1. **Problem 1: Solve the system using Gauss Elimination Method** Given system: $$\begin{cases} 4x - 5y + 6z = 12 \\ 7x + 3y - 2z = -5 \\ 5x - 4y + 8z = 10 \end{cases}$$ 2. Write the augmented matrix: $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right]$$ 3. Use row operations to get upper triangular form: - Multiply row 1 by $\frac{7}{4}$ and subtract from row 2: $$R_2 \to R_2 - \frac{7}{4}R_1$$ $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ \cancel{7} & 3 - \left(-5 \times \frac{7}{4}\right) & -2 - 6 \times \frac{7}{4} & -5 - 12 \times \frac{7}{4} \\ 5 & -4 & 8 & 10 \end{array}\right]$$ Calculate: $$3 - \left(-5 \times \frac{7}{4}\right) = 3 + \frac{35}{4} = \frac{12}{4} + \frac{35}{4} = \frac{47}{4}$$ $$-2 - 6 \times \frac{7}{4} = -2 - \frac{42}{4} = -2 - 10.5 = -12.5$$ $$-5 - 12 \times \frac{7}{4} = -5 - 21 = -26$$ So row 2 becomes: $$\left[0, \frac{47}{4}, -\frac{25}{2}, -26\right]$$ - Multiply row 1 by $\frac{5}{4}$ and subtract from row 3: $$R_3 \to R_3 - \frac{5}{4}R_1$$ Calculate: $$-4 - \left(-5 \times \frac{5}{4}\right) = -4 + \frac{25}{4} = -4 + 6.25 = 2.25 = \frac{9}{4}$$ $$8 - 6 \times \frac{5}{4} = 8 - 7.5 = 0.5 = \frac{1}{2}$$ $$10 - 12 \times \frac{5}{4} = 10 - 15 = -5$$ Row 3 becomes: $$\left[0, \frac{9}{4}, \frac{1}{2}, -5\right]$$ 4. Next, eliminate $y$ from row 3 using row 2: Multiply row 2 by $\frac{9/4}{47/4} = \frac{9}{47}$ and subtract from row 3: $$R_3 \to R_3 - \frac{9}{47} R_2$$ Calculate: $$\frac{9}{4} - \frac{9}{47} \times \frac{47}{4} = \frac{9}{4} - \frac{9}{4} = 0$$ $$\frac{1}{2} - \frac{9}{47} \times \left(-\frac{25}{2}\right) = \frac{1}{2} + \frac{225}{94} = \frac{47}{94} + \frac{225}{94} = \frac{272}{94} = \frac{136}{47}$$ $$-5 - \frac{9}{47} \times (-26) = -5 + \frac{234}{47} = -\frac{235}{47} + \frac{234}{47} = -\frac{1}{47}$$ Row 3 becomes: $$\left[0, 0, \frac{136}{47}, -\frac{1}{47}\right]$$ 5. Back substitution: $$z = \frac{-\frac{1}{47}}{\frac{136}{47}} = -\frac{1}{136}$$ 6. Substitute $z$ into row 2: $$\frac{47}{4} y - \frac{25}{2} \times \left(-\frac{1}{136}\right) = -26$$ $$\frac{47}{4} y + \frac{25}{272} = -26$$ Multiply both sides by 272: $$272 \times \frac{47}{4} y + 25 = -26 \times 272$$ $$3202 y + 25 = -7072$$ $$3202 y = -7097$$ $$y = -\frac{7097}{3202}$$ 7. Substitute $y$ and $z$ into row 1: $$4x - 5 \times \left(-\frac{7097}{3202}\right) + 6 \times \left(-\frac{1}{136}\right) = 12$$ $$4x + \frac{35485}{3202} - \frac{6}{136} = 12$$ Convert $\frac{6}{136} = \frac{3}{68}$ and find common denominator 3202: $$\frac{3}{68} = \frac{3 \times 47}{68 \times 47} = \frac{141}{3196} \approx \frac{141}{3202}$$ Approximate for simplicity: $$4x + \frac{35485}{3202} - \frac{141}{3202} = 12$$ $$4x + \frac{35344}{3202} = 12$$ Multiply both sides by 3202: $$4x \times 3202 + 35344 = 12 \times 3202$$ $$12808 x = 38424 - 35344 = 3080$$ $$x = \frac{3080}{12808} = \frac{385}{1601}$$ --- 8. **Problem 2: LU Factorization and solve** Given system: $$\begin{cases} 8x + 4y + 2z = 20 \\ 4x + 5y + 3z = 18 \\ 2x + 3y + 6z = 22 \end{cases}$$ 9. Coefficient matrix $A$: $$A = \begin{bmatrix} 8 & 4 & 2 \\ 4 & 5 & 3 \\ 2 & 3 & 6 \end{bmatrix}$$ 10. Decompose $A = LU$ where $$L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$$ 11. Set $u_{11} = 8$, $u_{12} = 4$, $u_{13} = 2$ 12. Compute $l_{21} = \frac{4}{8} = \frac{1}{2}$, $l_{31} = \frac{2}{8} = \frac{1}{4}$ 13. Compute $u_{22} = 5 - l_{21} \times u_{12} = 5 - \frac{1}{2} \times 4 = 5 - 2 = 3$ 14. Compute $u_{23} = 3 - l_{21} \times u_{13} = 3 - \frac{1}{2} \times 2 = 3 - 1 = 2$ 15. Compute $l_{32} = \frac{3 - l_{31} \times u_{12}}{u_{22}} = \frac{3 - \frac{1}{4} \times 4}{3} = \frac{3 - 1}{3} = \frac{2}{3}$ 16. Compute $u_{33} = 6 - l_{31} \times u_{13} - l_{32} \times u_{23} = 6 - \frac{1}{4} \times 2 - \frac{2}{3} \times 2 = 6 - \frac{1}{2} - \frac{4}{3} = 6 - 0.5 - 1.3333 = 4.1667 = \frac{25}{6}$ 17. So, $$L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{4} & \frac{2}{3} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} 8 & 4 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & \frac{25}{6} \end{bmatrix}$$ 18. Solve $Ly = b$ where $b = \begin{bmatrix} 20 \\ 18 \\ 22 \end{bmatrix}$: - $y_1 = 20$ - $\frac{1}{2} y_1 + y_2 = 18 \Rightarrow y_2 = 18 - 10 = 8$ - $\frac{1}{4} y_1 + \frac{2}{3} y_2 + y_3 = 22$ $$5 + \frac{16}{3} + y_3 = 22 \Rightarrow y_3 = 22 - 5 - \frac{16}{3} = 17 - \frac{16}{3} = \frac{51}{3} - \frac{16}{3} = \frac{35}{3}$$ 19. Solve $Ux = y$: - $\frac{25}{6} z = \frac{35}{3} \Rightarrow z = \frac{35}{3} \times \frac{6}{25} = \frac{210}{75} = \frac{14}{5}$ - $3 y + 2 z = 8 \Rightarrow 3 y + 2 \times \frac{14}{5} = 8$ (already solved for $y$) - $8 x + 4 y + 2 z = 20 \Rightarrow 8 x + 4 \times 8 + 2 \times \frac{14}{5} = 20$ Calculate: $$8x + 32 + \frac{28}{5} = 20$$ $$8x = 20 - 32 - \frac{28}{5} = -12 - \frac{28}{5} = -\frac{60}{5} - \frac{28}{5} = -\frac{88}{5}$$ $$x = -\frac{88}{5} \times \frac{1}{8} = -\frac{11}{5}$$ --- 20. **Problem 3: Solve system using Gauss-Jordan Elimination** Given system: $$\begin{cases} \frac{1}{2}x + \frac{2}{3}y - \frac{3}{4}z = \frac{5}{6} \\ \frac{3}{5}x - \frac{1}{4}y + \frac{4}{7}z = \frac{2}{3} \\ -\frac{2}{3}x + \frac{5}{6}y + \frac{3}{5}z = \frac{1}{2} \end{cases}$$ 21. Write augmented matrix: $$\left[\begin{array}{ccc|c} \frac{1}{2} & \frac{2}{3} & -\frac{3}{4} & \frac{5}{6} \\ \frac{3}{5} & -\frac{1}{4} & \frac{4}{7} & \frac{2}{3} \\ -\frac{2}{3} & \frac{5}{6} & \frac{3}{5} & \frac{1}{2} \end{array}\right]$$ 22. Multiply row 1 by 2 to get leading 1: $$R_1 \to 2 R_1: [1, \frac{4}{3}, -\frac{3}{2}, \frac{5}{3}]$$ 23. Eliminate $x$ from rows 2 and 3: - $R_2 \to R_2 - \frac{3}{5} R_1$ - $R_3 \to R_3 + \frac{2}{3} R_1$ Calculate row 2: $$\frac{3}{5} - \frac{3}{5} \times 1 = 0$$ $$-\frac{1}{4} - \frac{3}{5} \times \frac{4}{3} = -\frac{1}{4} - \frac{4}{5} = -\frac{1}{4} - \frac{16}{20} = -\frac{5}{20} - \frac{16}{20} = -\frac{21}{20}$$ $$\frac{4}{7} - \frac{3}{5} \times \left(-\frac{3}{2}\right) = \frac{4}{7} + \frac{9}{10} = \frac{40}{70} + \frac{63}{70} = \frac{103}{70}$$ $$\frac{2}{3} - \frac{3}{5} \times \frac{5}{3} = \frac{2}{3} - 1 = -\frac{1}{3}$$ Row 2 becomes: $$[0, -\frac{21}{20}, \frac{103}{70}, -\frac{1}{3}]$$ Calculate row 3: $$-\frac{2}{3} + \frac{2}{3} \times 1 = 0$$ $$\frac{5}{6} + \frac{2}{3} \times \frac{4}{3} = \frac{5}{6} + \frac{8}{9} = \frac{15}{18} + \frac{16}{18} = \frac{31}{18}$$ $$\frac{3}{5} + \frac{2}{3} \times \left(-\frac{3}{2}\right) = \frac{3}{5} - 1 = -\frac{2}{5}$$ $$\frac{1}{2} + \frac{2}{3} \times \frac{5}{3} = \frac{1}{2} + \frac{10}{9} = \frac{9}{18} + \frac{20}{18} = \frac{29}{18}$$ Row 3 becomes: $$[0, \frac{31}{18}, -\frac{2}{5}, \frac{29}{18}]$$ 24. Multiply row 2 by $-\frac{20}{21}$ to get leading 1: $$R_2 \to -\frac{20}{21} R_2: [0, 1, -\frac{20}{21} \times \frac{103}{70}, \frac{20}{21} \times \frac{1}{3}]$$ Calculate: $$-\frac{20}{21} \times \frac{103}{70} = -\frac{2060}{1470} = -\frac{206}{147}$$ $$\frac{20}{21} \times \frac{1}{3} = \frac{20}{63}$$ Row 2 becomes: $$[0, 1, -\frac{206}{147}, \frac{20}{63}]$$ 25. Eliminate $y$ from rows 1 and 3: - $R_1 \to R_1 - \frac{4}{3} R_2$ - $R_3 \to R_3 - \frac{31}{18} R_2$ Calculate row 1: $$1, 0, -\frac{3}{2} - \frac{4}{3} \times \left(-\frac{206}{147}\right), \frac{5}{3} - \frac{4}{3} \times \frac{20}{63}$$ Calculate: $$-\frac{3}{2} + \frac{824}{441} = -\frac{661.5}{441} + \frac{824}{441} = \frac{162.5}{441} = \frac{325}{882}$$ $$\frac{5}{3} - \frac{80}{189} = \frac{315}{189} - \frac{80}{189} = \frac{235}{189}$$ Row 1 becomes: $$[1, 0, \frac{325}{882}, \frac{235}{189}]$$ Calculate row 3: $$0, 0, -\frac{2}{5} - \frac{31}{18} \times \left(-\frac{206}{147}\right), \frac{29}{18} - \frac{31}{18} \times \frac{20}{63}$$ Calculate: $$-\frac{2}{5} + \frac{31 \times 206}{18 \times 147} = -\frac{2}{5} + \frac{6386}{2646} = -\frac{2}{5} + 2.414 = 1.994$$ $$\frac{29}{18} - \frac{31 \times 20}{18 \times 63} = \frac{29}{18} - \frac{620}{1134} = 1.611 - 0.547 = 1.064$$ Row 3 becomes: $$[0, 0, 1.994, 1.064]$$ 26. Divide row 3 by 1.994 to get leading 1: $$R_3 \to \frac{1}{1.994} R_3: [0, 0, 1, \frac{1.064}{1.994} = 0.533]$$ 27. Eliminate $z$ from rows 1 and 2: - $R_1 \to R_1 - \frac{325}{882} R_3$ - $R_2 \to R_2 + \frac{206}{147} R_3$ Calculate row 1: $$z: \frac{325}{882} - \frac{325}{882} = 0$$ $$rhs: \frac{235}{189} - \frac{325}{882} \times 0.533 = 1.243 - 0.196 = 1.047$$ Row 1 becomes: $$[1, 0, 0, 1.047]$$ Calculate row 2: $$z: -\frac{206}{147} + \frac{206}{147} = 0$$ $$rhs: \frac{20}{63} + \frac{206}{147} \times 0.533 = 0.317 + 0.747 = 1.064$$ Row 2 becomes: $$[0, 1, 0, 1.064]$$ 28. Final solution: $$x = 1.047, \quad y = 1.064, \quad z = 0.533$$