1. Problem No. 2: Solve the system of linear equations using the Gauss-Jordan method: $$\begin{cases} -x_1 + 4x_2 + x_3 = 8 \\ \frac{5}{3}x_1 + \frac{2}{3}x_2 + \frac{2}{3}x_3 = 1 \\ 2x_1 + x_2 + 4x_3 = 11 \end{cases}$$ 2. Write the augmented matrix: $$\left[\begin{array}{ccc|c} -1 & 4 & 1 & 8 \\ 5/3 & 2/3 & 2/3 & 1 \\ 2 & 1 & 4 & 11 \end{array}\right]$$ 3. Make the first pivot (row 1, column 1) 1 by multiplying row 1 by -1: $$\left[\begin{array}{ccc|c} 1 & -4 & -1 & -8 \\ 5/3 & 2/3 & 2/3 & 1 \\ 2 & 1 & 4 & 11 \end{array}\right]$$ 4. Eliminate $x_1$ from rows 2 and 3: - Row 2: $R_2 - \frac{5}{3}R_1 \to R_2$ $$\left[ 0, \frac{2}{3} - \frac{5}{3}(-4), \frac{2}{3} - \frac{5}{3}(-1), 1 - \frac{5}{3}(-8) \right] = [0, \frac{2}{3} + \frac{20}{3}, \frac{2}{3} + \frac{5}{3}, 1 + \frac{40}{3}] = [0, \frac{22}{3}, \frac{7}{3}, \frac{43}{3}]$$ - Row 3: $R_3 - 2R_1 \to R_3$ $$[0, 1 - 2(-4), 4 - 2(-1), 11 - 2(-8)] = [0, 1 + 8, 4 + 2, 11 + 16] = [0, 9, 6, 27]$$ So matrix becomes: $$\left[\begin{array}{ccc|c} 1 & -4 & -1 & -8 \\ 0 & \frac{22}{3} & \frac{7}{3} & \frac{43}{3} \\ 0 & 9 & 6 & 27 \end{array}\right]$$ 5. Make pivot in row 2, column 2 equal to 1 by multiplying row 2 by $\frac{3}{22}$: $$R_2 = \left[0, 1, \frac{7}{22}, \frac{43}{22}\right]$$ 6. Eliminate $x_2$ from rows 1 and 3: - Row 1: $R_1 + 4R_2 \to R_1$ $$[1, 0, -1 + 4(\frac{7}{22}), -8 + 4(\frac{43}{22})] = [1, 0, -1 + \frac{28}{22}, -8 + \frac{172}{22}] = [1, 0, -1 + \frac{14}{11}, -8 + \frac{86}{11}] = [1, 0, \frac{3}{11}, -\frac{2}{11}]$$ - Row 3: $R_3 - 9R_2 \to R_3$ $$[0, 0, 6 - 9(\frac{7}{22}), 27 - 9(\frac{43}{22})] = [0, 0, 6 - \frac{63}{22}, 27 - \frac{387}{22}] = [0, 0, \frac{69}{22}, \frac{207}{22}]$$ 7. Make pivot in row 3, column 3 equal to 1 by multiplying row 3 by $\frac{22}{69}$: $$R_3 = [0, 0, 1, 3]$$ 8. Eliminate $x_3$ from rows 1 and 2: - Row 1: $R_1 - \frac{3}{11}R_3 \to R_1$ $$[1, 0, 0, -\frac{2}{11} - \frac{3}{11} \times 3] = [1, 0, 0, -\frac{2}{11} - \frac{9}{11}] = [1, 0, 0, -1]$$ - Row 2: $R_2 - \frac{7}{22}R_3 \to R_2$ $$[0, 1, 0, \frac{43}{22} - \frac{7}{22} \times 3] = [0, 1, 0, \frac{43}{22} - \frac{21}{22}] = [0, 1, 0, 1]$$ 9. Final solution: $$x_1 = -1, \quad x_2 = 1, \quad x_3 = 3$$ --- Since the user asked only to solve Question 2, Question 3 and graph descriptions are ignored per instructions.