1. **State the problem:** Solve the system of mesh current equations using the Gauss Elimination Method: $$\begin{cases} 2i_1 + i_2 - i_3 + 4i_4 = 19 \\ -i_1 - 2i_2 + i_3 + 2i_4 = -3 \\ 2i_1 + 4i_2 + 2i_3 + i_4 = 25 \\ -i_1 + i_2 - i_3 - 2i_4 = -5 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{cccc|c} 2 & 1 & -1 & 4 & 19 \\ -1 & -2 & 1 & 2 & -3 \\ 2 & 4 & 2 & 1 & 25 \\ -1 & 1 & -1 & -2 & -5 \end{array}\right]$$ 3. **Step 1: Make the first pivot 1 by dividing row 1 by 2:** $$\left[\begin{array}{cccc|c} \cancel{2} & \cancel{1} & \cancel{-1} & \cancel{4} & \cancel{19} \\ -1 & -2 & 1 & 2 & -3 \\ 2 & 4 & 2 & 1 & 25 \\ -1 & 1 & -1 & -2 & -5 \end{array}\right] \to \left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ -1 & -2 & 1 & 2 & -3 \\ 2 & 4 & 2 & 1 & 25 \\ -1 & 1 & -1 & -2 & -5 \end{array}\right]$$ 4. **Step 2: Eliminate the first column entries below and above the pivot:** - Add row 1 to row 2: $$R_2 = R_2 + R_1: (-1+1, -2+\frac{1}{2}, 1 - \frac{1}{2}, 2+2, -3 + \frac{19}{2}) = (0, -\frac{3}{2}, \frac{1}{2}, 4, \frac{13}{2})$$ - Subtract 2 times row 1 from row 3: $$R_3 = R_3 - 2R_1: (2 - 2\cdot1, 4 - 2\cdot\frac{1}{2}, 2 - 2\cdot(-\frac{1}{2}), 1 - 2\cdot2, 25 - 2\cdot\frac{19}{2}) = (0, 3, 3, -3, 6)$$ - Add row 1 to row 4: $$R_4 = R_4 + R_1: (-1+1, 1 + \frac{1}{2}, -1 - \frac{1}{2}, -2 + 2, -5 + \frac{19}{2}) = (0, \frac{3}{2}, -\frac{3}{2}, 0, \frac{9}{2})$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & -\frac{3}{2} & \frac{1}{2} & 4 & \frac{13}{2} \\ 0 & 3 & 3 & -3 & 6 \\ 0 & \frac{3}{2} & -\frac{3}{2} & 0 & \frac{9}{2} \end{array}\right]$$ 5. **Step 3: Make pivot in row 2, column 2 equal to 1 by dividing row 2 by $-\frac{3}{2}$:** $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & \cancel{-\frac{3}{2}} & \cancel{\frac{1}{2}} & \cancel{4} & \cancel{\frac{13}{2}} \\ 0 & 3 & 3 & -3 & 6 \\ 0 & \frac{3}{2} & -\frac{3}{2} & 0 & \frac{9}{2} \end{array}\right] \to \left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 3 & 3 & -3 & 6 \\ 0 & \frac{3}{2} & -\frac{3}{2} & 0 & \frac{9}{2} \end{array}\right]$$ 6. **Step 4: Eliminate entries in column 2 for rows 3 and 4:** - Row 3: $R_3 = R_3 - 3R_2$ $$ (0, 3 - 3\cdot1, 3 - 3\cdot(-\frac{1}{3}), -3 - 3\cdot(-\frac{8}{3}), 6 - 3\cdot(-\frac{13}{3})) = (0, 0, 4, 5, 19)$$ - Row 4: $R_4 = R_4 - \frac{3}{2} R_2$ $$ (0, \frac{3}{2} - \frac{3}{2}\cdot1, -\frac{3}{2} - \frac{3}{2}\cdot(-\frac{1}{3}), 0 - \frac{3}{2}\cdot(-\frac{8}{3}), \frac{9}{2} - \frac{3}{2}\cdot(-\frac{13}{3})) = (0, 0, -1, 4, 11)$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & 4 & 5 & 19 \\ 0 & 0 & -1 & 4 & 11 \end{array}\right]$$ 7. **Step 5: Make pivot in row 3, column 3 equal to 1 by dividing row 3 by 4:** $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & \cancel{4} & \cancel{5} & \cancel{19} \\ 0 & 0 & -1 & 4 & 11 \end{array}\right] \to \left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & 1 & \frac{5}{4} & \frac{19}{4} \\ 0 & 0 & -1 & 4 & 11 \end{array}\right]$$ 8. **Step 6: Eliminate entry in row 4, column 3:** $$R_4 = R_4 + R_3: (0, 0, -1 + 1, 4 + \frac{5}{4}, 11 + \frac{19}{4}) = (0, 0, 0, \frac{21}{4}, \frac{63}{4})$$ Matrix now: $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & 1 & \frac{5}{4} & \frac{19}{4} \\ 0 & 0 & 0 & \frac{21}{4} & \frac{63}{4} \end{array}\right]$$ 9. **Step 7: Make pivot in row 4, column 4 equal to 1 by dividing row 4 by $\frac{21}{4}$:** $$\left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & 1 & \frac{5}{4} & \frac{19}{4} \\ 0 & 0 & 0 & \cancel{\frac{21}{4}} & \cancel{\frac{63}{4}} \end{array}\right] \to \left[\begin{array}{cccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 2 & \frac{19}{2} \\ 0 & 1 & -\frac{1}{3} & -\frac{8}{3} & -\frac{13}{3} \\ 0 & 0 & 1 & \frac{5}{4} & \frac{19}{4} \\ 0 & 0 & 0 & 1 & 3 \end{array}\right]$$ 10. **Step 8: Back substitution to find $i_4, i_3, i_2, i_1$:** - From row 4: $i_4 = 3$ - From row 3: $i_3 + \frac{5}{4}i_4 = \frac{19}{4} \Rightarrow i_3 = \frac{19}{4} - \frac{5}{4} \times 3 = \frac{19}{4} - \frac{15}{4} = 1$ - From row 2: $i_2 - \frac{1}{3}i_3 - \frac{8}{3}i_4 = -\frac{13}{3} \Rightarrow i_2 = -\frac{13}{3} + \frac{1}{3} \times 1 + \frac{8}{3} \times 3 = -\frac{13}{3} + \frac{1}{3} + 8 = \frac{-13 + 1 + 24}{3} = \frac{12}{3} = 4$ - From row 1: $i_1 + \frac{1}{2}i_2 - \frac{1}{2}i_3 + 2i_4 = \frac{19}{2} \Rightarrow i_1 = \frac{19}{2} - \frac{1}{2} \times 4 + \frac{1}{2} \times 1 - 2 \times 3 = \frac{19}{2} - 2 + \frac{1}{2} - 6 = \frac{19 + 1 - 4 - 12}{2} = \frac{4}{2} = 2$ **Final solution:** $$\boxed{i_1 = 2, \quad i_2 = 4, \quad i_3 = 1, \quad i_4 = 3}$$