Determine the rank of following matrices by reducing them to echlon form   

          1     2    -1    3

A=     3    -1     2    1

          2    -2     3    2

         1    -1    1    -1

We are tasked with determining the rank of matrix \( A \) by reducing it to its row echelon form. The rank of a matrix is the number of non-zero rows in its row echelon form. Let's proceed with Gaussian elimination (row reduction) on the matrix:

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

### Step 1: Make the first pivot entry 1 (already satisfied).
The first pivot is in the first row, first column. We will now eliminate the entries below this pivot.

#### R2 = R2 - 3R1
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
3 & -1 & 2 & 1 \\
2 & -2 & 3 & 2 \\
1 & -1 & 1 & -1
\end{pmatrix} 
\to 
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & -7 & 5 & -8 \\
2 & -2 & 3 & 2 \\
1 & -1 & 1 & -1
\end{pmatrix}
\]

#### R3 = R3 - 2R1
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & -7 & 5 & -8 \\
0 & -6 & 5 & -4 \\
1 & -1 & 1 & -1
\end{pmatrix}
\]

#### R4 = R4 - R1
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & -7 & 5 & -8 \\
0 & -6 & 5 & -4 \\
0 & -3 & 2 & -4
\end{pmatrix}
\]

### Step 2: Make the second pivot entry 1 (by dividing R2 by -7).
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & -6 & 5 & -4 \\
0 & -3 & 2 & -4
\end{pmatrix}
\]

### Step 3: Eliminate the entries below the second pivot.
#### R3 = R3 + 6R2
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & 0 & \frac{5}{7} & \frac{4}{7} \\
0 & -3 & 2 & -4
\end{pmatrix}
\]

#### R4 = R4 + 3R2
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & 0 & \frac{5}{7} & \frac{4}{7} \\
0 & 0 & \frac{1}{7} & \frac{4}{7}
\end{pmatrix}
\]

### Step 4: Make the third pivot entry 1 (by multiplying R3 by 7/5).
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & 0 & 1 & \frac{4}{5} \\
0 & 0 & \frac{1}{7} & \frac{4}{7}
\end{pmatrix}
\]

### Step 5: Eliminate the entry below the third pivot.
#### R4 = R4 - \(\frac{1}{7}\) R3
\[
\begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & 0 & 1 & \frac{4}{5} \\
0 & 0 & 0 & 0
\end{pmatrix}
\]

### Step 6: The matrix is now in row echelon form.
\[
\text{Echelon form of } A = \begin{pmatrix}
1 & 2 & -1 & 3 \\
0 & 1 & -\frac{5}{7} & \frac{8}{7} \\
0 & 0 & 1 & \frac{4}{5} \\
0 & 0 & 0 & 0
\end{pmatrix}
\]

### Conclusion:
The matrix has three non-zero rows, so the **rank of the matrix is 3**.

Would you like further details on any of these steps?

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Here are 5 related questions:
1. What is the definition of the rank of a matrix?
2. How does Gaussian elimination help in determining the rank?
3. What are the conditions for a matrix to have full rank?
4. What is the difference between row echelon form and reduced row echelon form?
5. Can you calculate the nullity of this matrix based on its rank?

**Tip:** The rank of a matrix can never be greater than the number of rows or columns, whichever is smaller.

Determine the rank of following matrices by reducing them to echelon form: 
1 2 -1 3
A= 3 -1 2 1
2 -2 3 2
1 -1 1 -1

Learn how to determine the rank of a matrix by reducing it to echelon form using Gaussian elimination. The problem involves a 4x4 matrix, and the step-by-step solution explains each transformation. Suitable for college-level linear algebra students.

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