1. **State the problem:** We are given the augmented matrix $$\begin{bmatrix} 2 & 3 & -1 & 9 \\ 1 & -1 & 2 & -3 \\ 3 & 1 & 3 & 2 \end{bmatrix}$$ and asked to find its reduced row echelon form (RREF). 2. **Recall the goal and rules:** The RREF of a matrix has leading 1s in each row, zeros below and above these leading 1s, and rows of zeros (if any) at the bottom. 3. **Step 1: Make the first pivot 1.** Divide row 2 by 1 (already 1), so keep row 2 as is. 4. **Step 2: Swap row 1 and row 2** to get a leading 1 at the top: $$\begin{bmatrix} 1 & -1 & 2 & -3 \\ 2 & 3 & -1 & 9 \\ 3 & 1 & 3 & 2 \end{bmatrix}$$ 5. **Step 3: Eliminate entries below the pivot in column 1:** - Replace row 2 by row 2 - 2*row 1: $$2 - 2(1) = 0, \quad 3 - 2(-1) = 3 + 2 = 5, \quad -1 - 2(2) = -1 - 4 = -5, \quad 9 - 2(-3) = 9 + 6 = 15$$ - Replace row 3 by row 3 - 3*row 1: $$3 - 3(1) = 0, \quad 1 - 3(-1) = 1 + 3 = 4, \quad 3 - 3(2) = 3 - 6 = -3, \quad 2 - 3(-3) = 2 + 9 = 11$$ Matrix becomes: $$\begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 5 & -5 & 15 \\ 0 & 4 & -3 & 11 \end{bmatrix}$$ 6. **Step 4: Make the pivot in row 2, column 2 equal to 1:** Divide row 2 by 5: $$\begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & 4 & -3 & 11 \end{bmatrix}$$ 7. **Step 5: Eliminate entries above and below the pivot in column 2:** - Replace row 1 by row 1 + row 2: $$1 + 0 = 1, \quad -1 + 1 = 0, \quad 2 + (-1) = 1, \quad -3 + 3 = 0$$ - Replace row 3 by row 3 - 4*row 2: $$0 - 4(0) = 0, \quad 4 - 4(1) = 0, \quad -3 - 4(-1) = -3 + 4 = 1, \quad 11 - 4(3) = 11 - 12 = -1$$ Matrix becomes: $$\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$ 8. **Step 6: Make the pivot in row 3, column 3 equal to 1 (already 1).** 9. **Step 7: Eliminate entries above the pivot in column 3:** - Replace row 1 by row 1 - row 3: $$1 - 0 = 1, \quad 0 - 0 = 0, \quad 1 - 1 = 0, \quad 0 - (-1) = 1$$ - Replace row 2 by row 2 + row 3: $$0 + 0 = 0, \quad 1 + 0 = 1, \quad -1 + 1 = 0, \quad 3 + (-1) = 2$$ Matrix becomes: $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$ 10. **Final RREF matrix:** $$\boxed{\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix}}$$ This corresponds to the system: $$x = 1, \quad y = 2, \quad z = -1$$ **Summary:** We transformed the original matrix into its reduced row echelon form by performing row operations to get leading 1s and zeros above and below pivots.