1. **State the problem:** Prove that the vectors $\mathbf{u} = \langle 1, 1, 0 \rangle$, $\mathbf{v} = \langle 0, 1, 1 \rangle$, and $\mathbf{w} = \langle 1, 0, 1 \rangle$ are linearly independent. 2. **Recall the definition:** Vectors are linearly independent if the only solution to the equation $$c_1 \mathbf{u} + c_2 \mathbf{v} + c_3 \mathbf{w} = \mathbf{0}$$ is $c_1 = c_2 = c_3 = 0$. 3. **Set up the equation:** $$c_1 \langle 1, 1, 0 \rangle + c_2 \langle 0, 1, 1 \rangle + c_3 \langle 1, 0, 1 \rangle = \langle 0, 0, 0 \rangle$$ 4. **Write component-wise equations:** $$\begin{cases} c_1 + 0 \cdot c_2 + c_3 = 0 \\ c_1 + c_2 + 0 \cdot c_3 = 0 \\ 0 \cdot c_1 + c_2 + c_3 = 0 \end{cases}$$ 5. **Simplify the system:** $$\begin{cases} c_1 + c_3 = 0 \\ c_1 + c_2 = 0 \\ c_2 + c_3 = 0 \end{cases}$$ 6. **From the first equation:** $$c_3 = -c_1$$ 7. **From the second equation:** $$c_2 = -c_1$$ 8. **Substitute $c_2$ and $c_3$ into the third equation:** $$(-c_1) + (-c_1) = 0 \implies -2c_1 = 0$$ 9. **Solve for $c_1$:** $$c_1 = 0$$ 10. **Back-substitute to find $c_2$ and $c_3$:** $$c_2 = -0 = 0, \quad c_3 = -0 = 0$$ 11. **Conclusion:** The only solution is $c_1 = c_2 = c_3 = 0$, so the vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are linearly independent.