1. **Problem:** Determine if the vectors \(\begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix}\), \(\begin{bmatrix}0 \\ 5 \\ -8\end{bmatrix}\), and \(\begin{bmatrix}-3 \\ 4 \\ 1\end{bmatrix}\) are linearly independent. 2. **Step 1: Set up the linear independence equation.** We want to see if the equation $$c_1 \begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix} + c_2 \begin{bmatrix}0 \\ 5 \\ -8\end{bmatrix} + c_3 \begin{bmatrix}-3 \\ 4 \\ 1\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$ has only the trivial solution \(c_1 = c_2 = c_3 = 0\). 3. **Step 2: Write the system of equations from the vector equation:** From each component: - First component: $$0c_1 + 0c_2 - 3c_3 = 0 \implies -3c_3 = 0$$ - Second component: $$0c_1 + 5c_2 + 4c_3 = 0$$ - Third component: $$2c_1 - 8c_2 + 1c_3 = 0$$ 4. **Step 3: Solve the first equation:** $$-3c_3 = 0 \implies c_3 = 0$$ 5. **Step 4: Substitute \(c_3 = 0\) into the second and third equations:** - Second: $$5c_2 + 4(0) = 0 \implies 5c_2 = 0 \implies c_2 = 0$$ - Third: $$2c_1 - 8(0) + 0 = 0 \implies 2c_1 = 0 \implies c_1 = 0$$ 6. **Step 5: Conclusion:** The only solution is \(c_1 = c_2 = c_3 = 0\), so the vectors are **linearly independent**. **Final answer:** The vectors are linearly independent because the only solution to the linear combination equaling zero is the trivial solution where all coefficients are zero.