Find rank of the matrix A=[1 2 -1 3] [4 1 2 1][3 -1 1 2][1 2 0 1]

To find the rank of the matrix \( A \), we need to determine the maximum number of linearly independent rows or columns in the matrix. Let's go through the steps:

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

1. **Form the augmented matrix:** We'll perform row reduction (Gaussian elimination) to transform this matrix into its row echelon form (REF).

2. **Row reduction steps:**
   - Start with the first element in the first row (pivot = 1):
     \[
     R_2 \rightarrow R_2 - 4R_1, \quad R_3 \rightarrow R_3 - 3R_1, \quad R_4 \rightarrow R_4 - R_1
     \]
     After these operations:
     \[
     \begin{bmatrix}
     1 & 2 & -1 & 3 \\
     0 & -7 & 6 & -11 \\
     0 & -7 & 4 & -7 \\
     0 & 0 & 1 & -2
     \end{bmatrix}
     \]

   - Now, take the second row and normalize the pivot (second element of the second row = -7):
     \[
     R_2 \rightarrow \frac{R_2}{-7}
     \]
     After normalization:
     \[
     \begin{bmatrix}
     1 & 2 & -1 & 3 \\
     0 & 1 & -\frac{6}{7} & \frac{11}{7} \\
     0 & -7 & 4 & -7 \\
     0 & 0 & 1 & -2
     \end{bmatrix}
     \]

   - Make the elements below the pivot in column 2 zero:
     \[
     R_3 \rightarrow R_3 + 7R_2
     \]
     After this operation:
     \[
     \begin{bmatrix}
     1 & 2 & -1 & 3 \\
     0 & 1 & -\frac{6}{7} & \frac{11}{7} \\
     0 & 0 & -\frac{2}{7} & \frac{6}{7} \\
     0 & 0 & 1 & -2
     \end{bmatrix}
     \]

   - Now, work on the third row:
     \[
     R_3 \rightarrow \frac{R_3}{-\frac{2}{7}}
     \]
     After normalization:
     \[
     \begin{bmatrix}
     1 & 2 & -1 & 3 \\
     0 & 1 & -\frac{6}{7} & \frac{11}{7} \\
     0 & 0 & 1 & -3 \\
     0 & 0 & 1 & -2
     \end{bmatrix}
     \]

   - Finally, make the element below the third pivot (row 4) zero:
     \[
     R_4 \rightarrow R_4 - R_3
     \]
     After this operation:
     \[
     \begin{bmatrix}
     1 & 2 & -1 & 3 \\
     0 & 1 & -\frac{6}{7} & \frac{11}{7} \\
     0 & 0 & 1 & -3 \\
     0 & 0 & 0 & 1
     \end{bmatrix}
     \]

3. **Count the non-zero rows:** In the row echelon form, all rows are non-zero, indicating four linearly independent rows.

Thus, the rank of matrix \( A \) is **4**.

Would you like more details or have any questions?

Here are some related questions to expand your understanding:
1. How does one determine the linear independence of rows or columns in a matrix?
2. What is the significance of a matrix's rank in linear algebra?
3. How would you find the rank of a non-square matrix?
4. What role does the determinant play in finding the rank of a square matrix?
5. How does Gaussian elimination relate to solving systems of linear equations?

**Tip:** The rank of a matrix provides insight into the dimension of the vector space spanned by its rows or columns, and it's a crucial concept in understanding solutions to linear systems.

Find the rank of the matrix A = [1 2 -1 3] [4 1 2 1] [3 -1 1 2] [1 2 0 1]

Discover how to find the rank of a matrix using Gaussian elimination with a detailed, step-by-step solution. Learn key concepts of linear algebra like row reduction and matrix rank. This problem is ideal for students in Grades 10-12, exploring advanced algebra.

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