Solve the system of equations using matrices. Use the Gauss-Jordal \( \left\{\begin{array}{l}x+y+z=-1 \\ x-y-z=-1 \\ x-y+z=5\end{array}\right. \)

To solve the system using Gauss-Jordan elimination, we first write the augmented matrix for the system of equations: \[ \begin{bmatrix} 1 & 1 & 1 & | & -1 \\ 1 & -1 & -1 & | & -1 \\ 1 & -1 & 1 & | & 5 \end{bmatrix} \] Next, we perform row operations to achieve reduced row echelon form. 1. Subtract row 1 from rows 2 and 3: - Row 2: \( R_2 - R_1 \rightarrow R_2 \: \Rightarrow \:[0, -2, -2 | 0] \) - Row 3: \( R_3 - R_1 \rightarrow R_3 \: \Rightarrow \:[0, -2, 0 | 6] \) The augmented matrix now is: \[ \begin{bmatrix} 1 & 1 & 1 & | & -1 \\ 0 & -2 & -2 & | & 0 \\ 0 & -2 & 0 & | & 6 \end{bmatrix} \] 2. We can simplify the second row by dividing it by -2: - Row 2: \( \frac{1}{-2} R_2 \rightarrow R_2 \: \Rightarrow \:[0, 1, 1 | 0] \) 3. We'll do the same for Row 3: - Row 3: \( \frac{1}{-2} R_3 \rightarrow R_3 \: \Rightarrow \:[0, 1, 0 | -3] \) Now our matrix is: \[ \begin{bmatrix} 1 & 1 & 1 & | & -1 \\ 0 & 1 & 1 & | & 0 \\ 0 & 1 & 0 & | & -3 \end{bmatrix} \] 4. Next, we'll eliminate the second entry in Row 1 by subtracting row 2 from it: - Row 1: \( R_1 - R_2 \rightarrow R_1 \: \Rightarrow \:[1, 0, 0 | -1] \) 5. We'll eliminate the second entry from Row 3: - Row 3: \( R_3 - R_2 \rightarrow R_3 \: \Rightarrow \:[0, 0, -1 | -3] \) Our matrix now looks like: \[ \begin{bmatrix} 1 & 0 & 0 & | & -1 \\ 0 & 1 & 1 & | & 0 \\ 0 & 0 & -1 & | & -3 \end{bmatrix} \] 6. Finally, we can multiply Row 3 by -1 to simplify: - Row 3: \( -R_3 \rightarrow R_3 \: \Rightarrow \:[0, 0, 1 | 3] \) Now we back substitute from here: The system now reads: 1. \( x = -1 \) 2. \( y + z = 0 \Rightarrow y = -z = -3 \) 3. \( z = 3 \) So, the solution to the system is: \[ x = -1, \quad y = -3, \quad z = 3 \] The final result is: \((x, y, z) = (-1, -3, 3)\) And there you have it! Solving this system with matrices can be quite an adventure, just like unraveling a mystery!