1. **State the problem:** We need to find the rank of the matrix $$\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$. 2. **Recall the definition:** The rank of a matrix is the maximum number of linearly independent rows or columns. One way to find it is by reducing the matrix to row echelon form and counting the number of nonzero rows. 3. **Start with the matrix:** $$\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$ 4. **Perform row operations:** - 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. **New matrix:** $$\begin{bmatrix}3 & 2 & 1 \\ 0 & \frac{7}{3} & \frac{14}{3} \\ 0 & \frac{2}{3} & \frac{10}{3}\end{bmatrix}$$ 6. **Eliminate below pivot in column 2:** - Replace row 3 with row 3 minus $\frac{2/3}{7/3} = \frac{2}{7}$ times row 2: $$R_3 = R_3 - \frac{2}{7} R_2 = \begin{bmatrix}0, \frac{2}{3} - \frac{2}{7} \times \frac{7}{3}, \frac{10}{3} - \frac{2}{7} \times \frac{14}{3}\end{bmatrix} = \begin{bmatrix}0, 0, \frac{10}{3} - \frac{28}{21}\end{bmatrix} = \begin{bmatrix}0, 0, \frac{10}{3} - \frac{4}{3}\end{bmatrix} = \begin{bmatrix}0, 0, 2\end{bmatrix}$$ 7. **Final row echelon form:** $$\begin{bmatrix}3 & 2 & 1 \\ 0 & \frac{7}{3} & \frac{14}{3} \\ 0 & 0 & 2\end{bmatrix}$$ 8. **Count nonzero rows:** All three rows have nonzero entries, so the rank is 3. **Final answer:** The rank of the matrix is **3**.