APM3711 Assignment 3 solutions 2026

APM3711 Assignment 3 solutions 2026
UNISA

,Question 1

Given
4 1 1 1
(0)
𝐴 = [1 3 0 ] , 𝑥 = [1 ]
1 0 2 0


The Power Method consists of:

1. Compute

𝑦 (𝑘) = 𝐴𝑥 (𝑘)


2. Find the largest absolute component of 𝑦 (𝑘) :
(𝑘)
𝑚 = max⁡ ∣ 𝑦𝑖 ∣


3. Normalize:

(𝑘 +1)
𝑦 (𝑘)
𝑥 =
𝑚




(a) First Iteration

Multiply

𝑦 (0) = 𝐴𝑥 (0)
4 1 1 1
= [1 3 0 ] [ 1 ]
1 0 2 0


Calculate each entry:

First row

4(1) + 1(1) + 1(0) = 5


Second row

1(1) + 3(1) + 0(0) = 4


Third row

, 1(1) + 0(1) + 2(0) = 1


Therefore,
5
𝑦 (0) = [4]
1




Normalize

Largest component

𝑚1 = 5


Hence

1 5 1
(1)
𝑥 = [4] = [0.8000]
5
1 0.2000




First iteration result

1
𝑥 (1) = [0.8000] , 𝜆1 ≈ 5
0.2000




Second Iteration

Multiply

𝑦 (1) = 𝐴𝑥 (1)
4 1 1 1
= [1 3 0] [0.8]
1 0 2 0.2


First row

4(1) + 1(0.8) + 1(0.2) = 5.0000


Second row