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.) \( \square \) \( \square \) \( \square \) \( \square \) \( \square \)

Solve the system of equations \( x+2y+z=8;2x+y-5z=1;3x+y+2z=-1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y+z=8\\2x+y-5z=1\\3x+y+2z=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=8-2y-z\\2x+y-5z=1\\3x+y+2z=-1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}2\left(8-2y-z\right)+y-5z=1\\3\left(8-2y-z\right)+y+2z=-1\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}16-3y-7z=1\\24-5y-z=-1\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}16-3y-7z=1\\z=25-5y\end{array}\right.\) - step5: Substitute the value of \(z:\) \(16-3y-7\left(25-5y\right)=1\) - step6: Simplify: \(-159+32y=1\) - step7: Move the constant to the right side: \(32y=1+159\) - step8: Add the numbers: \(32y=160\) - step9: Divide both sides: \(\frac{32y}{32}=\frac{160}{32}\) - step10: Divide the numbers: \(y=5\) - step11: Substitute the value of \(y:\) \(z=25-5\times 5\) - step12: Calculate: \(z=0\) - step13: Substitute the values of \(y,z:\) \(x=8-2\times 5-0\) - step14: Substitute back: \(x=8-2\times 5\) - step15: Calculate: \(x=-2\) - step16: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=5\\z=0\end{array}\right.\) - step17: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=5\\z=0\end{array}\right.\) - step18: Rewrite: \(\left(x,y,z\right) = \left(-2,5,0\right)\) The solution to the system of equations is \( x = -2, y = 5, z = 0 \). To find the reduced row echelon form of the augmented matrix, we need to perform row operations on the augmented matrix. The augmented matrix for the system of equations is: \[ \begin{bmatrix} 1 & 2 & 1 & 8 \\ 2 & 1 & -5 & 1 \\ 3 & 1 & 2 & -1 \\ \end{bmatrix} \] Performing row operations to reduce the matrix to reduced row echelon form, we get: \[ \begin{bmatrix} 1 & 2 & 1 & 8 \\ 0 & -3 & -7 & -15 \\ 0 & -5 & -1 & -25 \\ \end{bmatrix} \] \[ \begin{bmatrix} 1 & 2 & 1 & 8 \\ 0 & 1 & \frac{7}{3} & 5 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \] \[ \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \] Therefore, the reduced row echelon form of the augmented matrix is: \[ \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \]