Use the Gauss-Jordan method to solve the given system of equations. \[ \begin{aligned} x+y-z & =8 \\ 4 x-y+z & =2 \\ x-3 y+2 z & =-26 \end{aligned} \] Select the correct choice below and till in any answer boxes within your choice. A. There is one solution. The solution is \( \square \) \( \square \) , ) (Type an exact answer in simplified form.) B. There are infinitely many solutions. The solutions are \( \square \) , , \( z \) ), where \( z \) is any real number. C. There is no solution.

To solve the given system of equations using the Gauss-Jordan method, we first represent the system as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & -1 & | & 8 \\ 4 & -1 & 1 & | & 2 \\ 1 & -3 & 2 & | & -26 \end{bmatrix} \] Now, we apply row operations to transform this matrix into reduced row echelon form. 1. Swap \( R_1 \) and \( R_3 \): \[ \begin{bmatrix} 1 & -3 & 2 & | & -26 \\ 4 & -1 & 1 & | & 2 \\ 1 & 1 & -1 & | & 8 \end{bmatrix} \] 2. Replace \( R_2 \) with \( R_2 - 4R_1 \) and \( R_3 \) with \( R_3 - R_1 \): \[ \begin{bmatrix} 1 & -3 & 2 & | & -26 \\ 0 & 11 & -7 & | & 106 \\ 0 & 4 & -3 & | & 34 \end{bmatrix} \] 3. Now simplify further by replacing \( R_2 \) and making \( R_3 \) a multiple of \( R_2 \): Scale \( R_2 \): \[ \begin{bmatrix} 1 & -3 & 2 & | & -26 \\ 0 & 1 & -\frac{7}{11} & | & \frac{106}{11} \\ 0 & 4 & -3 & | & 34 \end{bmatrix} \] 4. Replace \( R_3 \) with \( R_3 - 4R_2 \): \[ \begin{bmatrix} 1 & -3 & 2 & | & -26 \\ 0 & 1 & -\frac{7}{11} & | & \frac{106}{11} \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \] At this point, we interpret the last row of zeros as a dependent equation. We can express \( z \) in terms of a free variable. From \( R_2 \): \[ y - \frac{7}{11}z = \frac{106}{11} \Rightarrow y = \frac{106}{11} + \frac{7}{11}z \] Substituting into \( R_1 \): \[ x - 3y + 2z = -26 \Rightarrow x - 3\left(\frac{106}{11} + \frac{7}{11}z\right) + 2z = -26 \] Solving gives: \[ x = -26 + \frac{318}{11} - \frac{21}{11}z + 2z = \frac{-286 + 318 + (2 - 21)z}{11} = \frac{32 - 19z}{11} \] Thus, the general solution form is: \[ (x, y, z) = \left(\frac{32 - 19z}{11}, \frac{106 + 7z}{11}, z\right) \] This means there are infinitely many solutions. The final answer is: B. There are infinitely many solutions. The solutions are \(\left( \frac{32 - 19z}{11}, \frac{106 + 7z}{11}, z \right) \), where \( z \) is any real number.