defined by f (x) = ex . The image of the set S = {x ∈ R :
0 ≤ x2 ≤ 9} under f is:
(1) (−∞, e−3 ] ∪ [e3 , ∞) (2) [e−3 , e3 ] (3) [e3 , ∞) (4) None of the above
Answer:
Correct answer: (2) [e−3 , e3 ].
Step 1 — Determine S: 0 ≤ x2 ≤ 9 =⇒ |x| ≤ 3 =⇒ x ∈ [−3, 3].
Step 2 — Apply f : Since ex is strictly increasing,
f (S) = {ex : x ∈ [−3, 3]} = [e−3 , e3 ].
Exam Tip
Always simplify the domain set first (S), then apply the function. For monotone func-
tions, the image of an interval [a, b] is simply [f (a), f (b)].
Question 3 [5] marks
Ù Oct/Nov 2025
Question: The two distinct square roots of the complex number i are:
√ √ √ √ √ √ √ √
2 2 2 2 2 2 2 2
(1) + i and − − i (2) − i and − + i
2 2 2 2 2 2 2 2
(3) 1 + i and −1 − i (4) None of the above
Answer:
Correct answer: (1)
Method — Polar form: Write i = eiπ/2 . The square roots are:
√ √
√ iπ/4 π π 2 2
i=e = cos 4 + i sin 4 = + i
2 2
Page 3 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
and the second root is the negative:
√ √
√ 2 2
− i=− − i.
2 2
Key Concept
For z = reiθ , the n-th roots are zk = r1/n ei(θ+2πk)/n , k = 0, 1, . . . , n − 1.
Watch Out
√ √ 2
Option (2) gives the square roots of −i, not i. Verify by squaring: 2
2
+ 2
2 i = 12 (1 +
2i − 1) = i. ✓
Question 4 [5] marks
Ù Oct/Nov 2025
Question: The value of x ∈ R4 such that (4, −3, −3, 2) + 3x = (7, −3, 3, 2) is:
(1) x = (1, 0, 2, 0) (2) x = (4, −4, 3, 0) (3) x = (1, −2, 0, 0) (4) None of the above
Answer:
Correct answer: (1) x = (1, 0, 2, 0).
Solve 3x = (7, −3, 3, 2) − (4, −3, −3, 2) = (3, 0, 6, 0), hence x = 13 (3, 0, 6, 0) = (1, 0, 2, 0).
Question 5 [5] marks
Ù Oct/Nov 2025
Question: Let W be a subset of R3 defined as W = {(x, y, z) ∈ R3 : 2x + y − z − 1 = 0}.
Then:
(1) W is a subspace of R3 (2) W is closed under scalar multiplication (3) W is not a
subspace of R3 (4) W is closed under addition
Answer:
Correct answer: (3) W is not a subspace of R3 .
Page 4 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Key Concept
A subset W is a subspace only if it contains 0 and is closed under addition and scalar
multiplication.
Check zero vector: (0, 0, 0): 2(0) + 0 − 0 − 1 = −1 ̸= 0. So 0 ∈
/ W , and W is not a subspace.
Exam Tip
Always check the zero vector first — it is the fastest way to disprove the subspace
property. The equation 2x + y − z = 1 describes a plane that does not pass through the
origin.
Page 5 of 16 ⋆
, MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Question 6 [5] marks
Ù Oct/Nov 2025
Question: Which of the following sets of vectors in R3 forms a basis for R3 ?
(1) {(1, 0, 0), (0, 1, 0), (1, 1, 0)} (2) {(1, 2, 1), (0, 1, 1), (1, 3, 2)} (3)
{(1, 0, 1), (0, 1, 0), (2, −1, 2)} (4) {(1, 0, 0), (0, 1, 0), (0, 0, 1)}
Answer:
Correct answer: (4) (the standard basis for R3 ).
Check (1): Third vector = first + second, so linearly dependent.
