1. **Problem Statement:** Determine which of the given vectors (a) [3, 9, 0], (b) [-4, 0, -2], (c) [3, 2, 1], and (d) [3, 3, 1] are linearly dependent with the vectors \(\mathbf{v_1} = \begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}\) and \(\mathbf{v_2} = \begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}\). 2. **Concept:** Vectors are linearly dependent if one can be expressed as a linear combination of the others. Here, we check if each vector \(\mathbf{v}\) satisfies: $$\mathbf{v} = a \mathbf{v_1} + b \mathbf{v_2}$$ for some scalars \(a, b\). 3. **Method:** For each candidate vector \(\mathbf{v} = \begin{bmatrix}v_1 \\ v_2 \\ v_3\end{bmatrix}\), solve the system: $$\begin{cases} v_1 = a \cdot 1 + b \cdot 2 \\ v_2 = a \cdot 3 + b \cdot 0 \\ v_3 = a \cdot 0 + b \cdot 1 \end{cases}$$ 4. **Check each vector:** - (a) \(\begin{bmatrix}3 \\ 9 \\ 0\end{bmatrix}\): From second equation: \(9 = 3a \Rightarrow a = 3\). From third equation: \(0 = b \Rightarrow b = 0\). From first equation: \(3 = 3 + 0 = 3\) (true). So, \(\mathbf{v_a} = 3 \mathbf{v_1} + 0 \mathbf{v_2}\). Vector (a) is linearly dependent. - (b) \(\begin{bmatrix}-4 \\ 0 \\ -2\end{bmatrix}\): From second equation: \(0 = 3a \Rightarrow a = 0\). From third equation: \(-2 = b \Rightarrow b = -2\). From first equation: \(-4 = 0 + 2(-2) = -4\) (true). So, \(\mathbf{v_b} = 0 \mathbf{v_1} - 2 \mathbf{v_2}\). Vector (b) is linearly dependent. - (c) \(\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}\): From second equation: \(2 = 3a \Rightarrow a = \frac{2}{3}\). From third equation: \(1 = b \Rightarrow b = 1\). From first equation: \(3 = \frac{2}{3} + 2(1) = \frac{2}{3} + 2 = \frac{8}{3} \neq 3\). No solution, so vector (c) is not linearly dependent. - (d) \(\begin{bmatrix}3 \\ 3 \\ 1\end{bmatrix}\): From second equation: \(3 = 3a \Rightarrow a = 1\). From third equation: \(1 = b \Rightarrow b = 1\). From first equation: \(3 = 1 + 2(1) = 3\) (true). So, \(\mathbf{v_d} = 1 \mathbf{v_1} + 1 \mathbf{v_2}\). Vector (d) is linearly dependent. 5. **Final answer:** Vectors (a), (b), and (d) are linearly dependent with \(\mathbf{v_1}\) and \(\mathbf{v_2}\). Vector (c) is not.