1. **Problem Statement:** Solve the system of linear equations using Gaussian Elimination: $$\begin{cases} x_1 - 3x_2 + 2x_3 + 3x_4 = 2 \\ 2x_1 - 6x_2 + 2x_3 + 10x_4 = -2 \\ -3x_1 + 9x_2 - 7x_3 - 7x_4 = -9 \end{cases}$$ 2. **Gaussian Elimination Overview:** Gaussian elimination transforms the system's augmented matrix into row echelon form using elementary row operations, making it easier to solve by back substitution. 3. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 2 & -6 & 2 & 10 & -2 \\ -3 & 9 & -7 & -7 & -9 \end{array}\right]$$ 4. **Step 1: Use row 1 to eliminate $x_1$ terms in rows 2 and 3.** - Replace row 2 with $R_2 - 2R_1$: $$R_2 = (2, -6, 2, 10, -2) - 2(1, -3, 2, 3, 2) = (0, 0, -2, 4, -6)$$ - Replace row 3 with $R_3 + 3R_1$: $$R_3 = (-3, 9, -7, -7, -9) + 3(1, -3, 2, 3, 2) = (0, 0, -1, 2, -3)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 0 & 0 & -2 & 4 & -6 \\ 0 & 0 & -1 & 2 & -3 \end{array}\right]$$ 5. **Step 2: Use row 3 to eliminate $x_3$ term in row 2.** - Replace row 2 with $R_2 - 2R_3$: $$R_2 = (0, 0, -2, 4, -6) - 2(0, 0, -1, 2, -3) = (0, 0, 0, 0, 0)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 2 & -3 \end{array}\right]$$ 6. **Step 3: Back substitution from the last nonzero row:** From row 3: $$-1x_3 + 2x_4 = -3 \implies x_3 = 2x_4 + 3$$ From row 1: $$x_1 - 3x_2 + 2x_3 + 3x_4 = 2$$ Substitute $x_3$: $$x_1 - 3x_2 + 2(2x_4 + 3) + 3x_4 = 2$$ Simplify: $$x_1 - 3x_2 + 4x_4 + 6 + 3x_4 = 2$$ $$x_1 - 3x_2 + 7x_4 = 2 - 6 = -4$$ 7. **Express $x_1$ in terms of $x_2$ and $x_4$:** $$x_1 = 3x_2 - 7x_4 - 4$$ 8. **Summary of solution:** - $x_1 = 3x_2 - 7x_4 - 4$ - $x_3 = 2x_4 + 3$ - $x_2$ and $x_4$ are free variables (parameters). This system has infinitely many solutions parameterized by $x_2$ and $x_4$. **Final answer:** $$\boxed{\begin{cases} x_1 = 3x_2 - 7x_4 - 4 \\ x_2 = x_2 \text{ (free)} \\ x_3 = 2x_4 + 3 \\ x_4 = x_4 \text{ (free)} \end{cases}}$$