1. **State the problem:** Solve the system of equations using row operations to transform the augmented matrix \((A|b)\) into echelon or reduced echelon form. 2. **Write the system as an augmented matrix:** Suppose the system is \(\begin{cases} a_{11}x + a_{12}y + \cdots = b_1 \\ a_{21}x + a_{22}y + \cdots = b_2 \\ \vdots \end{cases}\), then the augmented matrix is \(\left[ A | b \right]\). 3. **Row operations allowed:** - Swap two rows. - Multiply a row by a nonzero scalar. - Add or subtract a multiple of one row to another row. 4. **Goal:** Use these operations to get zeros below the pivots (leading coefficients) to form an upper triangular matrix (echelon form) or further simplify to reduced echelon form where pivots are 1 and all other entries in pivot columns are zero. 5. **Example:** For the system $$\begin{cases} 2x + 3y = 5 \\ 4x + 9y = 15 \end{cases}$$ The augmented matrix is $$\left[ \begin{array}{cc|c} 2 & 3 & 5 \\ 4 & 9 & 15 \end{array} \right]$$ 6. **Step 1:** Make the first pivot 1 by dividing row 1 by 2: $$R_1 \to \frac{1}{2} R_1 = \left[ 1 \quad \frac{3}{2} \quad \frac{5}{2} \right]$$ 7. **Step 2:** Eliminate the entry below the pivot in column 1: $$R_2 \to R_2 - 4 \times R_1 = \left[ 4 - 4(1) \quad 9 - 4(\frac{3}{2}) \quad 15 - 4(\frac{5}{2}) \right] = \left[ 0 \quad 3 \quad 5 \right]$$ 8. **Step 3:** Make the second pivot 1 by dividing row 2 by 3: $$R_2 \to \frac{1}{3} R_2 = \left[ 0 \quad 1 \quad \frac{5}{3} \right]$$ 9. **Step 4:** Eliminate the entry above the second pivot: $$R_1 \to R_1 - \frac{3}{2} R_2 = \left[ 1 \quad 0 \quad \frac{5}{2} - \frac{3}{2} \times \frac{5}{3} \right] = \left[ 1 \quad 0 \quad \frac{5}{2} - \frac{5}{2} \right] = \left[ 1 \quad 0 \quad 0 \right]$$ 10. **Final matrix:** $$\left[ \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \end{array} \right]$$ 11. **Solution:** From the matrix, we read $$x = 0, \quad y = \frac{5}{3}$$ This completes the solution using row operations to reduced echelon form.