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MAT1503 ASSIGNMENT 4 2023
- Course
- MAT1503 (MAT1503)
- Institution
- University Of South Africa (Unisa)
This document contains MAT1503 ASSIGNMENT 4 2023 solutions. Clear step by step calculations are provided.
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MAT1503ASSIGNMENT 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
,ii).
𝑎33 = (𝑎33 )2
1 1 2
𝑘 4 = (𝑘 4 )
1 1
𝑘4 = 𝑘2
1 4 1 4
(𝑘 4 ) = (𝑘 2 )
𝑘 = 𝑘2
𝑘2 − 𝑘 = 0
𝑘(𝑘 − 1) = 0
𝑘 = 0 𝑜𝑟 𝑘 = 1
1.2).
1 0 0
[ 0 1 0]
0 0 1
↓ 𝑅1 ↔ 𝑅3
0 0 1
[0 1 0] = 𝐸2
1 0 0
0 0 1 1 0 0
[𝐸2 |𝐼3 ] = [0 1 0| 0 1 0]
1 0 0 0 0 1
↓ 𝑅1 ↔ 𝑅3
1 0 0 0 0 1
[0 1 0| 0 1 0] = [𝐼3 |𝐸2 −1 ]
0 0 1 1 0 0
0 0 1
𝐸2 −1 = [0 1 0]
1 0 0
𝐴 = 𝐸2 𝐶
𝐸2 −1 𝐴 = 𝐸2 −1 𝐸2 𝐶
, 𝐸2 −1 𝐴 = 𝐶
𝐶 = 𝐸2 −1 𝐴
0 0 1 2 −1 1
= [0 1 0] [3 1 −1]
1
1 0 0 1 −3 𝑘 4
1
0+0+1 0+0−3 0 + 0 + 𝑘4
= [0 + 3 + 0 0+1+0 0−1+0]
2+0+0 −1 + 0 + 0 1+0+0
1
1 −3 𝑘 4
= [3 10 −1]
2 −1 1
QUESTION 2
Solution:
2.1).
⃗ ∙ 𝑣 = 〈1,3, −2〉 ∙ 〈−5,3,2〉
𝑢
= (1)(−5) + (3)(3) + (−2)(2)
= −5 + 9 − 4
=0
𝑆𝑖𝑛𝑐𝑒 𝑢
⃗ ∙ 𝑣 = 0 , 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑢
⃗ 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙.
2.2).
⃗ ∙ 𝑣 = 〈1, −2,4〉 ∙ 〈5,3,7〉
𝑢
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