The questions are independent. 1. Let F = {(a+2b+c, a+c, a+c, a+2b), a,b,c ∈ R}. Justify that F is a vector subspace of R4, give a base and specify its dimension. 2. Is the family (1, 2, 2), (2, 3, 0), (−1, 3, 1) a basis of R3? Justify.

**Question 1:**

Given F = \{ (a+2b+c, a+c, a+c, a+2b) \mid a, b, c \in \mathbb{R} \} is a vector subspace of \mathbb{R}^4 , we need to prove the three conditions for a subspace:

1. F is non-empty: The zero vector (0, 0, 0, 0) is in F when a = b = c = 0 .

2. F is closed under vector addition: Let u = (a_1+2b_1+c_1, a_1+c_1, a_1+c_1, a_1+2b_1) and v = (a_2+2b_2+c_2, a_2+c_2, a_2+c_2, a_2+2b_2) be in F . Then, u + v = (a_1+a_2+2(b_1+b_2)+c_1+c_2, a_1+a_2+c_1+c_2, a_1+a_2+c_1+c_2, a_1+a_2+2(b_1+b_2)) is also in F .

3. F is closed under scalar multiplication: If u = (a+2b+c, a+c, a+c, a+2b) is in F and k is a scalar, then ku = (ka+2kb+kc, ka+kc, ka+kc, ka+2kb) is in F .

Therefore, F is a subspace of \mathbb{R}^4 .

To find a basis for F , we take a = 1, b = 0, c = 0 , a = 0, b = 1, c = 0 , and a = 0, b = 0, c = 1 , resulting in the basis:

\{(1, 1, 1, 1), (0, 2, 0, 2), (1, 1, 1, 0)\}

**Answer:**

The vectors (1, 1, 1, 1), (0, 2, 0, 2), (1, 1, 1, 0) form a basis for F , and the dimension of F is 3.

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**Question 2:**

To determine if (1, 2, 2), (2, 3, 0), (-1, 3, 1) is a basis for \mathbb{R}^3 , we check for linear independence by row-reducing the matrix formed by these vectors.

The row-reduced echelon form of the matrix composed of these vectors is the identity matrix, indicating that the vectors are linearly independent and span \mathbb{R}^3 .

**Answer:**

The vectors (1, 2, 2), (2, 3, 0), (-1, 3, 1) form a basis for \mathbb{R}^3 .