1. **State the problem:** Solve the system of linear equations using Gauss-Jordan elimination: $$\begin{cases} x_1 + 2x_2 - x_3 + x_4 = 1 \\ 3x_2 + 6x_3 - 3x_4 = -3 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 0 & 3 & 6 & -3 & -3 \end{array}\right]$$ Note: The second equation has no $x_1$ term, so its coefficient is 0. 3. **Goal:** Use row operations to get the matrix into reduced row echelon form (RREF), where the left side is the identity matrix or as close as possible. 4. **Step 1:** Make the leading coefficient of the first row 1 (already 1). 5. **Step 2:** Eliminate the $x_2$ term in the first row using the second row. First, divide the second row by 3 to simplify: $$R_2 \to \frac{1}{3} R_2 = [0, 1, 2, -1, -1]$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 0 & 1 & 2 & -1 & -1 \end{array}\right]$$ 6. **Step 3:** Eliminate the $x_2$ term in the first row by subtracting $2 \times R_2$ from $R_1$: $$R_1 \to R_1 - 2R_2 = [1, 2-2(1), -1-2(2), 1-2(-1), 1-2(-1)] = [1, 0, -1-4, 1+2, 1+2] = [1, 0, -5, 3, 3]$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & 0 & -5 & 3 & 3 \\ 0 & 1 & 2 & -1 & -1 \end{array}\right]$$ 7. **Step 4:** Express variables in terms of free variables $x_3$ and $x_4$ (since we have 4 variables but only 2 equations). From row 1: $$x_1 - 5x_3 + 3x_4 = 3 \implies x_1 = 3 + 5x_3 - 3x_4$$ From row 2: $$x_2 + 2x_3 - x_4 = -1 \implies x_2 = -1 - 2x_3 + x_4$$ 8. **Final solution:** $$\boxed{\begin{cases} x_1 = 3 + 5x_3 - 3x_4 \\ x_2 = -1 - 2x_3 + x_4 \\ x_3 = x_3 \text{ (free)} \\ x_4 = x_4 \text{ (free)} \end{cases}}$$ This means the system has infinitely many solutions parameterized by $x_3$ and $x_4$.