1. **Problem statement:** Given vectors $$u_1=\begin{bmatrix}1\\2\\0\end{bmatrix}, u_2=\begin{bmatrix}2\\1\\1\end{bmatrix}, u_3=\begin{bmatrix}3\\3\\1\end{bmatrix}, u_4=\begin{bmatrix}11\\2\\-1\end{bmatrix}, u_5=\begin{bmatrix}3\\2\\1\end{bmatrix}$$ We want to determine whether $\{u_1,u_2,u_3\}$ spans $\mathbb{R}^3$. 2. **Recall:** A set of vectors spans $\mathbb{R}^3$ if their linear combinations can produce any vector in $\mathbb{R}^3$. This is true if and only if the vectors are linearly independent and there are 3 of them. 3. **Method:** Check if $u_1, u_2, u_3$ are linearly independent by forming a matrix with these vectors as columns and computing its determinant: $$A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{bmatrix}$$ 4. **Calculate determinant:** $$\det(A) = 1 \cdot \begin{vmatrix}1 & 3 \\ 1 & 1\end{vmatrix} - 2 \cdot \begin{vmatrix}2 & 3 \\ 0 & 1\end{vmatrix} + 3 \cdot \begin{vmatrix}2 & 1 \\ 0 & 1\end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix}1 & 3 \\ 1 & 1\end{vmatrix} = 1 \times 1 - 3 \times 1 = 1 - 3 = -2$$ $$\begin{vmatrix}2 & 3 \\ 0 & 1\end{vmatrix} = 2 \times 1 - 3 \times 0 = 2$$ $$\begin{vmatrix}2 & 1 \\ 0 & 1\end{vmatrix} = 2 \times 1 - 1 \times 0 = 2$$ 5. Substitute back: $$\det(A) = 1 \times (-2) - 2 \times 2 + 3 \times 2 = -2 - 4 + 6 = 0$$ 6. Since $\det(A) = 0$, the vectors $u_1, u_2, u_3$ are linearly dependent and do not span $\mathbb{R}^3$. **Final answer:** The set $\{u_1, u_2, u_3\}$ does not span $\mathbb{R}^3$ because they are linearly dependent.