1. **State the problem:** Solve the system of equations using Gauss-Jordan elimination: $$\begin{cases} x + y - z = -1 \\ -x + 5y + 19z = 9 \\ -5x + 5y + 3z = 8 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & 1 & -1 & -1 \\ -1 & 5 & 19 & 9 \\ -5 & 5 & 3 & 8 \end{array}\right]$$ 3. **Step 1: Make the pivot in row 1, column 1 equal to 1 (already done).** 4. **Step 2: Eliminate the first column entries below and above the pivot:** - Add row 1 to row 2: $$R_2 \to R_2 + R_1: \left[\begin{array}{ccc|c} 1 & 1 & -1 & -1 \\ 0 & 6 & 18 & 8 \\ -5 & 5 & 3 & 8 \end{array}\right]$$ - Add 5 times row 1 to row 3: $$R_3 \to R_3 + 5R_1: \left[\begin{array}{ccc|c} 1 & 1 & -1 & -1 \\ 0 & 6 & 18 & 8 \\ 0 & 10 & -2 & 3 \end{array}\right]$$ 5. **Step 3: Make pivot in row 2, column 2 equal to 1 by dividing row 2 by 6:** $$R_2 \to \frac{1}{6} R_2: \left[\begin{array}{ccc|c} 1 & 1 & -1 & -1 \\ 0 & \cancel{6} \frac{1}{\cancel{6}} & 3 & \frac{4}{3} \\ 0 & 10 & -2 & 3 \end{array}\right]$$ 6. **Step 4: Eliminate the second column entries above and below the pivot:** - Subtract row 2 from row 1: $$R_1 \to R_1 - R_2: \left[\begin{array}{ccc|c} 1 & 0 & -4 & -\frac{7}{3} \\ 0 & 1 & 3 & \frac{4}{3} \\ 0 & 10 & -2 & 3 \end{array}\right]$$ - Subtract 10 times row 2 from row 3: $$R_3 \to R_3 - 10R_2: \left[\begin{array}{ccc|c} 1 & 0 & -4 & -\frac{7}{3} \\ 0 & 1 & 3 & \frac{4}{3} \\ 0 & 0 & -32 & -\frac{31}{3} \end{array}\right]$$ 7. **Step 5: Make pivot in row 3, column 3 equal to 1 by dividing row 3 by -32:** $$R_3 \to \frac{1}{-32} R_3: \left[\begin{array}{ccc|c} 1 & 0 & -4 & -\frac{7}{3} \\ 0 & 1 & 3 & \frac{4}{3} \\ 0 & 0 & \cancel{-32} \frac{1}{\cancel{-32}} & \frac{31}{96} \end{array}\right]$$ 8. **Step 6: Eliminate the third column entries above the pivot:** - Add 4 times row 3 to row 1: $$R_1 \to R_1 + 4R_3: \left[\begin{array}{ccc|c} 1 & 0 & 0 & -\frac{7}{3} + 4 \times \frac{31}{96} \\ 0 & 1 & 3 & \frac{4}{3} \\ 0 & 0 & 1 & \frac{31}{96} \end{array}\right]$$ Calculate: $$-\frac{7}{3} + \frac{124}{96} = -\frac{224}{96} + \frac{124}{96} = -\frac{100}{96} = -\frac{25}{24}$$ - Subtract 3 times row 3 from row 2: $$R_2 \to R_2 - 3R_3: \left[\begin{array}{ccc|c} 1 & 0 & 0 & -\frac{25}{24} \\ 0 & 1 & 0 & \frac{4}{3} - 3 \times \frac{31}{96} \\ 0 & 0 & 1 & \frac{31}{96} \end{array}\right]$$ Calculate: $$\frac{4}{3} - \frac{93}{96} = \frac{128}{96} - \frac{93}{96} = \frac{35}{96}$$ 9. **Final solution:** $$x = -\frac{25}{24}, \quad y = \frac{35}{96}, \quad z = \frac{31}{96}$$ **Answer:** $$\boxed{x = -\frac{25}{24}, \quad y = \frac{35}{96}, \quad z = \frac{31}{96}}$$