1. **State the problem:** Determine if the vectors $U_1 = (1, -2, 3, 4)$, $U_2 = (-2, 4, -1, -3)$, and $U_3 = (-1, 2, 7, 6)$ are linearly dependent or independent. 2. **Recall the definition:** Vectors are linearly dependent if there exist scalars $c_1, c_2, c_3$, not all zero, such that $$c_1 U_1 + c_2 U_2 + c_3 U_3 = 0.$$ Otherwise, they are linearly independent. 3. **Set up the equation:** $$c_1(1, -2, 3, 4) + c_2(-2, 4, -1, -3) + c_3(-1, 2, 7, 6) = (0, 0, 0, 0).$$ 4. **Write the system of equations:** $$ \begin{cases} 1 c_1 - 2 c_2 - 1 c_3 = 0 \\ -2 c_1 + 4 c_2 + 2 c_3 = 0 \\ 3 c_1 - 1 c_2 + 7 c_3 = 0 \\ 4 c_1 - 3 c_2 + 6 c_3 = 0 \end{cases} $$ 5. **Solve the system:** From the first equation, express $c_1$: $$c_1 = 2 c_2 + c_3.$$ Substitute into the second: $$-2(2 c_2 + c_3) + 4 c_2 + 2 c_3 = 0 \Rightarrow -4 c_2 - 2 c_3 + 4 c_2 + 2 c_3 = 0,$$ which simplifies to $$0=0,$$ always true. 6. Substitute $c_1$ into the third equation: $$3(2 c_2 + c_3) - c_2 + 7 c_3 = 0 \Rightarrow 6 c_2 + 3 c_3 - c_2 + 7 c_3 = 0 \Rightarrow 5 c_2 + 10 c_3 = 0,$$ which simplifies to $$c_2 = -2 c_3.$$ 7. Substitute $c_1$ and $c_2$ into the fourth equation: $$4(2 c_2 + c_3) - 3 c_2 + 6 c_3 = 0 \Rightarrow 8 c_2 + 4 c_3 - 3 c_2 + 6 c_3 = 0 \Rightarrow 5 c_2 + 10 c_3 = 0,$$ which is the same as before, confirming consistency. 8. **Express all variables in terms of $c_3$:** $$c_2 = -2 c_3, \quad c_1 = 2(-2 c_3) + c_3 = -4 c_3 + c_3 = -3 c_3.$$ 9. Since $c_3$ can be any scalar (not necessarily zero), there exist nontrivial solutions. Therefore, the vectors are **linearly dependent**. **Final answer:** The vectors $U_1, U_2,$ and $U_3$ are linearly dependent.