1. **State the problem:** We have a set of vectors $W$ defined as $$\begin{bmatrix} a - 4b \\ 6 \\ 6a + b \\ -a - b \end{bmatrix}$$ where $a$ and $b$ are arbitrary real numbers. We want to find a set of vectors $S$ that span $W$. 2. **Rewrite the vector in terms of $a$ and $b$: ** $$\begin{bmatrix} a - 4b \\ 6 \\ 6a + b \\ -a - b \end{bmatrix} = a \begin{bmatrix} 1 \\ 0 \\ 6 \\ -1 \end{bmatrix} + b \begin{bmatrix} -4 \\ 0 \\ 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 6 \\ 0 \\ 0 \end{bmatrix}$$ 3. **Interpretation:** The vector is a linear combination of vectors $$\begin{bmatrix} 1 \\ 0 \\ 6 \\ -1 \end{bmatrix}, \begin{bmatrix} -4 \\ 0 \\ 1 \\ -1 \end{bmatrix}$$ and a constant vector $$\begin{bmatrix} 0 \\ 6 \\ 0 \\ 0 \end{bmatrix}$$ 4. **Check if $W$ is a vector space:** Since the second component is always 6 (not zero), $W$ is not closed under addition or scalar multiplication, so $W$ is not a vector space. 5. **Conclusion:** Since $W$ is not a vector space, it cannot be spanned by vectors in the usual sense. **Final answer:** D. W is not a vector space