1. **Problem Statement:** Determine if the vectors $\mathbf{u}_1 = (2, -1, 3, 2)$, $\mathbf{u}_2 = (1, 3, 4, 2)$, and $\mathbf{u}_3 = (3, -5, 2, 2)$ are linearly dependent or independent. 2. **Concept:** Vectors are linearly dependent if there exist scalars $c_1, c_2, c_3$, not all zero, such that $$c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3 = \mathbf{0}.$$ Otherwise, they are linearly independent. 3. **Set up the equation:** $$c_1 (2, -1, 3, 2) + c_2 (1, 3, 4, 2) + c_3 (3, -5, 2, 2) = (0, 0, 0, 0).$$ 4. **Write the system of equations:** $$\begin{cases} 2c_1 + c_2 + 3c_3 = 0 \\ -1c_1 + 3c_2 - 5c_3 = 0 \\ 3c_1 + 4c_2 + 2c_3 = 0 \\ 2c_1 + 2c_2 + 2c_3 = 0 \end{cases}$$ 5. **Solve the system:** From the fourth equation: $$2c_1 + 2c_2 + 2c_3 = 0 \implies c_1 + c_2 + c_3 = 0 \implies c_1 = -c_2 - c_3.$$ Substitute $c_1 = -c_2 - c_3$ into the first three equations: - First equation: $$2(-c_2 - c_3) + c_2 + 3c_3 = 0 \implies -2c_2 - 2c_3 + c_2 + 3c_3 = 0 \implies -c_2 + c_3 = 0 \implies c_3 = c_2.$$ - Second equation: $$-1(-c_2 - c_3) + 3c_2 - 5c_3 = 0 \implies c_2 + c_3 + 3c_2 - 5c_3 = 0 \implies 4c_2 - 4c_3 = 0.$$ Substitute $c_3 = c_2$: $$4c_2 - 4c_2 = 0,$$ which is true. - Third equation: $$3(-c_2 - c_3) + 4c_2 + 2c_3 = 0 \implies -3c_2 - 3c_3 + 4c_2 + 2c_3 = 0 \implies c_2 - c_3 = 0.$$ Substitute $c_3 = c_2$: $$c_2 - c_2 = 0,$$ which is true. 6. **Conclusion:** We have $c_3 = c_2$ and $c_1 = -c_2 - c_3 = -c_2 - c_2 = -2c_2$. Let $c_2 = t$, then $$c_1 = -2t, \quad c_2 = t, \quad c_3 = t,$$ where $t$ is any scalar. Since there exists a nontrivial solution (not all zero) for $c_1, c_2, c_3$, the vectors are **linearly dependent**. **Final answer:** The vectors $\mathbf{u}_1, \mathbf{u}_2,$ and $\mathbf{u}_3$ are linearly dependent.