1. The problem involves solving the system of linear equations represented by the augmented matrix: $$\begin{bmatrix}1 & -1 & 2 & 4 & 6 & | & 2 \\ 0 & 1 & 2 & 1 & -1 & | & -1 \\ 0 & 0 & 0 & 1 & 0 & | & 1 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$$ 2. Assign variables to the columns (except the augmented one): $$x_1, x_2, x_3, x_4, x_5$$ 3. Write equations for each nonzero row: - Row 1: $x_1 - x_2 + 2x_3 + 4x_4 + 6x_5 = 2$ - Row 2: $x_2 + 2x_3 + x_4 - x_5 = -1$ - Row 3: $x_4 = 1$ - Row 4: $0 = 0$ (no new information) 4. Substitute $x_4 = 1$ directly into the first two equations: - From Row 1: $$x_1 - x_2 + 2x_3 + 4(1) + 6x_5 = 2 \implies x_1 - x_2 + 2x_3 + 6x_5 = 2 - 4 = -2$$ - From Row 2: $$x_2 + 2x_3 + 1 - x_5 = -1 \implies x_2 + 2x_3 - x_5 = -2$$ 5. Express $x_2$ from the second equation: $$x_2 = -2 - 2x_3 + x_5$$ 6. Substitute $x_2$ into the first equation: $$x_1 - (-2 - 2x_3 + x_5) + 2x_3 + 6x_5 = -2$$ Simplify: $$x_1 + 2 + 2x_3 - x_5 + 2x_3 + 6x_5 = -2$$ $$x_1 + 2 + 4x_3 + 5x_5 = -2$$ 7. Solve for $x_1$: $$x_1 = -2 - 4x_3 - 5x_5 - 2 = -4 - 4x_3 - 5x_5$$ 8. Variables $x_3$ and $x_5$ are free parameters. Let: $$x_3 = t, \quad x_5 = s$$ 9. The solution set is: $$\begin{cases} x_1 = -4 - 4t - 5s \\ x_2 = -2 - 2t + s \\ x_3 = t \\ x_4 = 1 \\ x_5 = s \end{cases}$$ where $t,s \in \mathbb{R}$. 10. This describes an infinite number of solutions parameterized by $t$ and $s$. Final answer: $$\boxed{(x_1,x_2,x_3,x_4,x_5) = (-4 - 4t - 5s , -2 - 2t + s , t , 1 , s)}$$ with $t,s \in \mathbb{R}$.