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The problem appears to ask for:

1. **Definition** of the rank of a matrix.
2. **Finding the rank** of the given matrix by reducing it to its row echelon form.

The given matrix appears to be:

\[
A = 
\begin{bmatrix}
2 & -1 & 1 & 2 \\
2 & 4 & 1 & 3 \\
0 & 2 & 0 & 2 \\
3 & -1 & 3 & -1
\end{bmatrix}
\]

### Step 1: Definition of the Rank of a Matrix
The **rank of a matrix** is the maximum number of linearly independent rows or columns of the matrix. It is equivalent to the number of non-zero rows in its row echelon form.

### Step 2: Reduce the Matrix to Row Echelon Form
We will perform elementary row operations to reduce the matrix \( A \) to its row echelon form.

#### Initial Matrix:
\[
\begin{bmatrix}
2 & -1 & 1 & 2 \\
2 & 4 & 1 & 3 \\
0 & 2 & 0 & 2 \\
3 & -1 & 3 & -1
\end{bmatrix}
\]

#### Step 2.1: Make the first pivot element 1 (Divide Row 1 by 2)
\[
\begin{bmatrix}
1 & -\frac{1}{2} & \frac{1}{2} & 1 \\
2 & 4 & 1 & 3 \\
0 & 2 & 0 & 2 \\
3 & -1 & 3 & -1
\end{bmatrix}
\]

#### Step 2.2: Eliminate below the pivot
- \( R2 \to R2 - 2 \times R1 \)
- \( R4 \to R4 - 3 \times R1 \)

\[
\begin{bmatrix}
1 & -\frac{1}{2} & \frac{1}{2} & 1 \\
0 & 5 & 0 & 1 \\
0 & 2 & 0 & 2 \\
0 & \frac{5}{2} & \frac{3}{2} & -4
\end{bmatrix}
\]

#### Step 2.3: Make the second pivot element 1 (Divide Row 2 by 5)
\[
\begin{bmatrix}
1 & -\frac{1}{2} & \frac{1}{2} & 1 \\
0 & 1 & 0 & \frac{1}{5} \\
0 & 2 & 0 & 2 \\
0 & \frac{5}{2} & \frac{3}{2} & -4
\end{bmatrix}
\]

#### Step 2.4: Eliminate below and above the second pivot
- \( R3 \to R3 - 2 \times R2 \)
- \( R4 \to R4 - \frac{5}{2} \times R2 \)
- \( R1 \to R1 + \frac{1}{2} \times R2 \)

\[
\begin{bmatrix}
1 & 0 & \frac{1}{2} & \frac{6}{5} \\
0 & 1 & 0 & \frac{1}{5} \\
0 & 0 & 0 & \frac{8}{5} \\
0 & 0 & \frac{3}{2} & -\frac{23}{5}
\end{bmatrix}
\]

#### Step 2.5: Make the third pivot element 1 (Divide Row 4 by \( \frac{3}{2} \))
\[
\begin{bmatrix}
1 & 0 & \frac{1}{2} & \frac{6}{5} \\
0 & 1 & 0 & \frac{1}{5} \\
0 & 0 & 1 & -\frac{46}{15} \\
0 & 0 & 0 & \frac{8}{5}
\end{bmatrix}
\]

### Step 3: Count Non-Zero Rows
The final row echelon form has three non-zero rows. Thus, the **rank of the matrix is 3**.

---

Would you like a detailed explanation of any step or assistance with similar problems? 

### Related Questions:
1. What is the significance of rank in linear algebra?
2. How do elementary row operations affect the determinant of a matrix?
3. Can the rank of a matrix exceed its number of rows or columns?
4. What are the applications of rank in solving linear equations?
5. How is rank connected to the invertibility of a matrix?

### Tip:
When performing row operations, always focus on one pivot at a time to avoid errors in calculations.

Define the rank of a matrix. Obtain the rank of the given matrix by reducing it to row echelon form: 
A = [2 -1 1 2; 2 4 1 3; 0 2 0 2; 3 -1 3 -1]

Learn how to define and calculate the rank of a matrix by reducing it to row echelon form. The solution includes step-by-step row operations on the given matrix A = [2 -1 1 2; 2 4 1 3; 0 2 0 2; 3 -1 3 -1], leading to its rank. This guide is perfect for linear algebra students.

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