1. **Problem:** Find the rank of matrix $ A = \begin{bmatrix} 3 & 4 & 1 & 1 \\ 2 & 4 & 3 & 6 \\ -1 & -2 & 6 & 4 \\ 1 & -1 & 2 & -3 \end{bmatrix} $ using row echelon form. 2. **Formula and Rules:** - The rank of a matrix is the number of non-zero rows in its row echelon form. - Row echelon form is achieved by using elementary row operations to get zeros below pivots. 3. **Step-by-step solution:** - Start with $ A $: $$\begin{bmatrix} 3 & 4 & 1 & 1 \\ 2 & 4 & 3 & 6 \\ -1 & -2 & 6 & 4 \\ 1 & -1 & 2 & -3 \end{bmatrix}$$ - Make the first pivot 1 by dividing row 1 by 3: $$R_1 \to \frac{1}{3}R_1 = \begin{bmatrix} 1 & \frac{4}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}$$ - Eliminate below pivot in column 1: $$R_2 \to R_2 - 2R_1 = \begin{bmatrix} 0 & \frac{4}{3} & \frac{7}{3} & \frac{16}{3} \end{bmatrix}$$ $$R_3 \to R_3 + R_1 = \begin{bmatrix} 0 & \frac{2}{3} & \frac{19}{3} & \frac{13}{3} \end{bmatrix}$$ $$R_4 \to R_4 - R_1 = \begin{bmatrix} 0 & -\frac{7}{3} & \frac{5}{3} & -\frac{10}{3} \end{bmatrix}$$ - Multiply row 2 by $ \frac{3}{4} $ to get pivot 1: $$R_2 \to \frac{3}{4} R_2 = \begin{bmatrix} 0 & 1 & \frac{7}{4} & 4 \end{bmatrix}$$ - Eliminate below and above pivot in column 2: $$R_3 \to R_3 - \frac{2}{3} R_2 = \begin{bmatrix} 0 & 0 & 12 & 5 \end{bmatrix}$$ $$R_4 \to R_4 + \frac{7}{3} R_2 = \begin{bmatrix} 0 & 0 & 12 & 8 \end{bmatrix}$$ $$R_1 \to R_1 - \frac{4}{3} R_2 = \begin{bmatrix} 1 & 0 & -\frac{5}{4} & -\frac{13}{3} \end{bmatrix}$$ - Divide row 3 by 12: $$R_3 \to \frac{1}{12} R_3 = \begin{bmatrix} 0 & 0 & 1 & \frac{5}{12} \end{bmatrix}$$ - Eliminate below and above pivot in column 3: $$R_4 \to R_4 - 12 R_3 = \begin{bmatrix} 0 & 0 & 0 & 3 \end{bmatrix}$$ $$R_1 \to R_1 + \frac{5}{4} R_3 = \begin{bmatrix} 1 & 0 & 0 & -\frac{1}{4} \end{bmatrix}$$ $$R_2 \to R_2 - \frac{7}{4} R_3 = \begin{bmatrix} 0 & 1 & 0 & \frac{29}{12} \end{bmatrix}$$ - Divide row 4 by 3: $$R_4 \to \frac{1}{3} R_4 = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}$$ - Eliminate above pivot in column 4: $$R_1 \to R_1 + \frac{1}{4} R_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}$$ $$R_2 \to R_2 - \frac{29}{12} R_4 = \begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}$$ $$R_3 \to R_3 - \frac{5}{12} R_4 = \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}$$ 4. **Result:** The row echelon form has 4 non-zero rows, so $ \text{rank}(A) = 4 $. **Final answer:** $ \boxed{4} $