1. Stating the problem: Solve the system of linear equations using Gaussian elimination: $$\begin{cases} x_1 + 4x_2 + 2x_3 = 20 \\ -x_1 - 5x_2 + 3x_3 = -24 \\ 3x_1 - 3x_2 + 4x_3 = 0 \end{cases}$$ 2. Write the augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ -1 & -5 & 3 & -24 \\ 3 & -3 & 4 & 0 \end{array}\right]$$ 3. Add row 1 to row 2 to eliminate $x_1$ in row 2: $$R_2 \to R_2 + R_1: \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ \cancel{-1}+1 & \cancel{-5}+4 & 3+2 & -24+20 \\ 3 & -3 & 4 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & -1 & 5 & -4 \\ 3 & -3 & 4 & 0 \end{array}\right]$$ 4. Eliminate $x_1$ in row 3 by subtracting $3 \times$ row 1 from row 3: $$R_3 \to R_3 - 3R_1: \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & -1 & 5 & -4 \\ 3-3\times1 & -3-3\times4 & 4-3\times2 & 0-3\times20 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & -1 & 5 & -4 \\ 0 & -15 & -2 & -60 \end{array}\right]$$ 5. Multiply row 2 by $-1$ to simplify: $$R_2 \to -1 \times R_2: \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & 1 & -5 & 4 \\ 0 & -15 & -2 & -60 \end{array}\right]$$ 6. Add $15 \times$ row 2 to row 3 to eliminate $x_2$ in row 3: $$R_3 \to R_3 + 15R_2: \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & 1 & -5 & 4 \\ 0 & \cancel{-15}+15 & -2 + 15 \times (-5) & -60 + 15 \times 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 4 & 2 & 20 \\ 0 & 1 & -5 & 4 \\ 0 & 0 & -77 & 0 \end{array}\right]$$ 7. Solve for $x_3$ from row 3: $$-77x_3 = 0 \implies x_3 = 0$$ 8. Substitute $x_3=0$ into row 2: $$x_2 - 5 \times 0 = 4 \implies x_2 = 4$$ 9. Substitute $x_2=4$ and $x_3=0$ into row 1: $$x_1 + 4 \times 4 + 2 \times 0 = 20 \implies x_1 + 16 = 20 \implies x_1 = 4$$ Final solution: $$x_1 = 4, \quad x_2 = 4, \quad x_3 = 0$$