UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM1513: Applied Linear Algebra
Assignment 02 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM1513
Module Code:
Applied Linear Algebra
Module Name:
Assignment 02
Assignment:
8 June 2026
Due Date:
Submitted in partial fulfilment of the requirements for APM1513 — UNISA 2026
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Question 1: Matrix Operations on a 3x3 Complex Matrix
Question. Suppose that A is a 3 × 3 matrix with real numbers, complex numbers and functions
as elements. Compute (i) inverse of A, (ii) transpose of A−1 , (iii) trace of A−1 , (iv) determi-
nant of the transpose of A−1 .
1.1 Setting Up the Matrix
Let √
8+i e9−1 e10+i 2
A=
5 − 3i 2 + 7i 1
cos π 1 1
Step 1: Simplify known values.
e9−1 = e8 , cos π = −1
Therefore A becomes:
√
8+i e8 e10+i 2
A=
5 − 3i 2 + 7i 1
−1 1 1
1.2 Part (i): Inverse of A
Step 2: Apply the inverse formula.
1
A−1 = adj(A)
det(A)
Step 3: Compute det(A) by expansion along the first row.
√
det(A) = (8 + i) M11 − e8 M12 + e10+i 2
M13
where M11 , M12 , M13 are the minors.
Page 2 of 21
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 4: Compute the 2 × 2 minors.
2 + 7i 1
M11 = = (2 + 7i)(1) − (1)(1) = 2 + 7i − 1 = 1 + 7i
1 1
5 − 3i 1
M12 = = (5 − 3i)(1) − (1)(−1) = 5 − 3i + 1 = 6 − 3i
−1 1
5 − 3i 2 + 7i
M13 = = (5 − 3i)(1) − (2 + 7i)(−1) = 5 − 3i + 2 + 7i = 7 + 4i
−1 1
Step 5: Substitute back.
√
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(1) + 8(7i) + i(1) + i(7i) = 8 + 56i + i + 7i2
Since i2 = −1:
= 8 + 57i − 7 = 1 + 57i
Therefore:
√
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. The cofactors Cij = (−1)i+j Mij are
computed for all nine entries.
Row 1 cofactors (already found above):
C11 = +M11 = 1 + 7i, C12 = −M12 = −(6 − 3i) = −6 + 3i, C13 = +M13 = 7 + 4i
Page 3 of 21
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Row 2 cofactors:
√
e8 e10+i 2 √ √
C21 = − = −(e8 − e10+i 2
) = e10+i 2
− e8
1 1
√
8 + i e10+i 2 √ √
C22 = + = (8 + i)(1) − e10+i 2
(−1) = 8 + i + e10+i 2
−1 1
8 + i e8
= − (8 + i)(1) − e8 (−1) = −(8 + i + e8 )
C23 = −
−1 1
Row 3 cofactors:
√
e8 e10+i 2 √ √
C31 = + = e8 (1) − e10+i 2
(2 + 7i) = e8 − (2 + 7i)e10+i 2
2 + 7i 1
√
8+i e10+i 2 √ √
= − (8 + i)(1) − e10+i 2 (5 − 3i) = −(8 + i) + (5 − 3i)e10+i 2
C32 = −
5 − 3i 1
8+i e8
C33 = + = (8 + i)(2 + 7i) − e8 (5 − 3i)
5 − 3i 2 + 7i
Expanding (8 + i)(2 + 7i):
= 16 + 56i + 2i + 7i2 = 16 + 58i − 7 = 9 + 58i
Therefore C33 = (9 + 58i) − e8 (5 − 3i).
Step 8: Form the adjugate (transpose of cofactor matrix).
C11 C21 C31
adj(A) =
C12 C22 C32
C13 C23 C33
√ √
1 + 7i − e10+i 2
− (2 + e8 e8 7i)e10+i 2
√ √
= −6 + 3i 8 + i + e10+i 2 −(8 + i) + (5 − 3i)e10+i 2
7 + 4i −(8 + i + e8 ) (9 + 58i) − e8 (5 − 3i)
Page 4 of 21
, UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 9: Write the inverse.
√
Let D = (1 + 57i) − e8 (6 − 3i) + e10+i 2 (7 + 4i).
1
A−1 = adj(A)
D
. Critical Consideration
̸ 0, the inverse exists. If D = 0
The determinant D is a complex number. Provided D =
the matrix is singular and no inverse exists.
1.3 Part (ii): Transpose of A−1
Using the identity for invertible matrices:
(A−1 )T = (AT )−1
Step 10: Transpose each element of A−1 across the main diagonal. That is, swap row i, col-
umn j with row j, column i.
1 + 7i −6 + 3i 7 + 4i
1 √ √
(A−1 )T = e10+i 2 − e8 8 + i + e10+i 2 −(8 + i + e8 )
D
√ √
e8 − (2 + 7i)e10+i 2 −(8 + i) + (5 − 3i)e10+i 2 (9 + 58i) − e8 (5 − 3i)
1.4 Part (iii): Trace of A−1
The trace is the sum of the main diagonal entries.
