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].
