Answer :
To solve the system of equations using the Gauss-Jordan method, we first write the system as an augmented matrix:
[tex]\[ \begin{array}{l} y = -4 + x \\ y = -5 + z \\ z = 3 - x \end{array} \][/tex]
This can be rewritten in a more standard format:
[tex]\[ \begin{array}{l} x - y = 4 \\ y - z = -5 \\ x + z = 3 \end{array} \][/tex]
We convert this to an augmented matrix:
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 1 & 0 & 1 & \vert & 3 \end{pmatrix} \][/tex]
We will now use row operations to convert this to row-echelon form and eventually reduced row-echelon form.
1. Subtract Row 1 from Row 3:
[tex]\[ R3 = R3 - R1 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 1 & 1 & \vert & -1 \end{pmatrix} \][/tex]
2. Subtract Row 2 from Row 3:
[tex]\[ R3 = R3 - R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 0 & 2 & \vert & 4 \end{pmatrix} \][/tex]
3. Divide Row 3 by 2:
[tex]\[ R3 = \frac{R3}{2} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
4. Eliminate the z-term from Row 2 by adding Row 3 multiplied by 1:
[tex]\[ R2 = R2 + R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & 0 & \vert & -3 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
5. Finally, eliminate the y-term from Row 1 by adding Row 2:
[tex]\[ R1 = R1 + R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & \vert & 1 \\ 0 & 1 & 0 & \vert & -3 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
The resulting augmented matrix represents the system:
[tex]\[ \begin{cases} x = 1 \\ y = -3 \\ z = 2 \end{cases} \][/tex]
Thus, the solution to the system is:
[tex]\[ \boxed{1, -3, 2} \][/tex]
Therefore, the correct choice is:
A. There is one solution. The solution is [tex]\( (1, -3, 2) \)[/tex].
In the order [tex]\( x, y, z \)[/tex].
[tex]\[ \begin{array}{l} y = -4 + x \\ y = -5 + z \\ z = 3 - x \end{array} \][/tex]
This can be rewritten in a more standard format:
[tex]\[ \begin{array}{l} x - y = 4 \\ y - z = -5 \\ x + z = 3 \end{array} \][/tex]
We convert this to an augmented matrix:
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 1 & 0 & 1 & \vert & 3 \end{pmatrix} \][/tex]
We will now use row operations to convert this to row-echelon form and eventually reduced row-echelon form.
1. Subtract Row 1 from Row 3:
[tex]\[ R3 = R3 - R1 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 1 & 1 & \vert & -1 \end{pmatrix} \][/tex]
2. Subtract Row 2 from Row 3:
[tex]\[ R3 = R3 - R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 0 & 2 & \vert & 4 \end{pmatrix} \][/tex]
3. Divide Row 3 by 2:
[tex]\[ R3 = \frac{R3}{2} \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & -1 & \vert & -5 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
4. Eliminate the z-term from Row 2 by adding Row 3 multiplied by 1:
[tex]\[ R2 = R2 + R3 \][/tex]
[tex]\[ \begin{pmatrix} 1 & -1 & 0 & \vert & 4 \\ 0 & 1 & 0 & \vert & -3 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
5. Finally, eliminate the y-term from Row 1 by adding Row 2:
[tex]\[ R1 = R1 + R2 \][/tex]
[tex]\[ \begin{pmatrix} 1 & 0 & 0 & \vert & 1 \\ 0 & 1 & 0 & \vert & -3 \\ 0 & 0 & 1 & \vert & 2 \end{pmatrix} \][/tex]
The resulting augmented matrix represents the system:
[tex]\[ \begin{cases} x = 1 \\ y = -3 \\ z = 2 \end{cases} \][/tex]
Thus, the solution to the system is:
[tex]\[ \boxed{1, -3, 2} \][/tex]
Therefore, the correct choice is:
A. There is one solution. The solution is [tex]\( (1, -3, 2) \)[/tex].
In the order [tex]\( x, y, z \)[/tex].
