MAT1503 ASSIGNMENT 4 2023

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

,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