1. **Stating the problem:** We need to determine which of the given 3x3 matrices are in row-echelon form. 2. **Definition and rules of row-echelon form:** - All nonzero rows are above any rows of all zeros. - The leading entry (pivot) of each nonzero row is strictly to the right of the leading entry of the row above it. - All entries in a column below a leading entry are zeros. 3. **Check each matrix:** **a)** \(\begin{bmatrix}1 & -2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\) - The first row has leading 1 in column 1. - The second row has leading 1 in column 2, which is to the right of column 1. - The third row is all zeros and is at the bottom. - All entries below pivots are zero. - **This matrix is in row-echelon form.** **b)** \(\begin{bmatrix}1 & -5 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\) - Leading 1 in row 1 is in column 1. - Leading 1 in row 2 is in column 2, to the right of column 1. - Row 3 is zero row at bottom. - All entries below pivots are zero. - **This matrix is in row-echelon form.** **c)** \(\begin{bmatrix}1 & -9 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\) - Leading 1 in row 1 is column 1. - Leading 1 in row 2 is column 2, to the right. - Row 3 is zero row at bottom. - All entries below pivots zero. - **This matrix is in row-echelon form.** **d)** \(\begin{bmatrix}1 & -6 & -5 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\) - Row 2 is zero row but is above row 3 which is nonzero. - This violates the rule that zero rows must be at the bottom. - **This matrix is NOT in row-echelon form.** 4. **Summary:** Matrices a, b, and c are in row-echelon form. Matrix d is not. **Final answer:** a, b, c are in row-echelon form; d is not.