MAT1503
ASSIGNMENT 4
2023
, QUESTION 1
Solution:
1.1).
i).
1 0 0
[ 0 1 0]
0 0 1
↓ 𝑅2 ← 𝑅2 − 2𝑅1
1 0 0
[−2 1 0] = 𝐸1
0 0 1
𝐵 = 𝐸1 𝐴
1 0 0 2 −1 1
= [−2 1 0] [3 1 −1]
1
0 0 1 1 −3 𝑘 4
2 + 0 + 0 −1 + 0 + 0 1 + 0 + 0
= [−4 + 3 + 0 2 + 1 + 0 −2 − 1 + 0]
1
0+0+1 0 + 0 − 3 0 + 0 + 𝑘4
2 −1 1
= [−1 3 −3]
1
1 −3 𝑘4
ASSIGNMENT 4
2023
, QUESTION 1
Solution:
1.1).
i).
1 0 0
[ 0 1 0]
0 0 1
↓ 𝑅2 ← 𝑅2 − 2𝑅1
1 0 0
[−2 1 0] = 𝐸1
0 0 1
𝐵 = 𝐸1 𝐴
1 0 0 2 −1 1
= [−2 1 0] [3 1 −1]
1
0 0 1 1 −3 𝑘 4
2 + 0 + 0 −1 + 0 + 0 1 + 0 + 0
= [−4 + 3 + 0 2 + 1 + 0 −2 − 1 + 0]
1
0+0+1 0 + 0 − 3 0 + 0 + 𝑘4
2 −1 1
= [−1 3 −3]
1
1 −3 𝑘4