Let's analyze each part step-by-step as requested: --- ### 1. The set of all vectors \((x, y, z)\) in \(\mathbb{R}^3\) such that \(x + 2y + 3z = 0\) **Step 1: Check the three subspace conditions** 1. **Zero vector inclusion:** Plug \(x=0, y=0, z=0\) into the equation: \(0 + 2(0) + 3(0) = 0\), so **the zero vector is included**. 2. **Closed under addition:** Let \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) be in the set, i.e., \(x_1+2y_1+3z_1=0\) and \(x_2+2y_2+3z_2=0\). Their sum: \((x_1+x_2, y_1+y_2, z_1+z_2)\) Plug into the equation: \((x_1+x_2) + 2(y_1+y_2) + 3(z_1+z_2) = (x_1+2y_1+3z_1) + (x_2+2y_2+3z_2) = 0+0=0\) **Closed under addition.** **Explanation:** So far, the set contains the zero vector and is closed under addition. **Step 2: Closed under scalar multiplication** Let \(c\) be any scalar, and \((x, y, z)\) in the set. Then \(x+2y+3z=0\). Consider \(c(x, y, z) = (cx, cy, cz)\). Check: \(cx + 2cy + 3cz = c(x+2y+3z) = c \cdot 0 = 0\). **Closed under scalar multiplication.** **Explanation:** All three subspace conditions are satisfied. **Final Answer:** **This set is a subspace of \(\mathbb{R}^3\).** --- ### 2. The set of all vectors \((x, y, z)\) in \(\mathbb{R}^3\) such that \(x + 2y + 3z = 5\) **Step 1: Check for zero vector inclusion** Plug \(x=0, y=0, z=0\) into the equation: \(0 + 2(0) + 3(0) = 0 \neq 5\). So the zero vector is **not** in the set. **Explanation:** A subspace must contain the zero vector. This set does not. **Step 2: Counterexample for closure under addition** Let \((5, 0, 0)\) and \((0, 2.5, 0)\) both satisfy \(x+2y+3z=5\). Their sum: \((5, 2.5, 0)\), check: \(5 + 2 \times 2.5 + 3 \times 0 = 5 + 5 = 10 \neq 5\). So the set is **not closed under addition**. **Explanation:** The set fails to be a subspace because it does not contain the zero vector and is not closed under addition. **Final Answer:** **This set is NOT a subspace of \(\mathbb{R}^3\).** --- ### 3. The set of all continuous functions \(f\) on \([0, 1]\) such that \(f(0) = f(1)\) **Step 1: Zero vector and closure under addition** - **Zero function check:** \(f(x) = 0\) for all \(x\) is continuous and \(f(0) = f(1) = 0\), so included. - **Addition:** If \(f(0) = f(1)\) and \(g(0) = g(1)\), then \((f+g)(0) = f(0)+g(0)\), \((f+g)(1) = f(1)+g(1)\). So, \((f+g)(0) = (f+g)(1)\). **Explanation:** The set contains the zero function and is closed under addition. **Step 2: Closed under scalar multiplication** If \(f(0) = f(1)\), then for any scalar \(c\), \((cf)(0) = c f(0)\), \((cf)(1) = c f(1)\), so \((cf)(0) = (cf)(1)\). **Explanation:** All three subspace conditions are satisfied. **Final Answer:** **This set is a subspace of the vector space of continuous functions on \([0, 1]\).** --- ### 4. The set of all differentiable functions \(f\) on \(\mathbb{R}\) such that \(f'(x) = 3f(x)\) for all \(x\) in \(\mathbb{R}\) **Step 1: Zero function and closure under addition** - **Zero function:** \(f(x) = 0\) for all \(x\), \(f'(x) = 0 = 3 \cdot 0\), so included. - **Addition:** If \(f'(x) = 3f(x)\) and \(g'(x) = 3g(x)\), then \((f+g)'(x) = f'(x) + g'(x) = 3f(x) + 3g(x) = 3(f(x) + g(x))\). **Explanation:** Zero function is included and the set is closed under addition. **Step 2: Closed under scalar multiplication** If \(f'(x) = 3f(x)\), then for any scalar \(c\), \((cf)'(x) = c f'(x) = c \cdot 3 f(x) = 3 (c f(x))\). **Explanation:** All three subspace conditions are satisfied. **Final Answer:** **This set is a subspace of the vector space of differentiable functions on \(\mathbb{R}\).** --- ## **Summary Table** | Set Description | Subspace? | Reason (short) | |-----------------|-----------|---------------| | \(x + 2y + 3z = 0\) | Yes | All subspace conditions satisfied | | \(x + 2y + 3z = 5\) | No | Zero vector not included, not closed under addition | | \(f\) continuous, \(f(0) = f(1)\) | Yes | All subspace conditions satisfied | | \(f\) differentiable, \(f' = 3f\) | Yes | All subspace conditions satisfied |