Check (2): (1, 3, 2) = (1, 2, 1) + (0, 1, 1), linearly dependent.
1 0 1
Check (3): det
0 1 0 = 1(2 − 0) − 0 + 1(0 − 2) = 2 − 2 = 0, linearly dependent.
2 −1 2
Check (4): det(I3 ) = 1 ̸= 0 — linearly independent, spans R3 .
Key Concept
A set of n vectors in Rn is a basis ⇐⇒ the matrix formed by these vectors (as rows or
columns) has nonzero determinant.
Question 7 [5] marks
Ù Oct/Nov 2025
Question: Let T : R2 → R3 be defined by T (x, y) = (x + y, x − y, 2x). Then T is:
(1) Neither linear nor injective (2) Linear but not injective (3) Linear and injective
(4) Injective but not linear
Answer:
Correct answer: (3) Linear and injective.
Page 6 of 16 ⋆
]]
⋆
MAT2611: Linear Algebra II
OCT/NOV Examination 2026 Preparation
Covers Past Papers: Oct/Nov 2023 • Oct/Nov 2024 • Oct/Nov 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Mathematics – Linear Algebra [
_ Exam Revision Guide
MAT2611
Module Code:
Linear Algebra II
Module Name:
Oct/Nov 2023, 2024, 2025
Exam Papers:
Oct/Nov 2026 Examination
Preparation For:
20 MCQ Questions
Format:
100 Marks (5 per question)
Total Marks:
2 Hours
Duration:
Study for understanding. Each question is followed by a fully worked answer with
key concepts and exam tips.
Exam Revision Notes | MAT2611 | 2023–2026
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
UNIVERSITY EXAMINATIONS
OCT/NOV 2025 — MAT2611: Linear Algebra II
100 Marks | 2 Hours | 20 Questions (5 marks each)
All questions are multiple choice. Select the single best answer.
Question 1 [5] marks
Ù Oct/Nov 2025
Question: Consider the sets A = {x ∈ Z : x = 2(y − 3) for some y ∈ Z} and B = {x ∈ Z :
x = 2z for some z ∈ Z}. Which statement is true?
(1) A and B are equal (2) A ⊊ B (3) B ⊊ A (4) None of the above
Answer:
Correct answer: (1) A and B are equal.
Key Concept
Two sets are equal if and only if every element of one belongs to the other and vice
versa.
Proof:
• Any x ∈ A has the form x = 2(y − 3) = 2y − 6 = 2(y − 3). Since y ∈ Z, set z = y − 3 ∈ Z,
so x = 2z ∈ B. Thus A ⊆ B.
• Any x ∈ B has the form x = 2z for some z ∈ Z. Set y = z + 3 ∈ Z, then 2(y − 3) = 2z =
x ∈ A. Thus B ⊆ A.
Therefore A = B.
Exam Tip
When comparing sets defined by different expressions, always test both A ⊆ B and B ⊆
A by direct substitution.
Page 2 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Question 2 [5] marks
Ù Oct/Nov 2025
Question: Let f : R → R+ be defined by f (x) = ex . The image of the set S = {x ∈ R :
0 ≤ x2 ≤ 9} under f is:
(1) (−∞, e−3 ] ∪ [e3 , ∞) (2) [e−3 , e3 ] (3) [e3 , ∞) (4) None of the above
Answer:
Correct answer: (2) [e−3 , e3 ].
Step 1 — Determine S: 0 ≤ x2 ≤ 9 =⇒ |x| ≤ 3 =⇒ x ∈ [−3, 3].
Step 2 — Apply f : Since ex is strictly increasing,
f (S) = {ex : x ∈ [−3, 3]} = [e−3 , e3 ].
Exam Tip
Always simplify the domain set first (S), then apply the function. For monotone func-
tions, the image of an interval [a, b] is simply [f (a), f (b)].
