1. **State the problem:** Find the rank of the matrix $$\begin{bmatrix}1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 1\end{bmatrix}$$ using elementary row transformations. 2. **Recall:** The rank of a matrix is the number of non-zero rows in its row echelon form (REF). 3. **Start with the matrix:** $$A = \begin{bmatrix}1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 1\end{bmatrix}$$ 4. **Use row operations to get REF:** - Replace $R_2$ by $R_2 - 2R_1$: $$R_2 = \begin{bmatrix}2 & 1 & 2\end{bmatrix} - 2 \times \begin{bmatrix}1 & 3 & 4\end{bmatrix} = \begin{bmatrix}0 & -5 & -6\end{bmatrix}$$ - Replace $R_3$ by $R_3 - 3R_1$: $$R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix} - 3 \times \begin{bmatrix}1 & 3 & 4\end{bmatrix} = \begin{bmatrix}0 & -8 & -11\end{bmatrix}$$ Matrix becomes: $$\begin{bmatrix}1 & 3 & 4 \\ 0 & -5 & -6 \\ 0 & -8 & -11\end{bmatrix}$$ 5. **Next, eliminate below pivot in second column:** - Replace $R_3$ by $R_3 - \frac{8}{5} R_2$: $$R_3 = \begin{bmatrix}0 & -8 & -11\end{bmatrix} - \frac{8}{5} \times \begin{bmatrix}0 & -5 & -6\end{bmatrix} = \begin{bmatrix}0 & 0 & -11 + \frac{48}{5}\end{bmatrix} = \begin{bmatrix}0 & 0 & -\frac{7}{5}\end{bmatrix}$$ Matrix now: $$\begin{bmatrix}1 & 3 & 4 \\ 0 & -5 & -6 \\ 0 & 0 & -\frac{7}{5}\end{bmatrix}$$ 6. **Count non-zero rows:** All three rows have non-zero entries. 7. **Conclusion:** The rank of the matrix is $3$.