Let's solve the problem step by step: ### Step 1: Write the matrix and try to reduce it to row echelon form. The given matrix is: \[ \begin{bmatrix} 1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4 \\ \end{bmatrix} \] Let's use the first row as the pivot and eliminate the corresponding entries in the second and third rows. - Row 2: \( R_2 - 3R_1 \rightarrow (3, -1, 1) - 3(1, 1, 2) = (3-3, -1-3, 1-6) = (0, -4, -5) \) - Row 3: \( R_3 + R_1 \rightarrow (-1, 3, 4) + (1, 1, 2) = (0, 4, 6) \) New matrix: \[ \begin{bmatrix} 1 & 1 & 2 \\ 0 & -4 & -5 \\ 0 & 4 & 6 \\ \end{bmatrix} \] **Explanation:** We used row operations to create zeros below the pivot in the first column. --- ### Step 2: Continue row reduction to check for more zeros and find the rank. Now, let's eliminate the second element in the third row using the second row. - Row 3: \( R_3 + R_2 \rightarrow (0, 4, 6) + (0, -4, -5) = (0, 0, 1) \) So the matrix becomes: \[ \begin{bmatrix} 1 & 1 & 2 \\ 0 & -4 & -5 \\ 0 & 0 & 1 \\ \end{bmatrix} \] All rows are non-zero. **Explanation:** After further row reduction, we see that all three rows have non-zero values, meaning all are linearly independent. --- ### Final Answer **The rank of the matrix is \(\boxed{3}\).**