UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
MAT3700 ASSIGNMENT 03
Applied Mathematics — Semester 1, 2026
⋄
Module Code: MAT3700
Module Name: Applied Mathematics
Assignment No.: 03
Due Date: [Due Date]
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for MAT3700
at the University of South Africa.
, UNISA | MAT3700 Assignment 03 — 2026
Question 1 (5 marks)
2 4
Question: If B = , find the eigenvalues of B.
1 5
Step 1: Apply the characteristic equation det(B − λI) = 0.
Step 2: Subtract λ from the diagonal entries.
2−λ 4
B − λI =
1 5−λ
Step 3: Compute the determinant.
(2 − λ)(5 − λ) − (4)(1) = 0
Step 4: Expand (2 − λ)(5 − λ).
= 10 − 2λ − 5λ + λ2 = λ2 − 7λ + 10
Therefore:
λ2 − 7λ + 10 − 4 = 0 =⇒ λ2 − 7λ + 6 = 0
Step 5: Factorise. Two numbers that multiply to 6 and add to −7 are −6 and −1.
(λ − 6)(λ − 1) = 0
Step 6: Solve.
λ − 6 = 0 ⇒ λ1 = 6 λ − 1 = 0 ⇒ λ2 = 1
Final Answer
λ1 = 6 and λ2 = 1
Page 1 of 9
College of Science, Engineering and Technology
⋄
MAT3700 ASSIGNMENT 03
Applied Mathematics — Semester 1, 2026
⋄
Module Code: MAT3700
Module Name: Applied Mathematics
Assignment No.: 03
Due Date: [Due Date]
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for MAT3700
at the University of South Africa.
, UNISA | MAT3700 Assignment 03 — 2026
Question 1 (5 marks)
2 4
Question: If B = , find the eigenvalues of B.
1 5
Step 1: Apply the characteristic equation det(B − λI) = 0.
Step 2: Subtract λ from the diagonal entries.
2−λ 4
B − λI =
1 5−λ
Step 3: Compute the determinant.
(2 − λ)(5 − λ) − (4)(1) = 0
Step 4: Expand (2 − λ)(5 − λ).
= 10 − 2λ − 5λ + λ2 = λ2 − 7λ + 10
Therefore:
λ2 − 7λ + 10 − 4 = 0 =⇒ λ2 − 7λ + 6 = 0
Step 5: Factorise. Two numbers that multiply to 6 and add to −7 are −6 and −1.
(λ − 6)(λ − 1) = 0
Step 6: Solve.
λ − 6 = 0 ⇒ λ1 = 6 λ − 1 = 0 ⇒ λ2 = 1
Final Answer
λ1 = 6 and λ2 = 1
Page 1 of 9
