Find the rank of the matrix  A=1 0 -3 2           0 1 4 5    1 3 2 0    1 1 -2 0      by normal form

To find the rank of the matrix \( A \) using the normal form (or row-reduced echelon form), we need to perform Gaussian elimination on the matrix to convert it into a row echelon form, and then count the number of non-zero rows.

Given matrix \( A \):

\[
A = \begin{pmatrix}
1 & 0 & -3 & 2 \\
0 & 1 & 4 & 5 \\
1 & 3 & 2 & 0 \\
1 & 1 & -2 & 0
\end{pmatrix}
\]

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

1. **Row 1** is already in a good form, so we leave it as is.
2. We can eliminate the first column elements below the first row (i.e., in Row 3 and Row 4).

   Subtract Row 1 from Row 3:
   \[
   R3 \rightarrow R3 - R1 = (1-1, 3-0, 2-(-3), 0-2) = (0, 3, 5, -2)
   \]

   Subtract Row 1 from Row 4:
   \[
   R4 \rightarrow R4 - R1 = (1-1, 1-0, -2-(-3), 0-2) = (0, 1, 1, -2)
   \]

   Matrix after these operations:
   \[
   A = \begin{pmatrix}
   1 & 0 & -3 & 2 \\
   0 & 1 & 4 & 5 \\
   0 & 3 & 5 & -2 \\
   0 & 1 & 1 & -2
   \end{pmatrix}
   \]

3. **Row 2** is also in a good form, so we proceed to eliminate elements below it in the second column.

   Subtract \(3 \times R2\) from \(R3\):
   \[
   R3 \rightarrow R3 - 3R2 = (0, 

Learn how to find the rank of matrix A using normal form with detailed steps and explanations. This problem involves matrix operations and Gaussian elimination to convert matrix A to row echelon form and determine its rank. Suitable for advanced high school students and anyone studying linear algebra.

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