1. **State the problem:** Solve the system of equations: $$\begin{cases} x + 2y + 3z = 14 \\ 2x - y + z = 5 \\ 3x + y + 2z = 13 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 2 & -1 & 1 & 5 \\ 3 & 1 & 2 & 13 \end{array}\right]$$ 3. **Use row operations to get upper triangular form:** - Replace $R_2$ by $R_2 - 2R_1$: $$R_2 = \left[2, -1, 1, 5\right] - 2 \times \left[1, 2, 3, 14\right] = \left[2-2, -1-4, 1-6, 5-28\right] = \left[0, -5, -5, -23\right]$$ - Replace $R_3$ by $R_3 - 3R_1$: $$R_3 = \left[3, 1, 2, 13\right] - 3 \times \left[1, 2, 3, 14\right] = \left[3-3, 1-6, 2-9, 13-42\right] = \left[0, -5, -7, -29\right]$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 0 & -5 & -5 & -23 \\ 0 & -5 & -7 & -29 \end{array}\right]$$ 4. **Eliminate below pivot in column 2:** - Replace $R_3$ by $R_3 - R_2$: $$R_3 = \left[0, -5, -7, -29\right] - \left[0, -5, -5, -23\right] = \left[0, 0, -2, -6\right]$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 0 & -5 & -5 & -23 \\ 0 & 0 & -2 & -6 \end{array}\right]$$ 5. **Back substitution:** - From $R_3$: $$-2z = -6 \implies z = \frac{-6}{-2} = 3$$ - From $R_2$: $$-5y - 5z = -23 \implies -5y - 5(3) = -23 \implies -5y - 15 = -23$$ $$-5y = -23 + 15 = -8 \implies y = \frac{-8}{-5} = \frac{8}{5}$$ - From $R_1$: $$x + 2y + 3z = 14 \implies x + 2\left(\frac{8}{5}\right) + 3(3) = 14$$ $$x + \frac{16}{5} + 9 = 14 \implies x + \frac{16}{5} + \frac{45}{5} = 14$$ $$x + \frac{61}{5} = 14 \implies x = 14 - \frac{61}{5} = \frac{70}{5} - \frac{61}{5} = \frac{9}{5}$$ 6. **Final answer:** $$x = \frac{9}{5}, \quad y = \frac{8}{5}, \quad z = 3$$