1. **Solve the system using Gauss Elimination Method:** Given system: $$\begin{cases} 4x - 5y + 6z = 12 \\ 7x + 3y - 2z = -5 \\ 5x - 4y + 8z = 10 \end{cases}$$ Step 1: Write the augmented matrix: $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right]$$ Step 2: Make the first pivot 1 by dividing row 1 by 4: $$\left[\begin{array}{ccc|c} \cancel{4} & \cancel{-5} & \cancel{6} & \cancel{12} \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right]$$ Step 3: Eliminate $x$ from rows 2 and 3: Row 2: $R_2 - 7R_1$: $$7 - 7(1) = 0, \quad 3 - 7(-\frac{5}{4}) = 3 + \frac{35}{4} = \frac{47}{4}, \quad -2 - 7(\frac{3}{2}) = -2 - \frac{21}{2} = -\frac{25}{2}, \quad -5 - 7(3) = -5 - 21 = -26$$ Row 3: $R_3 - 5R_1$: $$5 - 5(1) = 0, \quad -4 - 5(-\frac{5}{4}) = -4 + \frac{25}{4} = \frac{9}{4}, \quad 8 - 5(\frac{3}{2}) = 8 - \frac{15}{2} = \frac{1}{2}, \quad 10 - 5(3) = 10 - 15 = -5$$ New matrix: $$\left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & \frac{47}{4} & -\frac{25}{2} & -26 \\ 0 & \frac{9}{4} & \frac{1}{2} & -5 \end{array}\right]$$ Step 4: Make pivot in row 2, column 2 equal to 1 by dividing row 2 by $\frac{47}{4}$: $$\left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & \cancel{\frac{47}{4}} & \cancel{-\frac{25}{2}} & \cancel{-26} \\ 0 & \frac{9}{4} & \frac{1}{2} & -5 \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & 1 & -\frac{50}{47} & -\frac{104}{47} \\ 0 & \frac{9}{4} & \frac{1}{2} & -5 \end{array}\right]$$ Step 5: Eliminate $y$ from row 3: Row 3: $R_3 - \frac{9}{4} R_2$: $$0, \quad \frac{9}{4} - \frac{9}{4}(1) = 0, \quad \frac{1}{2} - \frac{9}{4}(-\frac{50}{47}) = \frac{1}{2} + \frac{450}{188} = \frac{1}{2} + \frac{225}{94} = \frac{47}{94} + \frac{225}{94} = \frac{272}{94} = \frac{136}{47},$$ $$-5 - \frac{9}{4}(-\frac{104}{47}) = -5 + \frac{936}{188} = -5 + \frac{468}{94} = -5 + 4.9787 = -0.0213 \approx -\frac{1}{47}$$ New matrix: $$\left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & 1 & -\frac{50}{47} & -\frac{104}{47} \\ 0 & 0 & \frac{136}{47} & -\frac{1}{47} \end{array}\right]$$ Step 6: Make pivot in row 3, column 3 equal to 1 by dividing row 3 by $\frac{136}{47}$: $$\left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & 1 & -\frac{50}{47} & -\frac{104}{47} \\ 0 & 0 & \cancel{\frac{136}{47}} & \cancel{-\frac{1}{47}} \end{array}\right] \to \left[\begin{array}{ccc|c} 1 & -\frac{5}{4} & \frac{3}{2} & 3 \\ 0 & 1 & -\frac{50}{47} & -\frac{104}{47} \\ 0 & 0 & 1 & -\frac{1}{136} \end{array}\right]$$ Step 7: Back substitution: From row 3: $z = -\frac{1}{136}$ Row 2: $y - \frac{50}{47}z = -\frac{104}{47} \Rightarrow y = -\frac{104}{47} + \frac{50}{47} \times \frac{1}{136} = -\frac{104}{47} + \frac{50}{6392} = -\frac{104}{47} + \frac{25}{3196}$ Calculate common denominator and sum: $$y = -\frac{104 \times 68}{3196} + \frac{25}{3196} = -\frac{7072}{3196} + \frac{25}{3196} = -\frac{7047}{3196}$$ Row 1: $x - \frac{5}{4}y + \frac{3}{2}z = 3$ Substitute $y$ and $z$: $$x = 3 + \frac{5}{4}y - \frac{3}{2}z = 3 + \frac{5}{4} \times \left(-\frac{7047}{3196}\right) - \frac{3}{2} \times \left(-\frac{1}{136}\right)$$ $$= 3 - \frac{5 \times 7047}{4 \times 3196} + \frac{3}{272} = 3 - \frac{35235}{12784} + \frac{3}{272}$$ Convert 3 to fraction with denominator 12784: $$3 = \frac{38352}{12784}$$ Convert $\frac{3}{272}$ to denominator 12784: $$\frac{3}{272} = \frac{141}{12784}$$ Sum: $$x = \frac{38352}{12784} - \frac{35235}{12784} + \frac{141}{12784} = \frac{38352 - 35235 + 141}{12784} = \frac{3258}{12784} = \frac{813}{3196}$$ **Solution:** $$x = \frac{813}{3196}, \quad y = -\frac{7047}{3196}, \quad z = -\frac{1}{136}$$ --- 2. **LU Factorization and solve:** Given system: $$\begin{cases} 8x + 4y + 2z = 20 \\ 4x + 5y + 3z = 18 \\ 2x + 3y + 6z = 22 \end{cases}$$ Coefficient matrix $A$: $$\begin{bmatrix} 8 & 4 & 2 \\ 4 & 5 & 3 \\ 2 & 3 & 6 \end{bmatrix}$$ Step 1: Decompose $A = LU$ where $L$ is lower triangular with 1s on diagonal, $U$ upper triangular. Calculate multipliers: $$l_{21} = \frac{4}{8} = \frac{1}{2}, \quad l_{31} = \frac{2}{8} = \frac{1}{4}$$ Compute $U$ first row: $$u_{11} = 8, u_{12} = 4, u_{13} = 2$$ Compute second row of $U$: $$u_{22} = 5 - l_{21} \times u_{12} = 5 - \frac{1}{2} \times 4 = 5 - 2 = 3$$ $$u_{23} = 3 - l_{21} \times u_{13} = 3 - \frac{1}{2} \times 2 = 3 - 1 = 2$$ Compute third row of $U$: $$u_{32} = 3 - l_{31} \times u_{12} = 3 - \frac{1}{4} \times 4 = 3 - 1 = 2$$ $$u_{33} = 6 - l_{31} \times u_{13} = 6 - \frac{1}{4} \times 2 = 6 - \frac{1}{2} = \frac{11}{2}$$ Matrix $L$ and $U$: $$L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{4} & l_{32} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} 8 & 4 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & \frac{11}{2} \end{bmatrix}$$ Find $l_{32}$: $$l_{32} = \frac{u_{32}}{u_{22}} = \frac{2}{3}$$ So, $$L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{4} & \frac{2}{3} & 1 \end{bmatrix}$$ Step 2: Solve $Ly = b$ where $b = \begin{bmatrix} 20 \\ 18 \\ 22 \end{bmatrix}$ Equations: $$y_1 = 20$$ $$\frac{1}{2} y_1 + y_2 = 18 \Rightarrow y_2 = 18 - \frac{1}{2} \times 20 = 18 - 10 = 8$$ $$\frac{1}{4} y_1 + \frac{2}{3} y_2 + y_3 = 22 \Rightarrow y_3 = 22 - \frac{1}{4} \times 20 - \frac{2}{3} \times 8 = 22 - 5 - \frac{16}{3} = 17 - \frac{16}{3} = \frac{51}{3} - \frac{16}{3} = \frac{35}{3}$$ Step 3: Solve $Ux = y$ Equations: $$8x + 4y + 2z = 20$$ $$3y + 2z = 8$$ $$\frac{11}{2} z = \frac{35}{3}$$ From last: $$z = \frac{35}{3} \times \frac{2}{11} = \frac{70}{33}$$ From second: $$3y + 2 \times \frac{70}{33} = 8 \Rightarrow 3y = 8 - \frac{140}{33} = \frac{264}{33} - \frac{140}{33} = \frac{124}{33} \Rightarrow y = \frac{124}{99}$$ From first: $$8x + 4 \times \frac{124}{99} + 2 \times \frac{70}{33} = 20$$ $$8x + \frac{496}{99} + \frac{140}{33} = 20$$ Convert $\frac{140}{33}$ to denominator 99: $$\frac{140}{33} = \frac{420}{99}$$ Sum: $$8x + \frac{496}{99} + \frac{420}{99} = 20 \Rightarrow 8x + \frac{916}{99} = 20$$ $$8x = 20 - \frac{916}{99} = \frac{1980}{99} - \frac{916}{99} = \frac{1064}{99} \Rightarrow x = \frac{1064}{792} = \frac{266}{198} = \frac{133}{99}$$ **Solution:** $$x = \frac{133}{99}, \quad y = \frac{124}{99}, \quad z = \frac{70}{33}$$ --- 3. **Solve system using Gauss-Jordan Elimination:** Given system: $$\begin{cases} \frac{1}{2}x + \frac{2}{3}y - \frac{3}{4}z = \frac{5}{6} \\ \frac{3}{5}x - \frac{1}{4}y + \frac{4}{7}z = \frac{2}{3} \\ -\frac{2}{3}x + \frac{5}{6}y + \frac{3}{5}z = \frac{1}{2} \end{cases}$$ Step 1: Write augmented matrix: $$\left[\begin{array}{ccc|c} \frac{1}{2} & \frac{2}{3} & -\frac{3}{4} & \frac{5}{6} \\ \frac{3}{5} & -\frac{1}{4} & \frac{4}{7} & \frac{2}{3} \\ -\frac{2}{3} & \frac{5}{6} & \frac{3}{5} & \frac{1}{2} \end{array}\right]$$ Step 2: Make pivot in row 1, column 1 equal to 1 by multiplying row 1 by 2: $$\left[\begin{array}{ccc|c} 1 & \frac{4}{3} & -\frac{3}{2} & \frac{5}{3} \\ \frac{3}{5} & -\frac{1}{4} & \frac{4}{7} & \frac{2}{3} \\ -\frac{2}{3} & \frac{5}{6} & \frac{3}{5} & \frac{1}{2} \end{array}\right]$$ Step 3: Eliminate $x$ from rows 2 and 3: Row 2: $R_2 - \frac{3}{5} R_1$: $$0, -\frac{1}{4} - \frac{3}{5} \times \frac{4}{3} = -\frac{1}{4} - \frac{4}{5} = -\frac{1}{4} - \frac{16}{20} = -\frac{1}{4} - \frac{4}{5} = -\frac{21}{20},$$ $$\frac{4}{7} - \frac{3}{5} \times (-\frac{3}{2}) = \frac{4}{7} + \frac{9}{10} = \frac{40}{70} + \frac{63}{70} = \frac{103}{70},$$ $$\frac{2}{3} - \frac{3}{5} \times \frac{5}{3} = \frac{2}{3} - 1 = -\frac{1}{3}$$ Row 3: $R_3 + \frac{2}{3} R_1$: $$0, \frac{5}{6} + \frac{2}{3} \times \frac{4}{3} = \frac{5}{6} + \frac{8}{9} = \frac{15}{18} + \frac{16}{18} = \frac{31}{18},$$ $$\frac{3}{5} + \frac{2}{3} \times (-\frac{3}{2}) = \frac{3}{5} - 1 = -\frac{2}{5},$$ $$\frac{1}{2} + \frac{2}{3} \times \frac{5}{3} = \frac{1}{2} + \frac{10}{9} = \frac{9}{18} + \frac{20}{18} = \frac{29}{18}$$ New matrix: $$\left[\begin{array}{ccc|c} 1 & \frac{4}{3} & -\frac{3}{2} & \frac{5}{3} \\ 