Use the Gauss-Jordan method to solve the system of equations:

[tex]\[
\begin{array}{l}
y = -4 + x \\
y = -5 + z \\
z = 3 - x
\end{array}
\][/tex]

Select the correct choice below and fill in any answer boxes within your choice.

A. There is one solution. The solution is [tex]\(\square, \square, \square\)[/tex] (in the order [tex]\(x, y, z\)[/tex]).

B. There are infinitely many solutions. The solution is [tex]\((\square, \square, z)\)[/tex], where [tex]\(z\)[/tex] is any real number.

C. There is no solution.



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].