1. **State the problem:** Solve the system of linear equations using Gaussian elimination: $$\begin{cases} x_1 + 3x_2 - x_3 = 2 \\ 2x_1 - x_2 + x_3 + 3x_4 = 14 \\ -3x_1 - 2x_2 + 4x_3 + x_4 = -12 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 1 & 3 & -1 & 0 & 2 \\ 2 & -1 & 1 & 3 & 14 \\ -3 & -2 & 4 & 1 & -12 \end{array}\right]$$ 3. **Use row operations to get upper triangular form:** - Replace $R_2$ by $R_2 - 2R_1$: $$R_2 = [2, -1, 1, 3, 14] - 2 \times [1, 3, -1, 0, 2] = [0, -7, 3, 3, 10]$$ - Replace $R_3$ by $R_3 + 3R_1$: $$R_3 = [-3, -2, 4, 1, -12] + 3 \times [1, 3, -1, 0, 2] = [0, 7, 1, 1, -6]$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 3 & -1 & 0 & 2 \\ 0 & -7 & 3 & 3 & 10 \\ 0 & 7 & 1 & 1 & -6 \end{array}\right]$$ 4. **Eliminate $x_2$ from $R_3$:** - Replace $R_3$ by $R_3 + R_2$: $$R_3 = [0, 7, 1, 1, -6] + [0, -7, 3, 3, 10] = [0, 0, 4, 4, 4]$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 3 & -1 & 0 & 2 \\ 0 & -7 & 3 & 3 & 10 \\ 0 & 0 & 4 & 4 & 4 \end{array}\right]$$ 5. **Back substitution:** - From $R_3$: $4x_3 + 4x_4 = 4 \Rightarrow x_3 + x_4 = 1 \Rightarrow x_3 = 1 - x_4$ - From $R_2$: $-7x_2 + 3x_3 + 3x_4 = 10$ Substitute $x_3$: $$-7x_2 + 3(1 - x_4) + 3x_4 = 10 \Rightarrow -7x_2 + 3 = 10 \Rightarrow -7x_2 = 7 \Rightarrow x_2 = -1$$ - From $R_1$: $x_1 + 3x_2 - x_3 = 2$ Substitute $x_2$ and $x_3$: $$x_1 + 3(-1) - (1 - x_4) = 2 \Rightarrow x_1 - 3 - 1 + x_4 = 2 \Rightarrow x_1 + x_4 = 6 \Rightarrow x_1 = 6 - x_4$$ 6. **Solution:** The system has infinitely many solutions parameterized by $x_4$: $$x_1 = 6 - x_4, \quad x_2 = -1, \quad x_3 = 1 - x_4, \quad x_4 = x_4$$ where $x_4$ is any real number.