Solve using​ Gauss-Jordan elimination.

33x 1x1minus−

7 x 27x2

minus−

5 x 3 equals5x3=

2626

x 1x1minus−

4 x 24x2

equals=

1212

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Part 1

Select the correct choice below and fill in the answer​ box(es) within your choice.

A.

The unique solution is 

x 1x1equals=enter your response here​, 

x 2x2equals=enter your response here​, 

and 

x 3x3equals=enter your response here.

B.

The system has infinitely many solutions. The solution is 

x 1x1equals=enter your response here​, 

x 2x2equals=enter your response here​, 

and 

x 3x3equals=t.

​(Simplify your answers. Type expressions using t as the​ variable.)

C.

The system has infinitely many solutions. The solution is 

x 1x1equals=enter your response here​, 

x 2x2equals=​s, 

and 

x 3x3equals=t.

​(Simplify your answer. Type an expression using s and t as the​ variables.)

D.

There is no solution.

Let's first organize the given system of equations and solve it using Gauss-Jordan elimination.

### Given System of Equations:

1. \( 33x_1 - 7x_2 - 5x_3 = 26 \)
2. \( x_1 - 4x_2 = 12 \)

### Step 1: Write the augmented matrix

We write the augmented matrix for the system of equations:

\[
\begin{pmatrix}
33 & -7 & -5 & | & 26 \\
1 & -4 & 0 & | & 12
\end{pmatrix}
\]

### Step 2: Perform row operations to reduce to row-echelon form

Our goal is to reduce the matrix to the form where we can easily read off the solutions. We will apply Gauss-Jordan elimination.

1. We already have a 1 in the second row, first column. We'll use that to eliminate the first entry of the first row.
   
   Subtract \( 33 \times \text{(row 2)} \) from row 1:
   \[
   R_1 \rightarrow R_1 - 33R_2
   \]
   Perform the calculation:
   \[
   (33, -7, -5 | 26) - 33 \times (1, -4, 0 | 12)
   \]
   \[
   = (33, -7, -5 | 26) - (33, -132, 0 | 396)
   \]
   \[
   = (0, 125, -5 | -370)
   \]

So the augmented matrix now looks like:
\[
\begin{pmatrix}
0 & 125 & -5 & | & -370 \\
1 & -4 & 0 & | & 12
\end{pmatrix}
\]

### Step 3: Make the coefficient of \(x_2\) in row 1 a 1

Divide the first row by 125:
\[
R_1 \rightarrow \frac{R_1}{125}
\]
\[
\left( 0, 125, -5 | -370 \right) \rightarrow \left( 0, 1, -\frac{1}{25} | -\frac{74}{25} \right)
\]

The new augmented matrix is:
\[
\begin{pmatrix}
0 & 1 & -\frac{1}{25} & | & -\frac{74}{25} \\
1 & -4 & 0 & | & 12
\end{pmatrix}
\]

### Step 4: Eliminate the \(x_2\) term from the second row

Add \(4 \times \text{(row 1)}\) to row 2:
\[
R_2 \rightarrow R_2 + 4R_1
\]
Perform the calculation:
\[
(1, -4, 0 | 12) + 4 \times (0, 1, -\frac{1}{25} | -\frac{74}{25})
\]
\[
= (1, -4, 0 | 12) + (0, 4, -\frac{4}{25} | -\frac{296}{25})
\]
\[
= (1, 0, -\frac{4}{25} | \frac{4}{25})
\]

So the augmented matrix is now:
\[
\begin{pmatrix}
0 & 1 & -\frac{1}{25} & | & -\frac{74}{25} \\
1 & 0 & -\frac{4}{25} & | & \frac{4}{25}
\end{pmatrix}
\]

### Step 5: Solve for the variables

From row 1:
\[
x_2 - \frac{1}{25}x_3 = -\frac{74}{25}
\]
\[
x_2 = -\frac{74}{25} + \frac{1}{25}x_3
\]

From row 2:
\[
x_1 - \frac{4}{25}x_3 = \frac{4}{25}
\]
\[
x_1 = \frac{4}{25} + \frac{4}{25}x_3
\]

### Step 6: Conclusion

Since \(x_3\) is free (it can take any value), the system has infinitely many solutions.

The solution is:
\[
x_1 = \frac{4}{25} + \frac{4}{25}t, \quad x_2 = -\frac{74}{25} + \frac{1}{25}t, \quad x_3 = t
\]
where \(t\) is a free variable.

### Final Answer:

The correct choice is **B**:
\[
x_1 = \frac{4}{25} + \frac{4}{25}t, \quad x_2 = -\frac{74}{25} + \frac{1}{25}t, \quad x_3 = t
\]

---

Would you like more details or have any questions? Here are five related problems to try:

1. Solve a 3x3 system of linear equations using Gauss-Jordan elimination.
2. What are the advantages of Gauss-Jordan elimination over Gaussian elimination?
3. How do free variables impact the solution set of a system of linear equations?
4. Can every system of linear equations be solved using Gauss-Jordan elimination?
5. What is the geometric interpretation of a system of linear equations with infinitely many solutions?

**Tip**: Always check if your matrix is in reduced row-echelon form before concluding the solution.

Solve using Gauss-Jordan elimination. 
33x1 - 7x2 - 5x3 = 26
x1 - 4x2 = 12

Select the correct choice:
A. The unique solution is x1 = ?, x2 = ?, x3 = ?
B. The system has infinitely many solutions. The solution is x1 = ?, x2 = ?, x3 = t.
C. The system has infinitely many solutions. The solution is x1 = ?, x2 = s, x3 = t.
D. There is no solution.

Learn how to solve a system of linear equations using Gauss-Jordan elimination. This problem involves reducing an augmented matrix and interpreting the results to determine if the system has one solution, infinitely many solutions, or no solution. Ideal for college-level algebra students.

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