Find the solution set of the system of linear equations represented by the augmented matrix. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set 
x3 = t
 and solve for x1 and x2 in terms of t.)

2	1	−1	0	

1	−1	1	0
0	1	2	3
(x1, x2, x3) = ?

 
 

We are given the augmented matrix:

\[
\begin{pmatrix}
2 & 1 & -1 & 0 \\
1 & -1 & 1 & 0 \\
0 & 1 & 2 & 3
\end{pmatrix}
\]

### Step 1: Convert to Row Echelon Form (REF)

We start by performing row operations to simplify the matrix.

The matrix is:

\[
\begin{pmatrix}
2 & 1 & -1 & 0 \\
1 & -1 & 1 & 0 \\
0 & 1 & 2 & 3
\end{pmatrix}
\]

#### Operation 1: Eliminate below the pivot in the first column.

- To make the first pivot a 1, divide the first row by 2.

\[
R_1 \rightarrow \frac{1}{2} R_1
\]

\[
\begin{pmatrix}
1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
1 & -1 & 1 & 0 \\
0 & 1 & 2 & 3
\end{pmatrix}
\]

- Subtract \( R_1 \) from \( R_2 \) to eliminate the first element in \( R_2 \).

\[
R_2 \rightarrow R_2 - R_1
\]

\[
\begin{pmatrix}
1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
0 & -\frac{3}{2} & \frac{3}{2} & 0 \\
0 & 1 & 2 & 3
\end{pmatrix}
\]

#### Operation 2: Simplify the second row.

- Multiply \( R_2 \) by \( -\frac{2}{3} \) to make the pivot 1.

\[
R_2 \rightarrow -\frac{2}{3} R_2
\]

\[
\begin{pmatrix}
1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
0 & 1 & -1 & 0 \\
0 & 1 & 2 & 3
\end{pmatrix}
\]

#### Operation 3: Eliminate below and above the second pivot.

- Subtract \( R_2 \) from \( R_3 \) to eliminate the 1 in \( R_3 \).

\[
R_3 \rightarrow R_3 - R_2
\]

\[
\begin{pmatrix}
1 & \frac{1}{2} & -\frac{1}{2} & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 3 & 3
\end{pmatrix}
\]

- Subtract \( \frac{1}{2} R_2 \) from \( R_1 \) to eliminate the \( \frac{1}{2} \) in \( R_1 \).

\[
R_1 \rightarrow R_1 - \frac{1}{2} R_2
\]

\[
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 3 & 3
\end{pmatrix}
\]

#### Operation 4: Simplify the third row.

- Divide \( R_3 \) by 3 to make the pivot 1.

\[
R_3 \rightarrow \frac{1}{3} R_3
\]

\[
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix}
\]

### Step 2: Interpret the Solution

The matrix is now in reduced row echelon form (RREF). The system of equations is:

1. \( x_1 = 0 \)
2. \( x_2 - x_3 = 0 \) → \( x_2 = x_3 \)
3. \( x_3 = 1 \)

Thus, the solution is:

\[
x_1 = 0, \quad x_2 = 1, \quad x_3 = 1
\]

The solution set is \( (x_1, x_2, x_3) = (0, 1, 1) \).

---

Would you like more details on any of these steps?

Here are 5 related questions:
1. How do row operations affect the solution set of a system of equations?
2. What is the difference between row echelon form and reduced row echelon form?
3. How can you tell if a system of equations has no solution?
4. What do free variables represent in an infinite solution system?
5. How can you solve a system of equations using matrix inverses?

**Tip:** Always check your row operations carefully to avoid mistakes in simplification!

Find the solution set of the system of linear equations represented by the augmented matrix. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x3 = t and solve for x1 and x2 in terms of t.)

2  1  −1  0

1  −1  1  0

0  1  2  3

(x1, x2, x3) = ?

Learn how to find the solution set of a system of linear equations using an augmented matrix. This problem covers key linear algebra concepts, including matrix operations, row reduction, and Gaussian elimination, suitable for students in Grades 10-12.

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