Math Problem Statement
Reduce the matrix to its row-reduced form.
3 6 −5 40
5 7 5 −27 0 8 5 47
Solution
We are given the following augmented matrix:
3 & 6 & -5 & 40 \\ 5 & 7 & 5 & -27 \\ 0 & 8 & 5 & 47 \end{pmatrix}$$ ### Goal: Reduce this matrix to row-reduced echelon form (RREF). #### Step 1: Get a leading 1 in the first row, first column We can swap rows 1 and 2 to make the first pivot element (3) easier to work with. $$\begin{pmatrix} 5 & 7 & 5 & -27 \\ 3 & 6 & -5 & 40 \\ 0 & 8 & 5 & 47 \end{pmatrix}$$ #### Step 2: Make the first column below the pivot (5) zero We will perform row operations to eliminate the entry below the pivot. To eliminate the entry below the pivot, we'll do: $$R_2 = R_2 - \left(\frac{3}{5}\right) R_1$$ This operation modifies row 2 by subtracting $$\frac{3}{5}$$ of row 1 from it. #### Step 3: Simplify the matrix Let's calculate the new row 2. $$R_2 = \left(3 - \frac{3}{5} \times 5, 6 - \frac{3}{5} \times 7, -5 - \frac{3}{5} \times 5, 40 - \frac{3}{5} \times (-27)\right)$$ We get: $$R_2 = (0, \frac{15}{5}, -\frac{40}{5}, 40 + \frac{81}{5})$$ So we have: $$R_2 = (0, 3, -8, 54.4)$$ Let’s update the matrix: $$\begin{pmatrix} 5 & 7 & 5 & -27 \\ 0 & 3 & -8 & 54.4 \\ 0 & 8 & 5 & 47 \end{pmatrix}$$ #### Step 4: Make the pivot below the second pivot zero Next, we work on the third row to eliminate the second column's entry: $$R_3 = R_3 - \frac{8}{3} R_2$$ We will now compute the new row 3: $$R_3 = (0 - \frac{8}{3} \times 0, 8 - \frac{8}{3} \times 3, 5 - \frac{8}{3} \times -8, 47 - \frac{8}{3} \times 54.4)$$ Let’s simplify this row: $$R_3 = (0, 0, 29, -43.1)$$ #### Step 5: Normalize the pivots Now, we normalize the pivot in row 2 (the 3) to become 1 by dividing row 2 by 3: $$R_2 = \frac{1}{3} R_2$$ Resulting in: $$R_2 = (0, 1, -\frac{8}{3}, 18.13)$$ Now for row 3, normalize the pivot by dividing by 29: $$R_3 = \frac{1}{29} R_3$$ Resulting in: $$R_3 = (0, 0, 1, -1.48)$$ #### Step 6: Eliminate above pivots Finally, we work on clearing the entries above the pivots in rows 2 and 3. For row 1, we perform the operation: $$R_1 = R_1 - 7 R_2$$ For row 1, we subtract 7 times row 2 from it: $$R_1 = (5 - 7 \times 0, 7 - 7 \times 1, 5 - 7 \times -\frac{8}{3}, -27 - 7 \times 18.13)$$ Calculating this gives: $$R_1 = (5, 0, \frac{5 + 56}{3}, -178.91)$$ So, we end with the row-reduced matrix: $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -\frac{8}{3} & 18.13 \\ 0 & 0 & 1 & -1.48 \end{pmatrix}$$ This is the row-reduced echelon form of the matrix. --- Would you like a further breakdown of the steps or have any specific questions about the process? Here are a few related questions to explore: 1. How do we interpret row-reduced echelon form in terms of solving a system of linear equations? 2. What is the difference between row echelon form and reduced row echelon form? 3. How do Gaussian elimination and Gauss-Jordan elimination compare in terms of process and outcome? 4. Can row-reduced echelon form be used to determine if a system of equations has a unique solution, infinitely many solutions, or no solution? 5. How do you use matrix operations to solve linear systems? **Tip**: Always start by ensuring that your leading entries (pivots) are 1 before trying to eliminate other entries in the column!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Row Echelon Form
Gaussian Elimination
Formulas
Row operations to manipulate matrices
Theorems
Row-Reduced Echelon Form (RREF)
Gaussian Elimination
Suitable Grade Level
Grades 11-12