⋆ ˆ ⋆
]]
MAT1503: LINEAR ALGEBRA I
OCT/NOV Examination 2026 Revision Guide
Covering Past Papers: Oct/Nov 2023 • Oct/Nov 2024 • Oct/Nov 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
Mathematics — Science, Engineering & Technology
Exam Revision Guide
MAT1503
Module Code:
Linear Algebra I
Module Name:
Oct/Nov 2023, 2024, 2025
Papers Covered:
Oct/Nov 2026 Examination
Prepared For:
100 per paper
Total Marks:
2 Hours
Duration:
Elementary Linear Algebra (Anton &
Textbook: Rorres)
Comprehensive question-and-answer revision covering all major topics. Focus on
understanding the method, not just memorising answers.
Exam Revision Notes | MAT1503 | 2023–2025 Coverage
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
PAPER 1: OCT/NOV 2025
University of South Africa — MAT1503 Linear Algebra I
Duration: 2 Hours Total Marks: 100 Closed Book
Page 2 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Question 1 [24 marks]
(a) [10 marks]
Question: Consider the following system of linear equations:
x1 + x 2 − x 3 = 2
3x1 + 2x2 − x3 = 3
−x1 − x2 + 2x3 = −1
Write down the augmented matrix, reduce it to generalized row echelon form (GREF),
and determine the solution of the system.
Answer:
Step 1: Write the augmented matrix.
1 1 −1 2
[A|b] =
3 2 −1 3
−1 −1 2 −1
Step 2: Row reduce to GREF.
Apply R2 ← R2 − 3R1 and R3 ← R3 + R1 :
1 1 −1 2
0 −1 2 −3
0 0 1 1
Apply R2 ← −R2 :
1 1 −1 2
0 1 −2 3
0 0 1 1
This is in GREF. Back-substitute:
• From Row 3: x3 = 1
• From Row 2: x2 − 2(1) = 3 ⇒ x2 = 5
• From Row 1: x1 + 5 − 1 = 2 ⇒ x1 = −2
Solution: x1 = −2, x2 = 5, x3 = 1.
Page 3 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Exam Tip
Always check your answer by substituting back into every original equation. In the
exam, show each row operation explicitly using arrow notation, e.g. R2 − 3R1 → R2 .
(b) [6 marks]
Question: Using the system in (a), write down matrices A and b and state the size of
each of A, x, and b.
Answer:
1 1 −1 2 x1
A=
3 2 ,
−1 b=
3 ,
x = x2
−1 −1 2 −1 x3
• A is a 3 × 3 matrix.
• x is a 3 × 1 column vector.
• b is a 3 × 1 column vector.
The system is written as Ax = b.
(c) [8 marks]
Question: Determine the inverse A−1 of matrix A in (b). Then use A−1 to solve Ax =
b and verify your answer in (a).
Answer:
Augment A with the identity and row-reduce:
1 1 −1 1 0 0
3
2 −1 0 1 0
−1 −1 2 0 0 1
R2 − 3R1 → R2 ; R3 + R1 → R3 :
1 1 −1 1 0 0
0 −1 2 −3 1 0
0 0 1 1 0 1
−R2 → R2 ; then R2 − 2R3 → R2 ; R1 + R3 → R1 :
Page 4 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
1 1 0 2 0 1
0 1 0 5 −1 2
0 0 1 1 0 1
R1 − R2 → R1 :
1 0 0 −3 1 −1
0 1 0
5 −1 2
0 0 1 1 0 1
Therefore:
−3 1 −1
A−1
=
5 −1 2
1 0 1
Verification: x = A−1 b:
−3 1 −1 2 −6 + 3 + 1 −2
x=
5 −1 2 3 = 10 − 3 − 2 = 5
1 0 1 −1 2+0−1 1
This confirms x1 = −2, x2 = 5, x3 = 1. ✓
Page 5 of 36
, MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Question 2 [20 marks]
(a) [8 marks]
Question: Find all values of k for which the matrix
−1 1 k
A=
0 1 3
k 4 7
is singular (i.e. not invertible).
Answer:
A matrix is singular when its determinant equals zero. Expand det(A) along the first row:
" # " # " #
1 3 0 3 0 1
det(A) = −1 det − 1 det + k det
4 7 k 7 k 4
= −1(7 − 12) − 1(0 − 3k) + k(0 − k)
= −1(−5) − 1(−3k) + k(−k)
= 5 + 3k − k 2
Set det(A) = 0:
−k 2 + 3k + 5 = 0 =⇒ k 2 − 3k − 5 ̸= 0 (use quadratic formula)
Wait – let us re-examine: 5 + 3k − k 2 = 0 ⇒ k 2 − 3k − 5 = 0.
√ √
3± 9 + 20 3 ± 29
k= =
2 2
However, for the standard UNISA quiz version where the matrix entries give integer solutions,
the determinant gives:
5 + 3k − k 2 = 0 ⇒ k 2 − 3k − 5 = 0
√ √
3 + 29 3 − 29
The matrix is singular for k = or k = .
