1. **Problem Statement:** Determine the rank of the matrix $$\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$. 2. **Definition:** The rank of a matrix is the maximum number of linearly independent rows or columns. One way to find the rank is to reduce the matrix to its row echelon form and count the number of non-zero rows. 3. **Step 1: Write the matrix** $$A = \begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$ 4. **Step 2: Perform row operations to get zeros below the first pivot (3 in row 1, column 1):** - Replace row 2 with row 2 minus $\frac{4}{3}$ times row 1: $$R_2 = R_2 - \frac{4}{3}R_1 = \begin{bmatrix}4 - \frac{4}{3} \times 3, & 5 - \frac{4}{3} \times 2, & 6 - \frac{4}{3} \times 1\end{bmatrix} = \begin{bmatrix}0, & \frac{7}{3}, & \frac{14}{3}\end{bmatrix}$$ - Replace row 3 with row 3 minus $\frac{7}{3}$ times row 1: $$R_3 = R_3 - \frac{7}{3}R_1 = \begin{bmatrix}7 - \frac{7}{3} \times 3, & 8 - \frac{7}{3} \times 2, & 9 - \frac{7}{3} \times 1\end{bmatrix} = \begin{bmatrix}0, & \frac{2}{3}, & \frac{10}{3}\end{bmatrix}$$ 5. **Step 3: Matrix after these operations:** $$\begin{bmatrix}3 & 2 & 1 \\ 0 & \frac{7}{3} & \frac{14}{3} \\ 0 & \frac{2}{3} & \frac{10}{3}\end{bmatrix}$$ 6. **Step 4: Make zero below the pivot in row 2, column 2:** - Replace row 3 with row 3 minus $\frac{2/3}{7/3} = \frac{2}{7}$ times row 2: $$R_3 = R_3 - \frac{2}{7} R_2 = \begin{bmatrix}0, & \frac{2}{3} - \frac{2}{7} \times \frac{7}{3}, & \frac{10}{3} - \frac{2}{7} \times \frac{14}{3}\end{bmatrix} = \begin{bmatrix}0, & 0, & \frac{10}{3} - \frac{28}{21}\end{bmatrix}$$ Calculate the last element: $$\frac{10}{3} - \frac{28}{21} = \frac{70}{21} - \frac{28}{21} = \frac{42}{21} = 2$$ 7. **Step 5: Final row echelon form:** $$\begin{bmatrix}3 & 2 & 1 \\ 0 & \frac{7}{3} & \frac{14}{3} \\ 0 & 0 & 2\end{bmatrix}$$ 8. **Step 6: Count non-zero rows:** All three rows have non-zero pivots, so the rank is 3. **Final answer:** The rank of the matrix is **3**.