Math Problem Statement
I want to transform a matrix with a first row of 1 0 -2 0, second row of -1 0 1 0, and final row of 3 2 6 0 into rref form.
Solution
We are tasked with transforming the matrix
1 & 0 & -2 & 0 \\ -1 & 0 & 1 & 0 \\ 3 & 2 & 6 & 0 \end{pmatrix}$$ into **Reduced Row Echelon Form (RREF)**. ### Step 1: Start with the matrix $$\begin{pmatrix} 1 & 0 & -2 & 0 \\ -1 & 0 & 1 & 0 \\ 3 & 2 & 6 & 0 \end{pmatrix}$$ ### Step 2: Make the first pivot element 1 (which is already the case) The first pivot is already 1 at position (1,1), so no changes are needed here. ### Step 3: Eliminate the first column below the pivot We want to eliminate the entries in the first column below the pivot (in positions (2,1) and (3,1)) by adding multiples of row 1 to rows 2 and 3: - Row 2: Add Row 1 to Row 2: $$\text{Row 2} \rightarrow \text{Row 2} + \text{Row 1} = (-1, 0, 1, 0) + (1, 0, -2, 0) = (0, 0, -1, 0)$$ - Row 3: Subtract 3 times Row 1 from Row 3: $$\text{Row 3} \rightarrow \text{Row 3} - 3 \times \text{Row 1} = (3, 2, 6, 0) - 3 \times (1, 0, -2, 0) = (0, 2, 12, 0)$$ The updated matrix now looks like this: $$\begin{pmatrix} 1 & 0 & -2 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 2 & 12 & 0 \end{pmatrix}$$ ### Step 4: Make the second pivot 1 We want to make the second pivot (position (2,3)) a 1. To do that, we multiply Row 2 by -1: $$\text{Row 2} \rightarrow -1 \times \text{Row 2} = (0, 0, 1, 0)$$ Now the matrix is: $$\begin{pmatrix} 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 12 & 0 \end{pmatrix}$$ ### Step 5: Eliminate the third column entries We want to eliminate the entries in the third column (position (1,3) and (3,3)) by using Row 2. - Row 1: Add 2 times Row 2 to Row 1: $$\text{Row 1} \rightarrow \text{Row 1} + 2 \times \text{Row 2} = (1, 0, -2, 0) + 2 \times (0, 0, 1, 0) = (1, 0, 0, 0)$$ - Row 3: Subtract 12 times Row 2 from Row 3: $$\text{Row 3} \rightarrow \text{Row 3} - 12 \times \text{Row 2} = (0, 2, 12, 0) - 12 \times (0, 0, 1, 0) = (0, 2, 0, 0)$$ The matrix now becomes: $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \end{pmatrix}$$ ### Step 6: Make the second pivot column entry 1 To finish the process, we want to eliminate the 2 in position (3,2) by using Row 3. Divide Row 3 by 2 to make the pivot a 1: $$\text{Row 3} \rightarrow \frac{1}{2} \times \text{Row 3} = (0, 1, 0, 0)$$ Now the matrix is: $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$$ ### Step 7: Final Cleanup Now, we use Row 3 to eliminate the 1 in position (2,2): - Row 2: Subtract Row 3 from Row 2: $$\text{Row 2} \rightarrow \text{Row 2} - \text{Row 3} = (0, 0, 1, 0) - (0, 1, 0, 0) = (0, -1, 0, 0)$$ So, now we have: $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$ This is the **Reduced Row Echelon Form (RREF)** of the given matrix. --- Would you like more details on any of the steps or have additional questions? ### Further Questions: 1. How would you perform the same row reduction process on a 5x5 matrix? 2. How does the rank of a matrix relate to its RREF form? 3. Can a matrix always be reduced to RREF form using elementary row operations? 4. What is the difference between row echelon form (REF) and reduced row echelon form (RREF)? 5. How do you interpret the RREF matrix in terms of the system of equations it represents? ### Tip: When performing row reductions, keep an eye on the pivot positions, as they guide how you eliminate variables.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Row Operations
Row Echelon Form
Reduced Row Echelon Form (RREF)
Formulas
Row operations: Addition, Subtraction, Scalar Multiplication
Theorems
Row Echelon Form
Reduced Row Echelon Form
Suitable Grade Level
Grades 10-12