It appears that you're working on determining the rank of a matrix using its canonical form. Let's walk through the solution steps presented in the image:

1. Matrix $A$: The matrix you're working with is:

   2 & 1 & 3 & 4 \\ 0 & 3 & 4 & 1 \\ 2 & 3 & 7 & 5 \\ 2 & 5 & 11 & 6 \end{bmatrix}$$

2. Row Reduction: The steps show a process of performing row operations to transform the matrix into row echelon form.

3. Row Operations:

   • First, $R_1$ is divided by 2.
   • Then $R_3$ is replaced by $R_3 - 2R_1$.
   • $R_4$ is replaced by $R_4 - 2R_1$.
   • Further row reductions are performed to simplify the matrix.

4. Final Matrix: After performing all the row operations, the matrix becomes:

   1 & 0.5 & 1.5 & 2 \\ 0 & 2 & 4 & 1 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 4 & 8 & 2 \end{bmatrix}$$ From this, you can see that there are 3 non-zero rows, indicating that the rank of the matrix is **3**.

The conclusion seems correct based on the row-reduction process, and the matrix's rank is indeed 3.

Let me know if you want further details on the row operations or need help with a related topic.

Here are 5 related questions to deepen your understanding:

1. What is the definition of the rank of a matrix?
2. Why is the canonical form useful for finding the rank of a matrix?
3. Can a matrix have a rank larger than the number of rows or columns?
4. How can you determine the rank of a matrix using determinants?
5. What is the relationship between the rank of a matrix and its nullity?

Tip: When performing row operations, always aim to simplify the leading entries of each row (pivot points) while keeping the rows above and below them zero.