1. **State the problem:** Solve the system of linear equations using Gauss elimination with partial pivoting: $$\begin{cases} x_1 - 3x_2 + x_3 = 9 \\ 2x_1 + x_2 - x_3 = 8 \\ 3x_1 - x_2 + 2x_3 = 13 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & -3 & 1 & 9 \\ 2 & 1 & -1 & 8 \\ 3 & -1 & 2 & 13 \end{array}\right]$$ 3. **Partial pivoting step 1:** Find the largest absolute value in column 1 from rows 1 to 3. The largest is 3 in row 3. Swap row 1 and row 3: $$\left[\begin{array}{ccc|c} 3 & -1 & 2 & 13 \\ 2 & 1 & -1 & 8 \\ 1 & -3 & 1 & 9 \end{array}\right]$$ 4. **Eliminate entries below pivot in column 1:** - For row 2: multiplier $m_{21} = \frac{2}{3}$ - For row 3: multiplier $m_{31} = \frac{1}{3}$ Update rows: Row 2: $R_2 - m_{21} R_1$: $$\begin{aligned} 2 - \cancel{\frac{2}{3} \times 3} &= 2 - 2 = 0 \\ 1 - \frac{2}{3} \times (-1) &= 1 + \frac{2}{3} = \frac{5}{3} \\ -1 - \frac{2}{3} \times 2 &= -1 - \frac{4}{3} = -\frac{7}{3} \\ 8 - \frac{2}{3} \times 13 &= 8 - \frac{26}{3} = \frac{24}{3} - \frac{26}{3} = -\frac{2}{3} \end{aligned}$$ Row 3: $R_3 - m_{31} R_1$: $$\begin{aligned} 1 - \cancel{\frac{1}{3} \times 3} &= 1 - 1 = 0 \\ -3 - \frac{1}{3} \times (-1) &= -3 + \frac{1}{3} = -\frac{8}{3} \\ 1 - \frac{1}{3} \times 2 &= 1 - \frac{2}{3} = \frac{1}{3} \\ 9 - \frac{1}{3} \times 13 &= 9 - \frac{13}{3} = \frac{27}{3} - \frac{13}{3} = \frac{14}{3} \end{aligned}$$ Updated matrix: $$\left[\begin{array}{ccc|c} 3 & -1 & 2 & 13 \\ 0 & \frac{5}{3} & -\frac{7}{3} & -\frac{2}{3} \\ 0 & -\frac{8}{3} & \frac{1}{3} & \frac{14}{3} \end{array}\right]$$ 5. **Partial pivoting step 2:** Look at column 2 from rows 2 and 3. Largest absolute value is $\left| -\frac{8}{3} \right| = \frac{8}{3}$ in row 3. Swap row 2 and row 3: $$\left[\begin{array}{ccc|c} 3 & -1 & 2 & 13 \\ 0 & -\frac{8}{3} & \frac{1}{3} & \frac{14}{3} \\ 0 & \frac{5}{3} & -\frac{7}{3} & -\frac{2}{3} \end{array}\right]$$ 6. **Eliminate entry below pivot in column 2:** Multiplier $m_{32} = \frac{\frac{5}{3}}{-\frac{8}{3}} = -\frac{5}{8}$ Update row 3: $R_3 - m_{32} R_2$: $$\begin{aligned} 0 - \cancel{-\frac{5}{8} \times 0} &= 0 \\ \frac{5}{3} - \cancel{-\frac{5}{8} \times -\frac{8}{3}} &= \frac{5}{3} - \frac{5}{3} = 0 \\ -\frac{7}{3} - \left(-\frac{5}{8}\right) \times \frac{1}{3} &= -\frac{7}{3} + \frac{5}{24} = -\frac{56}{24} + \frac{5}{24} = -\frac{51}{24} = -\frac{17}{8} \\ -\frac{2}{3} - \left(-\frac{5}{8}\right) \times \frac{14}{3} &= -\frac{2}{3} + \frac{70}{24} = -\frac{16}{24} + \frac{70}{24} = \frac{54}{24} = \frac{9}{4} \end{aligned}$$ Updated matrix: $$\left[\begin{array}{ccc|c} 3 & -1 & 2 & 13 \\ 0 & -\frac{8}{3} & \frac{1}{3} & \frac{14}{3} \\ 0 & 0 & -\frac{17}{8} & \frac{9}{4} \end{array}\right]$$ 7. **Back substitution:** From row 3: $$-\frac{17}{8} x_3 = \frac{9}{4} \implies x_3 = \frac{9}{4} \div -\frac{17}{8} = \frac{9}{4} \times -\frac{8}{17} = -\frac{72}{68} = -\frac{18}{17}$$ From row 2: $$-\frac{8}{3} x_2 + \frac{1}{3} x_3 = \frac{14}{3}$$ Multiply both sides by 3: $$-8 x_2 + x_3 = 14$$ Substitute $x_3$: $$-8 x_2 - \frac{18}{17} = 14 \implies -8 x_2 = 14 + \frac{18}{17} = \frac{238}{17}$$ Divide both sides by -8: $$x_2 = \frac{\cancel{\frac{238}{17}}}{\cancel{-8}} = -\frac{238}{136} = -\frac{119}{68}$$ From row 1: $$3 x_1 - x_2 + 2 x_3 = 13$$ Substitute $x_2$ and $x_3$: $$3 x_1 - \left(-\frac{119}{68}\right) + 2 \left(-\frac{18}{17}\right) = 13$$ Simplify: $$3 x_1 + \frac{119}{68} - \frac{36}{17} = 13$$ Convert $\frac{36}{17}$ to denominator 68: $$\frac{36}{17} = \frac{144}{68}$$ So: $$3 x_1 + \frac{119}{68} - \frac{144}{68} = 13 \implies 3 x_1 - \frac{25}{68} = 13$$ Add $\frac{25}{68}$ to both sides: $$3 x_1 = 13 + \frac{25}{68} = \frac{884}{68} + \frac{25}{68} = \frac{909}{68}$$ Divide both sides by 3: $$x_1 = \frac{909}{68} \times \frac{1}{3} = \frac{909}{204} = \frac{303}{68}$$ 8. **Final solution:** $$\boxed{\left(x_1, x_2, x_3\right) = \left(\frac{303}{68}, -\frac{119}{68}, -\frac{18}{17}\right)}$$ This completes the Gauss elimination with partial pivoting solution.