Question 3 [5] marks
Ù Oct/Nov 2025
Question: The two distinct square roots of the complex number i are:
√ √ √ √ √ √ √ √
2 2 2 2 2 2 2 2
(1) + i and − − i (2) − i and − + i
2 2 2 2 2 2 2 2
(3) 1 + i and −1 − i (4) None of the above
Answer:
Correct answer: (1)
Method — Polar form: Write i = eiπ/2 . The square roots are:
√ √
√ iπ/4 π π 2 2
i=e = cos 4 + i sin 4 = + i
2 2
Page 3 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
and the second root is the negative:
√ √
√ 2 2
− i=− − i.
2 2
Key Concept
For z = reiθ , the n-th roots are zk = r1/n ei(θ+2πk)/n , k = 0, 1, . . . , n − 1.
Watch Out
√ √ 2
Option (2) gives the square roots of −i, not i. Verify by squaring: 2
2
+ 2
2 i = 12 (1 +
2i − 1) = i. ✓
Question 4 [5] marks
Ù Oct/Nov 2025
Question: The value of x ∈ R4 such that (4, −3, −3, 2) + 3x = (7, −3, 3, 2) is:
(1) x = (1, 0, 2, 0) (2) x = (4, −4, 3, 0) (3) x = (1, −2, 0, 0) (4) None of the above
Answer:
Correct answer: (1) x = (1, 0, 2, 0).
Solve 3x = (7, −3, 3, 2) − (4, −3, −3, 2) = (3, 0, 6, 0), hence x = 13 (3, 0, 6, 0) = (1, 0, 2, 0).
Question 5 [5] marks
Ù Oct/Nov 2025
Question: Let W be a subset of R3 defined as W = {(x, y, z) ∈ R3 : 2x + y − z − 1 = 0}.
Then:
(1) W is a subspace of R3 (2) W is closed under scalar multiplication (3) W is not a
subspace of R3 (4) W is closed under addition
Answer:
Correct answer: (3) W is not a subspace of R3 .
Page 4 of 16 ⋆
,MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Key Concept
A subset W is a subspace only if it contains 0 and is closed under addition and scalar
multiplication.
Check zero vector: (0, 0, 0): 2(0) + 0 − 0 − 1 = −1 ̸= 0. So 0 ∈
/ W , and W is not a subspace.
Exam Tip
Always check the zero vector first — it is the fastest way to disprove the subspace
property. The equation 2x + y − z = 1 describes a plane that does not pass through the
origin.
Page 5 of 16 ⋆
, MAT2611 | Exam Revision 2023–2025 Linear Algebra II
Question 6 [5] marks
Ù Oct/Nov 2025
Question: Which of the following sets of vectors in R3 forms a basis for R3 ?
(1) {(1, 0, 0), (0, 1, 0), (1, 1, 0)} (2) {(1, 2, 1), (0, 1, 1), (1, 3, 2)} (3)
{(1, 0, 1), (0, 1, 0), (2, −1, 2)} (4) {(1, 0, 0), (0, 1, 0), (0, 0, 1)}
Answer:
Correct answer: (4) (the standard basis for R3 ).
Check (1): Third vector = first + second, so linearly dependent.
Check (2): (1, 3, 2) = (1, 2, 1) + (0, 1, 1), linearly dependent.
1 0 1
Check (3): det
0 1 0 = 1(2 − 0) − 0 + 1(0 − 2) = 2 − 2 = 0, linearly dependent.
2 −1 2
Check (4): det(I3 ) = 1 ̸= 0 — linearly independent, spans R3 .
Key Concept
A set of n vectors in Rn is a basis ⇐⇒ the matrix formed by these vectors (as rows or
columns) has nonzero determinant.
Question 7 [5] marks
Ù Oct/Nov 2025
Question: Let T : R2 → R3 be defined by T (x, y) = (x + y, x − y, 2x). Then T is:
(1) Neither linear nor injective (2) Linear but not injective (3) Linear and injective
(4) Injective but not linear
Answer:
Correct answer: (3) Linear and injective.
Page 6 of 16 ⋆
