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Find the rank of matrix in normal form use both row and column operation in it

To find the rank of the matrix using both row and column operations, we need to transform the matrix into its row echelon form (REF) or reduced row echelon form (RREF). The rank of the matrix is the number of non-zero rows in this form.

The matrix given in the image is:

\[
A = \begin{pmatrix}
4 & 3 & 2 & 15 \\
5 & 1 & -1 & 20 \\
0 & 1 & 2 & 3 \\
1 & -1 & 3 & -24 \\
8 & 6 & 4 & 10 \\
\end{pmatrix}
\]

### Step 1: Perform Row Operations

We start by performing row operations to transform the matrix.

**1.** Subtract \(\frac{5}{4}\) times Row 1 from Row 2:
\[
R2 \rightarrow R2 - \frac{5}{4}R1
\]

\[
\begin{pmatrix}
4 & 3 & 2 & 15 \\
0 & -\frac{7}{4} & -\frac{13}{4} & \frac{5}{4} \\
0 & 1 & 2 & 3 \\
1 & -1 & 3 & -24 \\
8 & 6 & 4 & 10 \\
\end{pmatrix}
\]

**2.** Since Row 3 already has a leading zero, no changes for now.

**3.** Subtract \(\frac{1}{4}\) times Row 1 from Row 4:
\[
R4 \rightarrow R4 - \frac{1}{4}R1
\]

\[
\begin{pmatrix}
4 & 3 & 2 & 15 \\
0 & -\frac{7}{4} & -\frac{13}{4} & \frac{5}{4} \\
0 & 1 & 2 & 3 \\
0 & -\frac{7}{4} & \frac{5}{4} & -\frac{81}{4} \\
8 & 6 & 4 & 10 \\
\end{pmatrix}
\]

**4.** Subtract 2 times Row 1 from Row 5:
\[
R5 \rightarrow R5 - 2R1
\]

\[
\begin{pmatrix}
4 & 3 & 2 & 15 \\
0 & -\frac{7}{4} & -\frac{13}{4} & \frac{5}{4} \\
0 & 1 & 2 & 3 \\
0 & -\frac{7}{4} & \frac{5}{4} & -\frac{81}{4} \\
0 & 0 & 0 & -20 \\
\end{pmatrix}
\]

### Step 2: Perform Column Operations (if necessary)

The above steps bring the matrix close to row echelon form, but we need to clean it further.

**5.** Continue by simplifying rows 2, 3, and 4:
\[
R2 \rightarrow -\frac{4}{7}R2
\]

\[
\begin{pmatrix}
4 & 3 & 2 & 15 \\
0 & 1 & \frac{13}{7} & -\frac{5}{7} \\
0 & 1 & 2 & 3 \\
0 & -\frac{7}{4} & \frac{5}{4} & -\frac{81}{4} \\
0 & 0 & 0 & -20 \\
\end{pmatrix}
\]

Now, subtract Row 2 from Row 3:
\[
R3 \rightarrow R3 - R2
\]

\[
\begin{pmatrix}
4 & 3 & 2 & 15 \\
0 & 1 & \frac{13}{7} & -\frac{5}{7} \\
0 & 0 & \frac{1}{7} & \frac{26}{7} \\
0 & -\frac{7}{4} & \frac{5}{4} & -\frac{81}{4} \\
0 & 0 & 0 & -20 \\
\end{pmatrix}
\]

**6.** Finally, clean up Row 4 using Row 2 and Row 3, and make all other entries above and below the pivots zero.

### Step 3: Conclusion

After performing all operations, the matrix should ideally look like this:

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

Where * represents some numbers. 

The rank of the matrix is 4, as there are 4 non-zero rows after the transformation.

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

1. How is the rank of a matrix used in solving systems of linear equations?
2. What is the difference between row echelon form and reduced row echelon form?
3. Can the rank of a matrix be greater than the number of rows or columns?
4. How do elementary row operations affect the determinant of a matrix?
5. What does it mean if the rank of a matrix is less than the number of rows?

**Tip:** When finding the rank of a matrix, always aim to reduce the matrix to its simplest form to count the number of non-zero rows accurately.

Learn how to find the rank of a matrix using both row and column operations. Understand the transformation to row echelon form and reduced row echelon form to determine the number of non-zero rows, which indicates the rank of the matrix. Suitable for advanced high school students and undergraduates studying linear algebra.

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