MAT2611 ASSIGNMENT 2 2023

MAT2611
ASSIGNMENT 2
2023

,Solution:



a).



For 𝐮 = (1, −2) = (u1 , u2 ) means u1 = 1 and u2 = −2



And



For 𝐯 = (3,0) = (v1 , v2 ) means v1 = 3 and v2 = 0



So



𝐮 + 𝐯 = (u1 + v1 − 1, u2 + v2 − 1)
= (1 + 3 − 1, −2 + 0 − 1)
= (3, −3)



∴k=4



k𝐮 = (ku1 , ku2 )

= (4(1), 4(−2))

, = (4, −8)
b).



If 𝐎 = (0,0)



𝐎 + (𝐮 + 𝐯) = (0,0) + (u1 + v1 − 1, u2 + v2 − 1)
= (0 + u1 + v1 − 1,0 + u2 + v2 − 1)
= (u1 + v1 − 1, u2 + v2 − 1)
= (u1 , u2 ) + (v1 , v2 ) + (−1, −1)
= (𝐮 + 𝐯) + (−1, −1)



Since 𝐎 + (𝐮 + 𝐯) ≠ 𝐮 + 𝐯 , then (0,0) ≠ 𝟎



c).



If 𝐎 = (1,1)



𝐎 + (𝐮 + 𝐯) = (1,1) + (u1 + v1 − 1, u2 + v2 − 1)
= (1 + u1 + v1 − 1,1 + u2 + v2 − 1)
= (u1 + v1 , u2 + v2 )
= (u1 , u2 ) + (v1 , v2 )
=𝐮+𝐯



Since 𝐎 + (𝐮 + 𝐯) = 𝐮 + 𝐯 , then (1,1) = 𝟎



d).



Assuming for each 𝐮 in 𝐕 there exist − 𝐮 = (1 − u1 , 1 − u2 ) ∈ V for then u1 , u2 ∈ 𝐅



𝐮 + (−𝐮) = O
𝐮 + (−𝐮) = (1,1)