Linear Algebra

Notice that row 2 is $2 \times R_1$ and row 3 is $3 \times R_1$. Performing row operations $R_2 \to R_2 - 2 R_1$ and $R_3 \to R_3 - 3 R_1$ yields: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ There is only 1 non-zero row in echelon form, so $\text{rank}(A) = 1$.