This is an image containing vector space problems. Let’s solve them step by step:

Problem 1: Linear Combination in $\mathbb{R}^3$

Given:

• $\mathbf{v}_1 = (1, -1, 2)$,
• $\mathbf{v}_2 = (2, 1, 1)$,
• $\mathbf{v}_3 = (-1, 3, 1)$,
• $\mathbf{v}_4 = (-8, 7, 5)$.

Questions:

(a) Prove that $\mathbf{v}_3$ is not a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$.
(b) Prove that $\mathbf{v}_4$ is a linear combination of $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$.

Part (a): Linear Combination of $\mathbf{v}_1$ and $\mathbf{v}_2$ for $\mathbf{v}_3$

To check if $\mathbf{v}_3$ is a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$, we need to solve for $a, b \in \mathbb{R}$ in: $a\mathbf{v}_1 + b\mathbf{v}_2 = \mathbf{v}_3.$ This expands to: $a(1, -1, 2) + b(2, 1, 1) = (-1, 3, 1).$ Breaking into components:

1. $a + 2b = -1$,
2. $-a + b = 3$,
3. $2a + b = 1$.

Now solve this system of equations. Substituting $a$ and $b$ from equations $1$ and $2$ into equation $3$ will show if a solution exists. If no solution exists, $\mathbf{v}_3$ is not a linear combination.

Part (b): Linear Combination of $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ for $\mathbf{v}_4$

To check if $\mathbf{v}_4$ is a linear combination of $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$, solve for $a, b, c \in \mathbb{R}$ in: $a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 = \mathbf{v}_4.$ This expands to: $a(1, -1, 2) + b(2, 1, 1) + c(-1, 3, 1) = (-8, 7, 5).$ Breaking into components:

1. $a + 2b - c = -8$,
2. $-a + b + 3c = 7$,
3. $2a + b + c = 5$.

Solve this system of equations to find if $a, b, c$ exist. If a solution exists, then $\mathbf{v}_4$ is a linear combination.

Problem 2: Linear Independence/Dependence in $\mathbb{R}^2$

Given six sets of vectors:

(a) $(0, 3)$,
(b) $(0, 0)$,
(c) $(0, 2)$, $(0, 0)$,
(d) $(1, 0)$,
(e) $(0, 1)$, $(1, -1)$,
(f) $(0, 1)$, $(1, -1)$, $(2, 4)$.

Approach:

1. A set of vectors is linearly dependent if at least one vector can be written as a linear combination of the others.
2. Otherwise, they are linearly independent.

• (a) A single vector like $(0, 3)$ is always independent.
• (b) The zero vector $(0, 0)$ is always dependent.
• (c) Two vectors $(0, 2)$, $(0, 0)$: Dependent because of the zero vector.
• (d) A single vector $(1, 0)$ is independent.
• (e) Two vectors $(0, 1)$, $(1, -1)$: Solve the determinant or check for proportionality.
• (f) Three vectors $(0, 1)$, $(1, -1)$, $(2, 4)$: Use a matrix rank check.

Let me know if you'd like the solutions worked out in detail, or if you’d like me to solve specific subparts for clarity.

Expanding on Concepts

1. What is a basis in a vector space?
2. How can the rank of a matrix help determine linear independence?
3. Why is the zero vector always linearly dependent?
4. How does solving systems of equations relate to linear combinations?
5. What is the geometric interpretation of linear dependence in $\mathbb{R}^2$ or $\mathbb{R}^3$?

Tip: For linear independence, always check if the determinant of the matrix formed by the vectors is zero (for square matrices).