1. **State the problem:** We are given vectors \(u = \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}\), \(v = \begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}\), \(w = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\), and \(r = \begin{bmatrix}1 \\ 1 \\ 6\end{bmatrix}\). We need to determine if the set \(\{u, v, w\}\) is linearly independent or dependent, if it forms a basis of \(\mathbb{R}^3\), if it spans \(\mathbb{R}^3\), and if it spans the vector \(r\). 2. **Check linear independence:** To check if \(\{u, v, w\}\) is linearly independent, we form the matrix \(A = [u \ v \ w]\) and check its determinant: $$A = \begin{bmatrix}1 & -2 & 1 \\ -1 & 1 & 2 \\ 0 & 1 & 3\end{bmatrix}$$ Calculate \(\det(A)\): $$\det(A) = 1 \cdot \begin{vmatrix}1 & 2 \\ 1 & 3\end{vmatrix} - (-2) \cdot \begin{vmatrix}-1 & 2 \\ 0 & 3\end{vmatrix} + 1 \cdot \begin{vmatrix}-1 & 1 \\ 0 & 1\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}1 & 2 \\ 1 & 3\end{vmatrix} = 1 \times 3 - 2 \times 1 = 3 - 2 = 1$$ $$\begin{vmatrix}-1 & 2 \\ 0 & 3\end{vmatrix} = (-1) \times 3 - 2 \times 0 = -3 - 0 = -3$$ $$\begin{vmatrix}-1 & 1 \\ 0 & 1\end{vmatrix} = (-1) \times 1 - 1 \times 0 = -1 - 0 = -1$$ Substitute back: $$\det(A) = 1 \times 1 - (-2) \times (-3) + 1 \times (-1) = 1 - 6 - 1 = -6$$ Since \(\det(A) \neq 0\), the vectors \(u, v, w\) are linearly independent. 3. **Is \(\{u, v, w\}\) a basis of \(\mathbb{R}^3\)?** Since the vectors are linearly independent and there are 3 vectors in \(\mathbb{R}^3\), they form a basis of \(\mathbb{R}^3\). 4. **Does \(\{u, v, w\}\) span \(\mathbb{R}^3\)?** A basis spans the entire space, so \(\{u, v, w\}\) spans \(\mathbb{R}^3\). 5. **Does \(\{u, v, w\}\) span vector \(r\)?** Since \(r \in \mathbb{R}^3\) and \(\{u, v, w\}\) spans \(\mathbb{R}^3\), \(r\) can be expressed as a linear combination of \(u, v, w\). Thus, \(\{u, v, w\}\) spans \(r\). **Final answer:** (a) {u, v, w} is linearly independent, so it is a basis in \(\mathbb{R}^3\), so it spans \(\mathbb{R}^3\) and it spans \(r\) as a vector from \(\mathbb{R}^3\).