The matrix given is in reduced echelon form.

1

0

3

0

4

3

0

1

−1

0

5

5

0

0

0

1

3

6

Write the system of equations represented by the matrix. (Use *x* as your variable and label each *x* with its corresponding column. Enter x_1 for *x*1, x_2 for *x*2, x_3 for *x*3, x_4 for *x*4, and x_5 for *x*5.)

row 1

= 3

row 2

= 5

row 3

= 6

Find the solution, if possible. (If there are infinitely many solutions, express *x*1, *x*2, *x*3, *x*4, and *x*5 in terms of the parameters *s* and *t*. If there is no solution, enter NO SOLUTION.)

The matrix represents a system of equations. To write the system, each row corresponds to a linear equation, and each column corresponds to a variable (x₁, x₂, x₃, x₄, x₅).

We are given the following reduced echelon form matrix:
\[
\begin{pmatrix}
1 & 0 & 3 & 0 & 4 & | & 3 \\
0 & 1 & -1 & 0 & 5 & | & 5 \\
0 & 0 & 0 & 1 & 3 & | & 6
\end{pmatrix}
\]

This matrix represents the system of equations:

### Step 1: Writing the equations
- Row 1: \( 1x_1 + 0x_2 + 3x_3 + 0x_4 + 4x_5 = 3 \), which simplifies to:
  \[
  x_1 + 3x_3 + 4x_5 = 3
  \]
- Row 2: \( 0x_1 + 1x_2 - 1x_3 + 0x_4 + 5x_5 = 5 \), which simplifies to:
  \[
  x_2 - x_3 + 5x_5 = 5
  \]
- Row 3: \( 0x_1 + 0x_2 + 0x_3 + 1x_4 + 3x_5 = 6 \), which simplifies to:
  \[
  x_4 + 3x_5 = 6
  \]

### Step 2: Finding the solution
- **Equation 3**: Solve for \( x_4 \):
  \[
  x_4 = 6 - 3x_5
  \]
- **Equation 2**: Solve for \( x_2 \):
  \[
  x_2 = 5 + x_3 - 5x_5
  \]
- **Equation 1**: Solve for \( x_1 \):
  \[
  x_1 = 3 - 3x_3 - 4x_5
  \]

### Conclusion:
Since \( x_3 \) and \( x_5 \) are free variables (they can take any value), the system has **infinitely many solutions**. We can express the solution as:

\[
x_1 = 3 - 3x_3 - 4x_5
\]
\[
x_2 = 5 + x_3 - 5x_5
\]
\[
x_4 = 6 - 3x_5
\]
\[
x_3 = s \quad (\text{where } s \text{ is any real number})
\]
\[
x_5 = t \quad (\text{where } t \text{ is any real number})
\]

Thus, the solution is:
\[
x_1 = 3 - 3s - 4t, \quad x_2 = 5 + s - 5t, \quad x_3 = s, \quad x_4 = 6 - 3t, \quad x_5 = t
\]

### Summary:
There are infinitely many solutions, and they are expressed in terms of the parameters \( s \) and \( t \).

Would you like more details or have any questions? Here are some related questions to explore:

1. How do you convert a matrix into reduced row echelon form?
2. What are the conditions for a system of equations to have infinitely many solutions?
3. How does the concept of free variables affect the solution set of a system?
4. What is the significance of pivot positions in a matrix?
5. Can you explain how the back-substitution method works in solving systems of linear equations?

**Tip:** Always check if a matrix is in reduced row echelon form by verifying that each leading coefficient (pivot) is 1 and that all entries above and below it are zero.

The matrix given is in reduced echelon form.

1
0
3
0
4
3
0
1
−1
0
5
5
0
0
0
1
3
6
Write the system of equations represented by the matrix. (Use *x* as your variable and label each *x* with its corresponding column. Enter x_1 for *x*1, x_2 for *x*2, x_3 for *x*3, x_4 for *x*4, and x_5 for *x*5.) Find the solution, if possible. (If there are infinitely many solutions, express *x*1, *x*2, *x*3, *x*4, and *x*5 in terms of the parameters *s* and *t*. If there is no solution, enter NO SOLUTION.)

Learn how to solve a system of linear equations represented by a reduced row echelon form matrix with infinitely many solutions. Understand the concepts of free variables and how to express solutions in terms of parameters. Suitable for undergraduate students studying linear algebra.

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