0 & -\frac{21}{20} & \frac{103}{70} & -\frac{1}{3} \\ 0 & \frac{31}{18} & -\frac{2}{5} & \frac{29}{18} \end{array}\right]$$ Step 4: Make pivot in row 2, column 2 equal to 1 by multiplying row 2 by $-\frac{20}{21}$: $$\left[\begin{array}{ccc|c} 1 & \frac{4}{3} & -\frac{3}{2} & \frac{5}{3} \\ 0 & 1 & -\frac{103}{70} \times \frac{20}{21} & \frac{1}{3} \times \frac{20}{21} \\ 0 & \frac{31}{18} & -\frac{2}{5} & \frac{29}{18} \end{array}\right]$$ Calculate: $$-\frac{103}{70} \times \frac{20}{21} = -\frac{103 \times 20}{70 \times 21} = -\frac{2060}{1470} = -\frac{206}{147}$$ $$\frac{1}{3} \times \frac{20}{21} = \frac{20}{63}$$ Step 5: Eliminate $y$ from rows 1 and 3: Row 1: $R_1 - \frac{4}{3} R_2$: $$1, 0, -\frac{3}{2} - \frac{4}{3} \times (-\frac{206}{147}) = -\frac{3}{2} + \frac{824}{441} = -\frac{661.5}{441} + \frac{824}{441} = \frac{162.5}{441} = \frac{325}{882},$$ $$\frac{5}{3} - \frac{4}{3} \times \frac{20}{63} = \frac{5}{3} - \frac{80}{189} = \frac{315}{189} - \frac{80}{189} = \frac{235}{189}$$ Row 3: $R_3 - \frac{31}{18} R_2$: $$0, 0, -\frac{2}{5} - \frac{31}{18} \times (-\frac{206}{147}) = -\frac{2}{5} + \frac{6386}{2646} = -\frac{1058.4}{2646} + \frac{6386}{2646} = \frac{5327.6}{2646} = \frac{26638}{13230},$$ $$\frac{29}{18} - \frac{31}{18} \times \frac{20}{63} = \frac{29}{18} - \frac{620}{1134} = \frac{1827}{1134} - \frac{620}{1134} = \frac{1207}{1134}$$ Step 6: Make pivot in row 3, column 3 equal to 1 by dividing row 3 by $\frac{26638}{13230}$: $$z = \frac{1207}{1134} \div \frac{26638}{13230} = \frac{1207}{1134} \times \frac{13230}{26638} = \frac{1207 \times 13230}{1134 \times 26638}$$ Simplify numerator and denominator: $$z = \frac{15962610}{30225692} = \frac{795}{1507}$$ Step 7: Back substitute to find $y$: $$y = \frac{20}{63} - \frac{206}{147} z = \frac{20}{63} - \frac{206}{147} \times \frac{795}{1507}$$ Calculate: $$\frac{206}{147} \times \frac{795}{1507} = \frac{163770}{221529}$$ Convert $\frac{20}{63}$ to denominator 221529: $$\frac{20}{63} = \frac{7030}{221529}$$ Subtract: $$y = \frac{7030}{221529} - \frac{163770}{221529} = -\frac{156740}{221529}$$ Step 8: Back substitute to find $x$: $$x = \frac{235}{189} - \frac{325}{882} z = \frac{235}{189} - \frac{325}{882} \times \frac{795}{1507}$$ Calculate: $$\frac{325}{882} \times \frac{795}{1507} = \frac{258375}{1328154}$$ Convert $\frac{235}{189}$ to denominator 1328154: $$\frac{235}{189} = \frac{1650150}{1328154}$$ Subtract: $$x = \frac{1650150}{1328154} - \frac{258375}{1328154} = \frac{1391775}{1328154}$$ **Solution:** $$x = \frac{1391775}{1328154}, \quad y = -\frac{156740}{221529}, \quad z = \frac{795}{1507}$$