Use the Gauss-Jordan method to solve the system of equations. \[ \begin{array}{l} y=x-4 \\ y=-3+z \\ z=5-x \end{array} \] Select the correct choice below and fill in any answer boxes within your choice. A. There is one solution. The solution is \( (\square, \square) \), in the order \( x, y, z \). \( \square \) \( \square \) (Type an exact answer in simplified form.) B. There are infinitely many solutions. The solution is \( \square \) \( \square \) , \( z \) ), where \( z \) is any real number. (Type an exact answer in simplified form.) C. There is no solution.

To solve the system of equations using the Gauss-Jordan method, we first rewrite the equations in standard form: 1. \( y - x + 4 = 0 \) (which is \( -x + y + 4 = 0 \)) 2. \( y + 3 - z = 0 \) (which is \( y - z + 3 = 0 \)) 3. \( z + x - 5 = 0 \) (which is \( x + z - 5 = 0 \)) This can be expressed in augmented matrix form: \[ \begin{bmatrix} -1 & 1 & 0 & | & -4 \\ 0 & 1 & -1 & | & -3 \\ 1 & 0 & 1 & | & 5 \end{bmatrix} \] Now, we perform row operations to reach reduced row echelon form. First, let's interchange Row 1 and Row 3 for easier manipulation. \[ \begin{bmatrix} 1 & 0 & 1 & | & 5 \\ 0 & 1 & -1 & | & -3 \\ -1 & 1 & 0 & | & -4 \end{bmatrix} \] Next, we eliminate the first column of Row 3: - Row 3 + Row 1 (to eliminate the -1 in Row 3): \[ R3 = R3 + R1 \] \[ \begin{bmatrix} 1 & 0 & 1 & | & 5 \\ 0 & 1 & -1 & | & -3 \\ 0 & 1 & 1 & | & 1 \end{bmatrix} \] Next, eliminate Row 2 from Row 3: - Row 3 - Row 2: \[ R3 = R3 - R2 \] \[ \begin{bmatrix} 1 & 0 & 1 & | & 5 \\ 0 & 1 & -1 & | & -3 \\ 0 & 0 & 2 & | & 4 \end{bmatrix} \] Next, let's simplify the Row 3: \[ R3 = \frac{1}{2} R3 \] \[ \begin{bmatrix} 1 & 0 & 1 & | & 5 \\ 0 & 1 & -1 & | & -3 \\ 0 & 0 & 1 & | & 2 \end{bmatrix} \] Now, we will back substitute to eliminate \( z \) from Row 1 and Row 2: - Row 1 - Row 3 (using \( z \)): \[ R1 = R1 - R3 \] \[ \begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & -1 & | & -3 \\ 0 & 0 & 1 & | & 2 \end{bmatrix} \] Now, substitute \( z = 2 \) into Row 2: \[ R2 = R2 + R3 \] \[ \begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & -1 \\ 0 & 0 & 1 & | & 2 \end{bmatrix} \] So, we have: - \( x = 3 \) - \( y = -1 \) - \( z = 2 \) Putting it all together, the solution is \( (3, -1, 2) \). Therefore, the correct choice is: A. There is one solution. The solution is \( (3, -1, 2) \).