1. **State the problem:** We want to find all solutions of a linear system whose augmented matrix can be row-reduced to $$\begin{bmatrix}1 & 0 & 1 & 4 \\ -1 & 2 & 3 & 2\end{bmatrix}$$ with the condition that $x_3 = 2$. 2. **Understand the matrix:** The matrix represents the system of equations: $$\begin{cases} x_1 + 0 \cdot x_2 + 1 \cdot x_3 = 4 \\ -1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3 = 2 \end{cases}$$ 3. **Substitute $x_3 = 2$ into the system:** $$\begin{cases} x_1 + 2 = 4 \\ -x_1 + 2x_2 + 3 \cdot 2 = 2 \end{cases}$$ 4. **Simplify each equation:** - From the first equation: $x_1 = 4 - 2 = 2$ - From the second equation: $-2 + 2x_2 + 6 = 2 \implies 2x_2 + 4 = 2 \implies 2x_2 = 2 - 4 = -2 \implies x_2 = -1$ 5. **Write the solution:** $$\boxed{(x_1, x_2, x_3) = (2, -1, 2)}$$ This is the unique solution given the condition $x_3 = 2$. **Summary:** We substituted the given value of $x_3$ into the system, solved for $x_1$ and $x_2$, and found the unique solution.