Use the Gauss-Jordan method to solve the given system of equations. \[ \begin{aligned} x+2 y+z & =8 \\ 2 x+y-5 z & =1 \\ 3 x+y+2 z & =-1 \end{aligned} \] What is the reduced row echelon form of the augmented matrix? (Simplify your answers.) \( \square \) \( \square \) \( \square \) \( \square \) \( \square \)

To solve the given system of equations using the Gauss-Jordan method, we start by writing the augmented matrix: \[ \begin{bmatrix} 1 & 2 & 1 & | & 8 \\ 2 & 1 & -5 & | & 1 \\ 3 & 1 & 2 & | & -1 \end{bmatrix} \] Next, we perform row operations to achieve reduced row echelon form (RREF). 1. **R1** is already fine: \( R_1 \rightarrow R_1 \). 2. Eliminate \( x \) from **R2** and **R3**: \[ R_2 \rightarrow R_2 - 2R_1 \Rightarrow [0, -3, -7 | -15] \] \[ R_3 \rightarrow R_3 - 3R_1 \Rightarrow [0, -5, -1 | -25] \] The matrix now looks like: \[ \begin{bmatrix} 1 & 2 & 1 & | & 8 \\ 0 & -3 & -7 & | & -15 \\ 0 & -5 & -1 & | & -25 \end{bmatrix} \] 3. Next, we simplify **R2** and **R3**: Divide **R2** by -3: \[ R_2 \rightarrow -\frac{1}{3} R_2 \Rightarrow [0, 1, \frac{7}{3} | 5] \] 4. Then, make R3 dependent on R2: \[ R_3 \rightarrow R_3 + 5R_2 \Rightarrow [0, 0, \frac{8}{3} | 0] \] The matrix is now: \[ \begin{bmatrix} 1 & 2 & 1 & | & 8 \\ 0 & 1 & \frac{7}{3} & | & 5 \\ 0 & 0 & 1 & | & 0 \end{bmatrix} \] 5. Finally, we back substitute to achieve the leading 1s in each pivot column: \[ R_1 \rightarrow R_1 - R_3 \quad R_2 \rightarrow R_2 - \frac{7}{3}R_3 \] After performing these steps, our augmented matrix simplifies to: \[ \begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & 0 \end{bmatrix} \] So, the reduced row echelon form of the augmented matrix is: \[ \begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 5 \\ 0 & 0 & 1 & | & 0 \end{bmatrix} \] The final solutions for \( x, y, z \) are: \[ x = 1, \quad y = 5, \quad z = 0 \]