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MAT2611 Assignment 2 full solutions 2026
All questions are solved full all steps and
calculations are shown fully and logically
,Problem 1
Determine whether 𝑈and 𝑉are subspaces of ℝ4 .
(a) Subspace 𝑼
Given
𝑈 = {(𝑥, 𝑦, 𝑧, 𝑢) ∈ ℝ4 : 𝑥 + 𝑦 − 𝑧𝑢 = 0}
To be a subspace, the set must satisfy:
1. Contains the zero vector
2. Closed under addition
3. Closed under scalar multiplication
Step 1: Check zero vector
Zero vector:
(0, 0, 0, 0)
Substitute:
0 + 0 − (0)(0) = 0
True.
So the zero vector belongs to 𝑈.
Step 2: Check closure under addition
Take two vectors in 𝑈:
𝑎 = (𝑥1 , 𝑦1 , 𝑧1 , 𝑢1 )
and
, 𝑏 = (𝑥2 , 𝑦2 , 𝑧2 , 𝑢2 )
Since both are in 𝑈,
𝑥1 + 𝑦1 − 𝑧1 𝑢1 = 0
𝑥2 + 𝑦2 − 𝑧2 𝑢2 = 0
Add the vectors:
𝑎 + 𝑏 = (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2 , 𝑢1 + 𝑢2 )
Check the condition:
(𝑥1 + 𝑥2 ) + (𝑦1 + 𝑦2 ) − (𝑧1 + 𝑧2 )(𝑢1 + 𝑢2 )
Expand:
𝑥1 + 𝑥2 + 𝑦1 + 𝑦2 − 𝑧1 𝑢1 − 𝑧1 𝑢2 − 𝑧2 𝑢1 − 𝑧2 𝑢2
This is not generally equal to zero.
Hence 𝑈is not closed under addition.
Therefore:
𝑈 is not a subspace of ℝ4
(b) Subspace 𝑽
Given
𝑉 = {(𝑥, 𝑦, 𝑧, 𝑣) ∈ ℝ4 : 𝑥 − 𝑦 = 𝑧, 2𝑧 + 𝑣 = 0}
Rewrite equations:
𝑥−𝑦−𝑧 =0
2𝑧 + 𝑣 = 0
MAT2611 Assignment 2 full solutions 2026
All questions are solved full all steps and
calculations are shown fully and logically
,Problem 1
Determine whether 𝑈and 𝑉are subspaces of ℝ4 .
(a) Subspace 𝑼
Given
𝑈 = {(𝑥, 𝑦, 𝑧, 𝑢) ∈ ℝ4 : 𝑥 + 𝑦 − 𝑧𝑢 = 0}
To be a subspace, the set must satisfy:
1. Contains the zero vector
2. Closed under addition
3. Closed under scalar multiplication
Step 1: Check zero vector
Zero vector:
(0, 0, 0, 0)
Substitute:
0 + 0 − (0)(0) = 0
True.
So the zero vector belongs to 𝑈.
Step 2: Check closure under addition
Take two vectors in 𝑈:
𝑎 = (𝑥1 , 𝑦1 , 𝑧1 , 𝑢1 )
and
, 𝑏 = (𝑥2 , 𝑦2 , 𝑧2 , 𝑢2 )
Since both are in 𝑈,
𝑥1 + 𝑦1 − 𝑧1 𝑢1 = 0
𝑥2 + 𝑦2 − 𝑧2 𝑢2 = 0
Add the vectors:
𝑎 + 𝑏 = (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2 , 𝑢1 + 𝑢2 )
Check the condition:
(𝑥1 + 𝑥2 ) + (𝑦1 + 𝑦2 ) − (𝑧1 + 𝑧2 )(𝑢1 + 𝑢2 )
Expand:
𝑥1 + 𝑥2 + 𝑦1 + 𝑦2 − 𝑧1 𝑢1 − 𝑧1 𝑢2 − 𝑧2 𝑢1 − 𝑧2 𝑢2
This is not generally equal to zero.
Hence 𝑈is not closed under addition.
Therefore:
𝑈 is not a subspace of ℝ4
(b) Subspace 𝑽
Given
𝑉 = {(𝑥, 𝑦, 𝑧, 𝑣) ∈ ℝ4 : 𝑥 − 𝑦 = 𝑧, 2𝑧 + 𝑣 = 0}
Rewrite equations:
𝑥−𝑦−𝑧 =0
2𝑧 + 𝑣 = 0