1. **Problem:** Solve the system of linear equations by Gauss elimination method for (i): $$\begin{cases} 2x + 3y + 4z = 2 \\ 2x + y + z = 5 \\ 3x - 2y + z = -3 \end{cases}$$ 2. **Step 1:** Write the augmented matrix: $$\left[\begin{array}{ccc|c} 2 & 3 & 4 & 2 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3 \end{array}\right]$$ 3. **Step 2:** Make zeros below the pivot in first column: - Subtract row 1 from row 2: $R_2 \to R_2 - R_1$ gives $[0, -2, -3, 3]$ - Subtract $\frac{3}{2}$ times row 1 from row 3: $R_3 \to R_3 - \frac{3}{2}R_1$ gives $[0, -6.5, -5, -6]$ 4. **Step 3:** Make zero below pivot in second column: - Multiply row 2 by $-\frac{1}{2}$: $R_2 = [0,1,1.5,-1.5]$ - Add $6.5$ times row 2 to row 3: $R_3 = [0,0,4.75,-15.75]$ 5. **Step 4:** Back substitution: - From $R_3$: $4.75z = -15.75 \Rightarrow z = -3$ - From $R_2$: $y + 1.5z = -1.5 \Rightarrow y + 1.5(-3) = -1.5 \Rightarrow y = 3$ - From $R_1$: $2x + 3y + 4z = 2 \Rightarrow 2x + 3(3) + 4(-3) = 2 \Rightarrow 2x + 9 - 12 = 2 \Rightarrow 2x - 3 = 2 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}$ 6. **Answer for (i):** $x = \frac{5}{2}, y = 3, z = -3$ --- 1. **Problem:** Solve the system of linear equations by Gauss elimination method for (ii): $$\begin{cases} 5x - 2y + z = 2 \\ 2x + 2y + 6z = 1 \\ 3x - 4y - 5z = 3 \end{cases}$$ 2. **Step 1:** Augmented matrix: $$\left[\begin{array}{ccc|c} 5 & -2 & 1 & 2 \\ 2 & 2 & 6 & 1 \\ 3 & -4 & -5 & 3 \end{array}\right]$$ 3. **Step 2:** Eliminate below pivot in first column: - $R_2 \to R_2 - \frac{2}{5}R_1 = [0, 2.8, 5.6, 0.2]$ - $R_3 \to R_3 - \frac{3}{5}R_1 = [0, -2.8, -6.6, 1.8]$ 4. **Step 3:** Eliminate below pivot in second column: - Add $R_2$ and $R_3$: $R_3 \to R_3 + R_2 = [0, 0, -1, 2]$ 5. **Step 4:** Back substitution: - From $R_3$: $-1z = 2 \Rightarrow z = -2$ - From $R_2$: $2.8y + 5.6z = 0.2 \Rightarrow 2.8y + 5.6(-2) = 0.2 \Rightarrow 2.8y - 11.2 = 0.2 \Rightarrow 2.8y = 11.4 \Rightarrow y = \frac{11.4}{2.8} = \frac{57}{14}$ - From $R_1$: $5x - 2y + z = 2 \Rightarrow 5x - 2(\frac{57}{14}) - 2 = 2 \Rightarrow 5x - \frac{114}{14} - 2 = 2 \Rightarrow 5x = 2 + 2 + \frac{114}{14} = 4 + \frac{57}{7} = \frac{28}{7} + \frac{57}{7} = \frac{85}{7} \Rightarrow x = \frac{17}{7}$ 6. **Answer for (ii):** $x = \frac{17}{7}, y = \frac{57}{14}, z = -2$ --- 1. **Problem:** Solve the system of linear equations by Gauss elimination method for (iii): $$\begin{cases} 2x + z = 2 \\ 2y - z = 3 \\ x + 3y = 5 \end{cases}$$ 2. **Step 1:** Express $z$ from first two equations: - From (1): $z = 2 - 2x$ - From (2): $z = 2y - 3$ 3. **Step 2:** Equate expressions for $z$: $$2 - 2x = 2y - 3 \Rightarrow 2x + 2y = 5$$ 4. **Step 3:** Use equation (3): $x + 3y = 5$ 5. **Step 4:** Solve system: $$\begin{cases} 2x + 2y = 5 \\ x + 3y = 5 \end{cases}$$ - Multiply second by 2: $2x + 6y = 10$ - Subtract first: $(2x + 6y) - (2x + 2y) = 10 - 5 \Rightarrow 4y = 5 \Rightarrow y = \frac{5}{4}$ - Substitute $y$ into $x + 3y = 5$: $x + 3(\frac{5}{4}) = 5 \Rightarrow x + \frac{15}{4} = 5 \Rightarrow x = 5 - \frac{15}{4} = \frac{20}{4} - \frac{15}{4} = \frac{5}{4}$ 6. **Step 5:** Find $z$: - $z = 2 - 2x = 2 - 2(\frac{5}{4}) = 2 - \frac{10}{4} = 2 - 2.5 = -0.5 = -\frac{1}{2}$ 7. **Answer for (iii):** $x = \frac{5}{4}, y = \frac{5}{4}, z = -\frac{1}{2}$ --- 1. **Problem:** Solve the system of linear equations by Gauss elimination method for (iv): $$\begin{cases} x + 2y + 5z = 4 \\ 3x - 2y + 2z = 3 \\ 5x - 8y - 4z = 1 \end{cases}$$ 2. **Step 1:** Augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 2 & 5 & 4 \\ 3 & -2 & 2 & 3 \\ 5 & -8 & -4 & 1 \end{array}\right]$$ 3. **Step 2:** Eliminate below pivot in first column: - $R_2 \to R_2 - 3R_1 = [0, -8, -13, -9]$ - $R_3 \to R_3 - 5R_1 = [0, -18, -29, -19]$ 4. **Step 3:** Eliminate below pivot in second column: - Multiply $R_2$ by $-\frac{1}{8}$: $R_2 = [0,1,\frac{13}{8}, \frac{9}{8}]$ - $R_3 \to R_3 - (-18)R_2 = [0,0, -29 + 18 \times \frac{13}{8}, -19 + 18 \times \frac{9}{8}]$ - Calculate: - $-29 + 18 \times \frac{13}{8} = -29 + \frac{234}{8} = -29 + 29.25 = 0.25$ - $-19 + 18 \times \frac{9}{8} = -19 + \frac{162}{8} = -19 + 20.25 = 1.25$ 5. **Step 4:** Back substitution: - From $R_3$: $0.25z = 1.25 \Rightarrow z = \frac{1.25}{0.25} = 5$ - From $R_2$: $y + \frac{13}{8}z = \frac{9}{8} \Rightarrow y + \frac{13}{8} \times 5 = \frac{9}{8} \Rightarrow y + \frac{65}{8} = \frac{9}{8} \Rightarrow y = \frac{9}{8} - \frac{65}{8} = -\frac{56}{8} = -7$ - From $R_1$: $x + 2y + 5z = 4 \Rightarrow x + 2(-7) + 5(5) = 4 \Rightarrow x - 14 + 25 = 4 \Rightarrow x + 11 = 4 \Rightarrow x = -7$ 6. **Answer for (iv):** $x = -7, y = -7, z = 5$