Step 11:
1
Tr(A−1 ) =
C11 + C22 + C33
D
1 √
(1 + 7i) + (8 + i + e10+i 2 ) + (9 + 58i − e8 (5 − 3i))
=
D
Page 5 of 21
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM1513: Applied Linear Algebra
Assignment 02 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM1513
Module Code:
Applied Linear Algebra
Module Name:
Assignment 02
Assignment:
8 June 2026
Due Date:
Submitted in partial fulfilment of the requirements for APM1513 — UNISA 2026
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Question 1: Matrix Operations on a 3x3 Complex Matrix
Question. Suppose that A is a 3 × 3 matrix with real numbers, complex numbers and functions
as elements. Compute (i) inverse of A, (ii) transpose of A−1 , (iii) trace of A−1 , (iv) determi-
nant of the transpose of A−1 .
1.1 Setting Up the Matrix
Let √
8+i e9−1 e10+i 2
A=
5 − 3i 2 + 7i 1
cos π 1 1
Step 1: Simplify known values.
e9−1 = e8 , cos π = −1
Therefore A becomes:
√
8+i e8 e10+i 2
A=
5 − 3i 2 + 7i 1
−1 1 1
1.2 Part (i): Inverse of A
Step 2: Apply the inverse formula.
1
A−1 = adj(A)
det(A)
Step 3: Compute det(A) by expansion along the first row.
√
det(A) = (8 + i) M11 − e8 M12 + e10+i 2
M13
where M11 , M12 , M13 are the minors.
Page 2 of 21
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 4: Compute the 2 × 2 minors.
2 + 7i 1
M11 = = (2 + 7i)(1) − (1)(1) = 2 + 7i − 1 = 1 + 7i
1 1
5 − 3i 1
M12 = = (5 − 3i)(1) − (1)(−1) = 5 − 3i + 1 = 6 − 3i
−1 1
5 − 3i 2 + 7i
M13 = = (5 − 3i)(1) − (2 + 7i)(−1) = 5 − 3i + 2 + 7i = 7 + 4i
−1 1
Step 5: Substitute back.
√
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(1) + 8(7i) + i(1) + i(7i) = 8 + 56i + i + 7i2
Since i2 = −1:
= 8 + 57i − 7 = 1 + 57i
Therefore:
√
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. The cofactors Cij = (−1)i+j Mij are
computed for all nine entries.
Row 1 cofactors (already found above):
C11 = +M11 = 1 + 7i, C12 = −M12 = −(6 − 3i) = −6 + 3i, C13 = +M13 = 7 + 4i
Page 3 of 21
,UNISA | APM1513 Applied Linear Algebra — Assignment 02
Row 2 cofactors:
√
e8 e10+i 2 √ √
C21 = − = −(e8 − e10+i 2
) = e10+i 2
− e8
1 1
√
8 + i e10+i 2 √ √
C22 = + = (8 + i)(1) − e10+i 2
(−1) = 8 + i + e10+i 2
−1 1
8 + i e8
= − (8 + i)(1) − e8 (−1) = −(8 + i + e8 )
C23 = −
−1 1
Row 3 cofactors:
√
e8 e10+i 2 √ √
C31 = + = e8 (1) − e10+i 2
(2 + 7i) = e8 − (2 + 7i)e10+i 2
2 + 7i 1
√
8+i e10+i 2 √ √
= − (8 + i)(1) − e10+i 2 (5 − 3i) = −(8 + i) + (5 − 3i)e10+i 2
C32 = −
5 − 3i 1
8+i e8
C33 = + = (8 + i)(2 + 7i) − e8 (5 − 3i)
5 − 3i 2 + 7i
Expanding (8 + i)(2 + 7i):
= 16 + 56i + 2i + 7i2 = 16 + 58i − 7 = 9 + 58i
Therefore C33 = (9 + 58i) − e8 (5 − 3i).
Step 8: Form the adjugate (transpose of cofactor matrix).
C11 C21 C31
adj(A) =
C12 C22 C32
C13 C23 C33
√ √
1 + 7i − e10+i 2
− (2 + e8 e8 7i)e10+i 2
√ √
= −6 + 3i 8 + i + e10+i 2 −(8 + i) + (5 − 3i)e10+i 2
7 + 4i −(8 + i + e8 ) (9 + 58i) − e8 (5 − 3i)
Page 4 of 21
, UNISA | APM1513 Applied Linear Algebra — Assignment 02
Step 9: Write the inverse.
√
Let D = (1 + 57i) − e8 (6 − 3i) + e10+i 2 (7 + 4i).
1
A−1 = adj(A)
D
. Critical Consideration
̸ 0, the inverse exists. If D = 0
The determinant D is a complex number. Provided D =
the matrix is singular and no inverse exists.
1.3 Part (ii): Transpose of A−1
Using the identity for invertible matrices:
(A−1 )T = (AT )−1
Step 10: Transpose each element of A−1 across the main diagonal. That is, swap row i, col-
umn j with row j, column i.
1 + 7i −6 + 3i 7 + 4i
1 √ √
(A−1 )T = e10+i 2 − e8 8 + i + e10+i 2 −(8 + i + e8 )
D
√ √
e8 − (2 + 7i)e10+i 2 −(8 + i) + (5 − 3i)e10+i 2 (9 + 58i) − e8 (5 − 3i)
1.4 Part (iii): Trace of A−1
The trace is the sum of the main diagonal entries.
Step 11:
1
Tr(A−1 ) =
C11 + C22 + C33
D
1 √
(1 + 7i) + (8 + i + e10+i 2 ) + (9 + 58i − e8 (5 − 3i))
=
D
Page 5 of 21
