UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Applied Linear Algebra — APM1513 — Semester 1, 2026
⋄
Module Code: APM1513
Module Name: Applied Linear Algebra
Assignment No.: 02
Due Date: Monday, 8 June 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for APM1513
at the University of South Africa.
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Question 1: Matrix Operations on a Complex and Function-Valued 3x3 Matrix
Question: Suppose that A is a 3 × 3 matrix with real numbers, complex numbers and functions
as elements. Compute (i) the inverse of A, (ii) the transpose of A−1 , (iii) the trace of A−1 ,
and (iv) the determinant of the transpose of A−1 .
Let √
8+i e9−1 e10+i 2
A = 5 − 3i 2 + 7i
1
cos π 1 1
Step 1: Simplify the Known Elements
Evaluate each simplified entry:
e9−1 = e8 , cos π = −1
Therefore: √
8+i e8 e10+i 2
A = 5 − 3i 2 + 7i
1
−1 1 1
1(i): Inverse of A
Step 2: Recall the Inverse Formula
1
A−1 = adj(A)
det(A)
Step 3: Compute det(A) by Cofactor Expansion Along Row 1
2 + 7i 1 5 − 3i 1 √ 5 − 3i 2 + 7i
det(A) = (8 + i) − e8 + e10+i 2
1 1 −1 1 −1 1
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, UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 4: Evaluate Each 2 × 2 Minor
Minor M11 :
(2 + 7i)(1) − (1)(1) = 2 + 7i − 1 = 1 + 7i
Minor M12 :
(5 − 3i)(1) − (1)(−1) = 5 − 3i + 1 = 6 − 3i
Minor M13 :
(5 − 3i)(1) − (2 + 7i)(−1) = 5 − 3i + 2 + 7i = 7 + 4i
Step 5: Substitute Back into the Expansion
√
det(A) = (8 + i)(1 + 7i) − e8 (6 − 3i) + e10+i 2
(7 + 4i)
Step 6: Expand (8 + i)(1 + 7i)
(8 + i)(1 + 7i) = 8 + 56i + i + 7i2 = 8 + 57i + 7(−1) = 1 + 57i
Implying that:
√
det(A) = (1 + 57i) − e8 (6 − 3i) + e10+i 2
(7 + 4i)
Step 7: Compute adj(A)
The adjugate is the transpose of the cofactor matrix. Compute each cofactor Cij = (−1)i+j Mij
and arrange them, then transpose. The nine cofactors are:
Row 1 cofactors:
2 + 7i 1
C11 = + = 1 + 7i
1 1
5 − 3i 1
C12 = − = −(6 − 3i) = −6 + 3i
−1 1
5 − 3i 2 + 7i
C13 = + = 7 + 4i
−1 1
Page 2 of 21
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Applied Linear Algebra — APM1513 — Semester 1, 2026
⋄
Module Code: APM1513
Module Name: Applied Linear Algebra
Assignment No.: 02
Due Date: Monday, 8 June 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for APM1513
at the University of South Africa.
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Question 1: Matrix Operations on a Complex and Function-Valued 3x3 Matrix
Question: Suppose that A is a 3 × 3 matrix with real numbers, complex numbers and functions
as elements. Compute (i) the inverse of A, (ii) the transpose of A−1 , (iii) the trace of A−1 ,
and (iv) the determinant of the transpose of A−1 .
Let √
8+i e9−1 e10+i 2
A = 5 − 3i 2 + 7i
1
cos π 1 1
Step 1: Simplify the Known Elements
Evaluate each simplified entry:
e9−1 = e8 , cos π = −1
Therefore: √
8+i e8 e10+i 2
A = 5 − 3i 2 + 7i
1
−1 1 1
1(i): Inverse of A
Step 2: Recall the Inverse Formula
1
A−1 = adj(A)
det(A)
Step 3: Compute det(A) by Cofactor Expansion Along Row 1
2 + 7i 1 5 − 3i 1 √ 5 − 3i 2 + 7i
det(A) = (8 + i) − e8 + e10+i 2
1 1 −1 1 −1 1
Page 1 of 21
, UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 4: Evaluate Each 2 × 2 Minor
Minor M11 :
(2 + 7i)(1) − (1)(1) = 2 + 7i − 1 = 1 + 7i
Minor M12 :
(5 − 3i)(1) − (1)(−1) = 5 − 3i + 1 = 6 − 3i
Minor M13 :
(5 − 3i)(1) − (2 + 7i)(−1) = 5 − 3i + 2 + 7i = 7 + 4i
Step 5: Substitute Back into the Expansion
√
det(A) = (8 + i)(1 + 7i) − e8 (6 − 3i) + e10+i 2
(7 + 4i)
Step 6: Expand (8 + i)(1 + 7i)
(8 + i)(1 + 7i) = 8 + 56i + i + 7i2 = 8 + 57i + 7(−1) = 1 + 57i
Implying that:
√
det(A) = (1 + 57i) − e8 (6 − 3i) + e10+i 2
(7 + 4i)
Step 7: Compute adj(A)
The adjugate is the transpose of the cofactor matrix. Compute each cofactor Cij = (−1)i+j Mij
and arrange them, then transpose. The nine cofactors are:
Row 1 cofactors:
2 + 7i 1
C11 = + = 1 + 7i
1 1
5 − 3i 1
C12 = − = −(6 − 3i) = −6 + 3i
−1 1
5 − 3i 2 + 7i
C13 = + = 7 + 4i
−1 1
Page 2 of 21
