1. **State the problem:** We need to reduce the matrix $$\begin{bmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{bmatrix}$$ to its echelon form. 2. **Recall the goal:** Echelon form means the matrix is in upper triangular form with leading coefficients (pivots) of each row to the right of the pivots in the rows above. 3. **Start with the original matrix:** $$A = \begin{bmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{bmatrix}$$ 4. **Step 1: Use row 1 to eliminate below:** - Multiply row 1 by 6 and add to row 2: $$R_2 \to R_2 + 6R_1 = (-6 + 6\times1, 8 + 6\times3, 3 + 6\times5) = (0, 26, 33)$$ - Multiply row 1 by 4 and add to row 3: $$R_3 \to R_3 + 4R_1 = (-4 + 4\times1, 6 + 4\times3, 5 + 4\times5) = (0, 18, 25)$$ Matrix now: $$\begin{bmatrix} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{bmatrix}$$ 5. **Step 2: Use row 2 to eliminate below:** - Multiply row 2 by $\frac{18}{26} = \frac{9}{13}$ and subtract from row 3: $$R_3 \to R_3 - \frac{9}{13} R_2 = \left(0, 18 - \frac{9}{13} \times 26, 25 - \frac{9}{13} \times 33\right) = (0, 18 - 18, 25 - \frac{297}{13}) = (0, 0, \frac{28}{13})$$ Matrix now: $$\begin{bmatrix} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 0 & \frac{28}{13} \end{bmatrix}$$ 6. **Final echelon form:** $$\boxed{\begin{bmatrix} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 0 & \frac{28}{13} \end{bmatrix}}$$ This matrix is in echelon form with leading entries moving rightward down the rows.