2 2
Page 6 of 36
]]
MAT1503: LINEAR ALGEBRA I
OCT/NOV Examination 2026 Revision Guide
Covering Past Papers: Oct/Nov 2023 • Oct/Nov 2024 • Oct/Nov 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
Mathematics — Science, Engineering & Technology
Exam Revision Guide
MAT1503
Module Code:
Linear Algebra I
Module Name:
Oct/Nov 2023, 2024, 2025
Papers Covered:
Oct/Nov 2026 Examination
Prepared For:
100 per paper
Total Marks:
2 Hours
Duration:
Elementary Linear Algebra (Anton &
Textbook: Rorres)
Comprehensive question-and-answer revision covering all major topics. Focus on
understanding the method, not just memorising answers.
Exam Revision Notes | MAT1503 | 2023–2025 Coverage
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
PAPER 1: OCT/NOV 2025
University of South Africa — MAT1503 Linear Algebra I
Duration: 2 Hours Total Marks: 100 Closed Book
Page 2 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Question 1 [24 marks]
(a) [10 marks]
Question: Consider the following system of linear equations:
x1 + x 2 − x 3 = 2
3x1 + 2x2 − x3 = 3
−x1 − x2 + 2x3 = −1
Write down the augmented matrix, reduce it to generalized row echelon form (GREF),
and determine the solution of the system.
Answer:
Step 1: Write the augmented matrix.
1 1 −1 2
[A|b] =
3 2 −1 3
−1 −1 2 −1
Step 2: Row reduce to GREF.
Apply R2 ← R2 − 3R1 and R3 ← R3 + R1 :
1 1 −1 2
0 −1 2 −3
0 0 1 1
Apply R2 ← −R2 :
1 1 −1 2
0 1 −2 3
0 0 1 1
This is in GREF. Back-substitute:
• From Row 3: x3 = 1
• From Row 2: x2 − 2(1) = 3 ⇒ x2 = 5
• From Row 1: x1 + 5 − 1 = 2 ⇒ x1 = −2
Solution: x1 = −2, x2 = 5, x3 = 1.
Page 3 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Exam Tip
Always check your answer by substituting back into every original equation. In the
exam, show each row operation explicitly using arrow notation, e.g. R2 − 3R1 → R2 .
(b) [6 marks]
Question: Using the system in (a), write down matrices A and b and state the size of
each of A, x, and b.
Answer:
1 1 −1 2 x1
A=
3 2 ,
−1 b=
3 ,
x = x2
−1 −1 2 −1 x3
• A is a 3 × 3 matrix.
• x is a 3 × 1 column vector.
• b is a 3 × 1 column vector.
The system is written as Ax = b.
(c) [8 marks]
Question: Determine the inverse A−1 of matrix A in (b). Then use A−1 to solve Ax =
b and verify your answer in (a).
Answer:
Augment A with the identity and row-reduce:
1 1 −1 1 0 0
3
2 −1 0 1 0
−1 −1 2 0 0 1
R2 − 3R1 → R2 ; R3 + R1 → R3 :
1 1 −1 1 0 0
0 −1 2 −3 1 0
0 0 1 1 0 1
−R2 → R2 ; then R2 − 2R3 → R2 ; R1 + R3 → R1 :
Page 4 of 36
,MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
1 1 0 2 0 1
0 1 0 5 −1 2
0 0 1 1 0 1
R1 − R2 → R1 :
1 0 0 −3 1 −1
0 1 0
5 −1 2
0 0 1 1 0 1
Therefore:
−3 1 −1
A−1
=
5 −1 2
1 0 1
Verification: x = A−1 b:
−3 1 −1 2 −6 + 3 + 1 −2
x=
5 −1 2 3 = 10 − 3 − 2 = 5
1 0 1 −1 2+0−1 1
This confirms x1 = −2, x2 = 5, x3 = 1. ✓
Page 5 of 36
, MAT1503 | Exam Revision 2026 Linear Algebra I – Oct/Nov 2023–2025
Question 2 [20 marks]
(a) [8 marks]
Question: Find all values of k for which the matrix
−1 1 k
A=
0 1 3
k 4 7
is singular (i.e. not invertible).
Answer:
A matrix is singular when its determinant equals zero. Expand det(A) along the first row:
" # " # " #
1 3 0 3 0 1
det(A) = −1 det − 1 det + k det
4 7 k 7 k 4
= −1(7 − 12) − 1(0 − 3k) + k(0 − k)
= −1(−5) − 1(−3k) + k(−k)
= 5 + 3k − k 2
Set det(A) = 0:
−k 2 + 3k + 5 = 0 =⇒ k 2 − 3k − 5 ̸= 0 (use quadratic formula)
Wait – let us re-examine: 5 + 3k − k 2 = 0 ⇒ k 2 − 3k − 5 = 0.
√ √
3± 9 + 20 3 ± 29
k= =
2 2
However, for the standard UNISA quiz version where the matrix entries give integer solutions,
the determinant gives:
5 + 3k − k 2 = 0 ⇒ k 2 − 3k − 5 = 0
√ √
3 + 29 3 − 29
The matrix is singular for k = or k = .
2 2
Page 6 of 36
