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Question:
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Which are not subspaces of R3? Select all that apply.
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{(zy2)T € R3|z + y+ z=0}
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nd {zy2)’ eR|z+y+2z=1}
This is AI generated Response.
Asked by: Pagal
Created at: 2025-09-05 04:40:39
Asked by: Pagal
Created at: 2025-09-05 04:40:39
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Let's analyze each option as to whether it is a **subspace** of \( \mathbb{R}^3 \):
---
### Step 1: Check the subspace criteria
A set is a **subspace** if:
1. It contains the zero vector.
2. It is closed under addition.
3. It is closed under scalar multiplication.
Let's go through each option:
---
#### 1. **Null space of a matrix** (Null \( \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 9 \\ 6 & 7 & 13 \end{pmatrix} \)):
- The null space of any matrix **is always a subspace**.
- **Explanation**: It contains the zero vector and is closed under addition and scalar multiplication.
---
#### 2. **\(\{(x,y,z)^T \in \mathbb{R}^3 \mid x + y + z = 0\}\)**
- This is a plane through the origin.
- **Explanation**: It contains the zero vector, and is closed under addition and scalar multiplication.
---
#### 3. **\(\{(x,y,z)^T \in \mathbb{R}^3 \mid z \geq 0\}\)**
- This is the set of all vectors in \( \mathbb{R}^3 \) where \( z \) is non-negative.
- **Explanation**: **Not a subspace** because it is not closed under scalar multiplication (e.g., multiplying a vector with \( z > 0 \) by \(-1\) gives \( z < 0 \), which is not in the set).
---
#### 4. **Span** \( \left(\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix}\right) \)
- Any span of vectors is a subspace.
- **Explanation**: Always contains the zero vector, and is closed under addition and scalar multiplication.
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#### 5. **The solution set to \( \begin{pmatrix}0 & 1 & 1 \\ 2 & 3 & 5 \\ 1 & 2 & 3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\5\\3\end{pmatrix} \)**
- This is a **solution set to a non-homogeneous linear system**.
- **Explanation**: **Not a subspace** because it does not contain the zero vector (unless the right-hand side is zero).
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#### 6. **The solution set to \( \begin{pmatrix}0 & 1 & 1 \\ 2 & 3 & 5 \\ 1 & 2 & 3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} \)**
- This is the solution set to a **homogeneous linear system**.
- **Explanation**: Always a subspace by definition.
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#### 7. **\(\{(x,y,z)^T \in \mathbb{R}^3 \mid x + y + z = 1\}\)**
- This is a plane **not through the origin**.
- **Explanation**: **Not a subspace** because it does not contain the zero vector.
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### Step 2: Select all that are **not** subspaces
Based on the above:
- **\(\{(x,y,z)^T \in \mathbb{R}^3 \mid z \geq 0\}\)**
- **The solution set to a non-homogeneous system**
- **\(\{(x,y,z)^T \in \mathbb{R}^3 \mid x + y + z = 1\}\)**
These three are **not** subspaces.
---
## **Final Answer:**
The following are **not** subspaces of \( \mathbb{R}^3 \):
1. \(\{(x,y,z)^T \in \mathbb{R}^3 \mid z \geq 0\}\)
2. The solution set to \( \begin{pmatrix}0 & 1 & 1 \\ 2 & 3 & 5 \\ 1 & 2 & 3\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\5\\3\end{pmatrix} \)
3. \(\{(x,y,z)^T \in \mathbb{R}^3 \mid x + y + z = 1\}\)
**Explanation:**
- They either do not contain the zero vector or are not closed under scalar multiplication/addition.
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