1. **State the problem:** Solve the system of linear equations using Gaussian elimination, scaling, and pivoting: $$\begin{cases} 2x_2 + x_4 = 0 \\ 2x_1 + 2x_2 + 3x_3 + 2x_4 = -2 \\ 4x_1 - 3x_2 + x_4 = -7 \\ 6x_1 + x_2 - 6x_3 - 5x_4 = 6 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 0 & 2 & 0 & 1 & 0 \\ 2 & 2 & 3 & 2 & -2 \\ 4 & -3 & 0 & 1 & -7 \\ 6 & 1 & -6 & -5 & 6 \end{array}\right]$$ 3. **Apply partial pivoting:** - Find the largest absolute value in the first column to use as pivot. - The largest is 6 in row 4, so swap row 1 and row 4: $$\left[\begin{array}{cccc|c} 6 & 1 & -6 & -5 & 6 \\ 2 & 2 & 3 & 2 & -2 \\ 4 & -3 & 0 & 1 & -7 \\ 0 & 2 & 0 & 1 & 0 \end{array}\right]$$ 4. **Scale rows for numerical stability (optional but recommended):** - Row 1 max abs element: 6 - Row 2 max abs element: 3 - Row 3 max abs element: 4 - Row 4 max abs element: 2 Divide each row by its max element: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ \frac{2}{3} & \frac{2}{3} & 1 & \frac{2}{3} & -\frac{2}{3} \\ 1 & -\frac{3}{4} & 0 & \frac{1}{4} & -\frac{7}{4} \\ 0 & 1 & 0 & \frac{1}{2} & 0 \end{array}\right]$$ 5. **Perform Gaussian elimination to get upper triangular form:** - Eliminate below pivot in column 1: Row 3 = Row 3 - Row 1: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ \frac{2}{3} & \frac{2}{3} & 1 & \frac{2}{3} & -\frac{2}{3} \\ 0 & -\frac{11}{12} & 1 & \frac{7}{6} & -\frac{11}{4} \\ 0 & 1 & 0 & \frac{1}{2} & 0 \end{array}\right]$$ Row 2 = Row 2 - \frac{2}{3}*Row 1: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & \frac{7}{9} & \frac{5}{3} & \frac{3}{2} & -\frac{8}{3} \\ 0 & -\frac{11}{12} & 1 & \frac{7}{6} & -\frac{11}{4} \\ 0 & 1 & 0 & \frac{1}{2} & 0 \end{array}\right]$$ 6. **Pivot on second column:** - Swap row 4 and row 2 to bring the largest pivot (1) to the top of column 2: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & 1 & 0 & \frac{1}{2} & 0 \\ 0 & -\frac{11}{12} & 1 & \frac{7}{6} & -\frac{11}{4} \\ 0 & \frac{7}{9} & \frac{5}{3} & \frac{3}{2} & -\frac{8}{3} \end{array}\right]$$ 7. **Eliminate below pivot in column 2:** Row 3 = Row 3 + \frac{11}{12}*Row 2: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & 1 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{10}{12} & -\frac{11}{4} \\ 0 & \frac{7}{9} & \frac{5}{3} & \frac{3}{2} & -\frac{8}{3} \end{array}\right]$$ Row 4 = Row 4 - \frac{7}{9}*Row 2: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & 1 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{5}{6} & -\frac{11}{4} \\ 0 & 0 & \frac{5}{3} & \frac{4}{3} & -\frac{8}{3} \end{array}\right]$$ 8. **Pivot on third column:** - Divide row 4 by \frac{5}{3} to make pivot 1: Row 4 = Row 4 * \frac{3}{5}: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & 1 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{5}{6} & -\frac{11}{4} \\ 0 & 0 & 1 & \frac{4}{5} & -\frac{8}{5} \end{array}\right]$$ 9. **Eliminate above pivot in column 3:** Row 3 = Row 3 - Row 4: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{6} & -1 & -\frac{5}{6} & 1 \\ 0 & 1 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{30} & -\frac{27}{20} \\ 0 & 0 & 1 & \frac{4}{5} & -\frac{8}{5} \end{array}\right]$$ 10. **Back substitution:** From row 3: $$\frac{1}{30}x_4 = -\frac{27}{20} \implies x_4 = -\frac{27}{20} \times 30 = -40.5$$ From row 4: $$x_3 + \frac{4}{5}x_4 = -\frac{8}{5} \implies x_3 = -\frac{8}{5} - \frac{4}{5}(-40.5) = -1.6 + 32.4 = 30.8$$ From row 2: $$x_2 + \frac{1}{2}x_4 = 0 \implies x_2 = -\frac{1}{2}(-40.5) = 20.25$$ From row 1: $$x_1 + \frac{1}{6}x_2 - x_3 - \frac{5}{6}x_4 = 1$$ Substitute known values: $$x_1 + \frac{1}{6}(20.25) - 30.8 - \frac{5}{6}(-40.5) = 1$$ Calculate: $$x_1 + 3.375 - 30.8 + 33.75 = 1$$ $$x_1 + 6.325 = 1 \implies x_1 = 1 - 6.325 = -5.325$$ **Final solution:** $$\boxed{\begin{cases} x_1 = -5.325 \\ x_2 = 20.25 \\ x_3 = 30.8 \\ x_4 = -40.5 \end{cases}}$$