1. **State the problem:** We have two matrices: $$\begin{bmatrix} 1 & -3 & 3 & 0 \\ 5 & -2 & 2 & -5 \\ 0 & 4 & -2 & 5 \end{bmatrix} \quad \to \quad \begin{bmatrix} 1 & -3 & 3 & 0 \\ 0 & 13 & -13 & -5 \\ 0 & 4 & -2 & 5 \end{bmatrix}$$ We want to find the elementary row operation that transforms the first matrix into the second, focusing on the change in row 2. 2. **Recall the types of elementary row operations:** - Replace a row by the sum of itself and a multiple of another row. - Interchange two rows. - Multiply a row by a nonzero scalar. 3. **Analyze the change in row 2:** Original row 2: $[5, -2, 2, -5]$ New row 2: $[0, 13, -13, -5]$ 4. **Check if row 2 is replaced by its sum with a multiple of row 1:** Let the multiple be $k$. Then: $$\text{new row 2} = \text{old row 2} + k \times \text{row 1}$$ Set up equations for each element: $$5 + k \times 1 = 0 \implies k = -5$$ Check other elements with $k = -5$: $$-2 + (-5)(-3) = -2 + 15 = 13$$ $$2 + (-5)(3) = 2 - 15 = -13$$ $$-5 + (-5)(0) = -5 + 0 = -5$$ All match the new row 2. 5. **Therefore, the elementary row operation is:** Replace row 2 by its sum with $-5$ times row 1. 6. **Find the reverse operation:** To reverse, we start from the new matrix and want to get back the old matrix. We have: $$\text{old row 2} = \text{new row 2} + m \times \text{row 1}$$ Using the first element: $$0 + m \times 1 = 5 \implies m = 5$$ Check other elements: $$13 + 5(-3) = 13 - 15 = -2$$ $$-13 + 5(3) = -13 + 15 = 2$$ $$-5 + 5(0) = -5$$ All match the old row 2. So the reverse operation is: Replace row 2 by its sum with $5$ times row 1. **Final answers:** - Forward operation: Replace row 2 by its sum with $-5$ times row 1. - Reverse operation: Replace row 2 by its sum with $5$ times row 1.