Reduce the following matrix in normal form and hence find its rank and using row and column transformation

A =         -3    4     6    

5    -2     -3    

3    1     -4     

We are given a \( 3 \times 3 \) matrix \( A \) as follows:

\[
A = \begin{pmatrix}
-3 & 4 & 6 \\
5 & -2 & -3 \\
3 & 1 & -4
\end{pmatrix}
\]

We will reduce this matrix to its row echelon form (or normal form) by using **elementary row operations**. These row operations will help us simplify the matrix step by step and determine its **rank**, which is the number of non-zero rows in the row echelon form.

### Step 1: Make the first element of the first row (pivot) equal to 1 by dividing the entire row by \(-3\).

\[
R_1 \rightarrow \frac{1}{-3}R_1
\]

This gives us:

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

### Step 2: Make the elements below the pivot (first column) equal to zero by performing row operations:

- \( R_2 \rightarrow R_2 - 5R_1 \)
- \( R_3 \rightarrow R_3 - 3R_1 \)

Performing these operations:

- \( R_2 \rightarrow 5 - 5(1) = 0 \), \( -2 - 5\left(-\frac{4}{3}\right) = \frac{10}{3} \), \( -3 - 5(-2) = 7 \)
- \( R_3 \rightarrow 3 - 3(1) = 0 \), \( 1 - 3\left(-\frac{4}{3}\right) = 5 \), \( -4 - 3(-2) = 2 \)

Thus, we get:

\[
\begin{pmatrix}
1 & -\frac{4}{3} & -2 \\
0 & \frac{10}{3} & 7 \\
0 & 5 & 2
\end{pmatrix}
\]

### Step 3: Make the second pivot element (second row, second column) equal to 1 by dividing the second row by \( \frac{10}{3} \).

\[
R_2 \rightarrow \frac{3}{10}R_2
\]

This gives:

\[
\begin{pmatrix}
1 & -\frac{4}{3} & -2 \\
0 & 1 & \frac{21}{10} \\
0 & 5 & 2
\end{pmatrix}
\]

### Step 4: Eliminate the elements above and below the second pivot.

- \( R_1 \rightarrow R_1 + \frac{4}{3}R_2 \)
- \( R_3 \rightarrow R_3 - 5R_2 \)

Performing these operations:

- \( R_1 \rightarrow 1 + \frac{4}{3}(0) = 1 \), \( -\frac{4}{3} + \frac{4}{3}(1) = 0 \), \( -2 + \frac{4}{3}\left(\frac{21}{10}\right) = -\frac{38}{15} \)
- \( R_3 \rightarrow 0 - 5(0) = 0 \), \( 5 - 5(1) = 0 \), \( 2 - 5\left(\frac{21}{10}\right) = -\frac{53}{10} \)

Thus, we get:

\[
\begin{pmatrix}
1 & 0 & -\frac{38}{15} \\
0 & 1 & \frac{21}{10} \\
0 & 0 & -\frac{53}{10}
\end{pmatrix}
\]

### Step 5: Make the third pivot element (third row, third column) equal to 1 by dividing the third row by \( -\frac{53}{10} \).

\[
R_3 \rightarrow \frac{10}{-53}R_3
\]

This gives:

\[
\begin{pmatrix}
1 & 0 & -\frac{38}{15} \\
0 & 1 & \frac{21}{10} \\
0 & 0 & 1
\end{pmatrix}
\]

### Step 6: Eliminate the element above the third pivot:

- \( R_1 \rightarrow R_1 + \frac{38}{15}R_3 \)
- \( R_2 \rightarrow R_2 - \frac{21}{10}R_3 \)

Performing these operations:

- \( R_1 \rightarrow -\frac{38}{15}(1) = 0 \)
- \( R_2 \rightarrow \frac{21}{10}(1) = 0 \)

Thus, the matrix becomes:

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

### Final Step: Rank of the Matrix

The matrix is now in **reduced row echelon form** (identity matrix), and since all three rows are non-zero, the rank of the matrix \( A \) is **3**.

---

Would you like to ask any questions or need further clarification?

Here are five related questions:
1. What are the main differences between row echelon form and reduced row echelon form?
2. How does column transformation affect the process of finding the rank of a matrix?
3. What is the significance of the rank of a matrix in linear algebra?
4. Can a matrix have more pivots than its smallest dimension (rows or columns)?
5. What does it mean if a matrix has full rank?

**Tip**: The rank of a matrix represents the maximum number of linearly independent rows or columns in the matrix.

Reduce the following matrix in normal form and hence find its rank using row and column transformation: A = [[-3, 4, 6], [5, -2, -3], [3, 1, -4]]

This problem involves reducing a 3x3 matrix to its normal form using elementary row operations and determining its rank. Learn how to perform row and column transformations step by step, and understand the process of finding the matrix's rank. Suitable for undergraduate students studying linear algebra.

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