Use the​ Gauss-Jordan method to solve the following system of equations.
x
plus
2y
minus
w
equals
3
4x
plus
8z
plus
4w
equals
negative 12
x
plus
2y
minus
z
equals
6
4x
minus
y
plus
nothingz
plus
nothingw
equals
negative 3
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The unique solution is left parenthesis nothing comma nothing comma nothing comma nothing right parenthesis
  
  
  
  
​, in the order​ x, y,​ z, w.
​(Simplify your​ answers.)
B.
There are infinitely many solutions. The solution is left parenthesis nothing comma nothing comma nothing comma w right parenthesis
  
  
  
​, in the order​ x, y,​ z, w, where w is any real number.
​(Simplify your answers. Use integers or fractions for any numbers in the​ expressions.)
C.
There is no solution.

To solve the given system of equations using the Gauss-Jordan elimination method, we start by representing the system as an augmented matrix and then perform row operations to reduce it to its reduced row echelon form (RREF).

### The System of Equations
Given:
1. \( x + 2y - w = 3 \)
2. \( 4x + 8z + 4w = -12 \)
3. \( x + 2y - z = 6 \)
4. \( 4x - y = -3 \)

### Step 1: Write the Augmented Matrix
We represent the system in augmented matrix form:

\[
\begin{bmatrix}
1 & 2 & 0 & -1 & \vert & 3 \\
4 & 0 & 8 & 4 & \vert & -12 \\
1 & 2 & -1 & 0 & \vert & 6 \\
4 & -1 & 0 & 0 & \vert & -3 
\end{bmatrix}
\]

### Step 2: Perform Gauss-Jordan Elimination

We'll perform row operations to reduce this matrix to its reduced row echelon form.

1. **Normalize Row 1** (R1):

\[
\text{R1: } \frac{1}{1} \times \text{R1} \implies \begin{bmatrix} 1 & 2 & 0 & -1 & \vert & 3 \end{bmatrix}
\]

2. **Eliminate the 4 in R2** (using R1):

\[
\text{R2: } \text{R2} - 4 \times \text{R1} \implies \text{R2: } (4 - 4) = 0, (0 - 8) = -8, (8 - 0) = 8, (4 + 4) = 8, (-12 - 12) = -24
\]

So,

\[
\text{New R2: } 0, -8, 8, 8, -24.
\]

Proceed with rest steps to reduce the matrix to RREF format.
The reduced row echelon form (RREF) of the augmented matrix is:

\[
\begin{bmatrix}
1 & 0 & 0 & 0 & -\frac{3}{14} \\
0 & 1 & 0 & 0 & \frac{15}{7} \\
0 & 0 & 1 & 0 & -\frac{27}{14} \\
0 & 0 & 0 & 1 & \frac{15}{14}
\end{bmatrix}
\]

### Interpretation of the RREF
Each row corresponds to an equation in the system:

1. \( x = -\frac{3}{14} \)
2. \( y = \frac{15}{7} \)
3. \( z = -\frac{27}{14} \)
4. \( w = \frac{15}{14} \)

### Conclusion
The system has a unique solution:

\[
(x, y, z, w) = \left( -\frac{3}{14}, \frac{15}{7}, -\frac{27}{14}, \frac{15}{14} \right)
\]

So, the correct choice is:

**A. The unique solution is** \(\left( -\frac{3}{14}, \frac{15}{7}, -\frac{27}{14}, \frac{15}{14} \right)\).

Would you like more details, or do you have any questions?

### Related Questions:
1. How does the Gauss-Jordan elimination method work in general?
2. How would you solve the system of equations using the inverse matrix method?
3. What are some cases where a system of linear equations has infinitely many solutions?
4. Can you explain how to determine if a system is consistent or inconsistent?
5. What are the differences between Gaussian elimination and Gauss-Jordan elimination?

**Tip:** When performing Gauss-Jordan elimination, always start by making the pivot elements 1 and systematically eliminate the other elements in their respective columns.

Use the Gauss-Jordan method to solve the following system of equations: x + 2y - w = 3, 4x + 8z + 4w = -12, x + 2y - z = 6, 4x - y = -3. Select the correct choice below: A. The unique solution is (x, y, z, w). B. There are infinitely many solutions. C. There is no solution.

Learn how to solve a system of equations using the Gauss-Jordan elimination method. This step-by-step guide explains how to transform the system into reduced row echelon form (RREF) and interpret the results to find the unique solution. Perfect for college students studying linear algebra.

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