Math Problem Statement
Find the reduced echelon form of the given matrix. Start 3 By 3 Table 1st Row 1st Column 5 2nd Column 2 3rd Column 41 2nd Row 1st Column 0 2nd Column 1 3rd Column 3 3rd Row 1st Column 4 2nd Column 1 3rd Column 31 EndTable
Question content area bottom Part 1 The reduced echelon form is enter your response here. (Type an integer or decimal for each matrix element.)
Solution
We are asked to find the reduced echelon form (also known as reduced row echelon form (RREF)) of the given 3x3 matrix:
5 & 2 & 41 \\ 0 & 1 & 3 \\ 4 & 1 & 31 \end{pmatrix}$$ To solve this, we'll perform Gaussian elimination, followed by back substitution. ### Step 1: Make the leading entry of the first row equal to 1 (if it’s not already). - Divide the entire first row by 5: $$R_1 \to \frac{1}{5} R_1$$ Resulting matrix: $$\begin{pmatrix} 1 & \frac{2}{5} & \frac{41}{5} \\ 0 & 1 & 3 \\ 4 & 1 & 31 \end{pmatrix}$$ ### Step 2: Zero out the first column below the pivot. - To zero out the 4 in the first column of the third row, subtract $$4 \times$$ the first row from the third row: $$R_3 \to R_3 - 4R_1$$ Resulting matrix: $$\begin{pmatrix} 1 & \frac{2}{5} & \frac{41}{5} \\ 0 & 1 & 3 \\ 0 & -\frac{3}{5} & \frac{1}{5} \end{pmatrix}$$ ### Step 3: Zero out the second column above the pivot (second row, second column). - Add $$\frac{3}{5} \times$$ the second row to the third row: $$R_3 \to R_3 + \frac{3}{5} R_2$$ Resulting matrix: $$\begin{pmatrix} 1 & \frac{2}{5} & \frac{41}{5} \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{pmatrix}$$ ### Step 4: Make the third row's third column entry equal to 1. - Divide the third row by 2: $$R_3 \to \frac{1}{2} R_3$$ Resulting matrix: $$\begin{pmatrix} 1 & \frac{2}{5} & \frac{41}{5} \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}$$ ### Step 5: Zero out the third column entries in rows 1 and 2. - Subtract $$\frac{41}{5} \times$$ the third row from the first row: $$R_1 \to R_1 - \frac{41}{5} R_3$$ - Subtract $$3 \times$$ the third row from the second row: $$R_2 \to R_2 - 3 R_3$$ Resulting matrix: $$\begin{pmatrix} 1 & \frac{2}{5} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ ### Step 6: Zero out the second column entry in the first row. - Subtract $$\frac{2}{5} \times$$ the second row from the first row: $$R_1 \to R_1 - \frac{2}{5} R_2$$ Final reduced echelon form matrix: $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ This is the **reduced echelon form** (or reduced row echelon form) of the given matrix. Let me know if you need more details or have any further questions! Here are some related questions you might find helpful: 1. How do you perform row swaps in Gaussian elimination? 2. What is the difference between row echelon form and reduced row echelon form? 3. How can you use Gaussian elimination to solve a system of linear equations? 4. What happens if a matrix has no solution after applying Gaussian elimination? 5. How can Gaussian elimination help in finding the rank of a matrix? **Tip**: Always start with the pivot in the first row and first column, and ensure that the entries below it are zero before moving to the next pivot. This systematic approach makes it easier to apply the transformations step by step.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Matrix Operations
Formulas
Gaussian Elimination
Row Operations (RREF)
Theorems
Row Echelon Form
Reduced Row Echelon Form (RREF)
Suitable Grade Level
Grades 10-12
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