To determine the rank of the matrix $A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix}$, we'll reduce it to its row echelon form using elementary row operations. The rank is the number of non-zero rows in this form.

Step 1: Write Down the Matrix

$A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix}$

Step 2: Make the Leading Entry of Row 1 a 1

Divide the first row by 2: $\begin{bmatrix} 1 & 1 & \frac{1}{2} \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix}$

Step 3: Eliminate the Leading Entry Below Row 1

Subtract Row 1 from Rows 2 and 3 to zero out the entries below the leading 1 in the first column.

• Row 2: $R_2 - R_1 = \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 1 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 & \frac{3}{2} \end{bmatrix}$
• Row 3: $R_3 - R_1 = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 1 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 & \frac{5}{2} \end{bmatrix}$

The matrix is now: $\begin{bmatrix} 1 & 1 & \frac{1}{2} \\ 0 & 1 & \frac{3}{2} \\ 0 & 1 & \frac{5}{2} \end{bmatrix}$

Step 4: Make the Second Column’s Entries Consistent

Subtract Row 2 from Row 3 to zero out the entry below the leading 1 in Row 2:

• Row 3: $R_3 - R_2 = \begin{bmatrix} 0 & 1 & \frac{5}{2} \end{bmatrix} - \begin{bmatrix} 0 & 1 & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$

Now the matrix is: $\begin{bmatrix} 1 & 1 & \frac{1}{2} \\ 0 & 1 & \frac{3}{2} \\ 0 & 0 & 1 \end{bmatrix}$

Step 5: The Matrix is in Row Echelon Form

The resulting matrix has three non-zero rows, which indicates that the matrix has full rank.

Conclusion

The rank of the matrix $A$ is $\boxed{3}$.

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Would you like further details on any step?

Related Questions

1. How do you determine rank for larger matrices?
2. What if two rows are linearly dependent? How does it affect the rank?
3. Can the rank be more than the number of rows or columns?
4. How is the rank related to the determinant of a square matrix?
5. What does rank tell us about the solutions to a system of linear equations?

Tip: A matrix's rank is also the dimension of the column space, which represents the maximum number of linearly independent columns.