Math Problem Statement
Reduced row elecon form
Reduced row echelon form
Solution
The matrix shown in the image is:
\begin{pmatrix} 0 & 0 & 2 & 0 & 7 & 12 \\ 2 & 4 & -10 & 6 & 12 & 28 \\ 4 & 4 & -5 & 6 & -5 & -1 \end{pmatrix}$$ The goal is to bring this matrix to Reduced Row Echelon Form (RREF). Let's follow the steps: ### Step 1: Swap Rows (if necessary) Since the first pivot should be in the top-left position, we see that the first column in the first row has a zero. So, we need to swap rows to ensure we have a non-zero pivot. Swap row 1 and row 2: $$A = \begin{pmatrix} 2 & 4 & -10 & 6 & 12 & 28 \\ 0 & 0 & 2 & 0 & 7 & 12 \\ 4 & 4 & -5 & 6 & -5 & -1 \end{pmatrix}$$ ### Step 2: Create leading 1 in Row 1 To create a leading 1 in the first row, divide the entire first row by 2: $$A = \begin{pmatrix} 1 & 2 & -5 & 3 & 6 & 14 \\ 0 & 0 & 2 & 0 & 7 & 12 \\ 4 & 4 & -5 & 6 & -5 & -1 \end{pmatrix}$$ ### Step 3: Eliminate entries below the leading 1 in Row 1 Subtract 4 times Row 1 from Row 3 to eliminate the leading entry in the third row: $$R_3 = R_3 - 4R_1$$ $$A = \begin{pmatrix} 1 & 2 & -5 & 3 & 6 & 14 \\ 0 & 0 & 2 & 0 & 7 & 12 \\ 0 & -4 & 15 & -6 & -29 & -57 \end{pmatrix}$$ ### Step 4: Create leading 1 in Row 2 In the second row, divide by 2 to make the leading 1 for the third column: $$A = \begin{pmatrix} 1 & 2 & -5 & 3 & 6 & 14 \\ 0 & 0 & 1 & 0 & 3.5 & 6 \\ 0 & -4 & 15 & -6 & -29 & -57 \end{pmatrix}$$ ### Step 5: Eliminate entries above and below the leading 1 in Row 2 Subtract (-5) times Row 2 from Row 1 to eliminate the third column entry in the first row: $$R_1 = R_1 + 5R_2$$ $$A = \begin{pmatrix} 1 & 2 & 0 & 3 & 23.5 & 44 \\ 0 & 0 & 1 & 0 & 3.5 & 6 \\ 0 & -4 & 15 & -6 & -29 & -57 \end{pmatrix}$$ Now subtract 15 times Row 2 from Row 3 to eliminate the third column entry in the third row: $$R_3 = R_3 - 15R_2$$ $$A = \begin{pmatrix} 1 & 2 & 0 & 3 & 23.5 & 44 \\ 0 & 0 & 1 & 0 & 3.5 & 6 \\ 0 & -4 & 0 & -6 & -81.5 & -147 \end{pmatrix}$$ ### Step 6: Create leading 1 in Row 3 To eliminate the -4 entry in Row 3, first divide the third row by -4: $$A = \begin{pmatrix} 1 & 2 & 0 & 3 & 23.5 & 44 \\ 0 & 0 & 1 & 0 & 3.5 & 6 \\ 0 & 1 & 0 & 1.5 & 20.375 & 36.75 \end{pmatrix}$$ Subtract 2 times Row 3 from Row 1: $$R_1 = R_1 - 2R_3$$ Finally, subtract 2 times Row 3 from Row 2. This will leave the matrix fully reduced. Would you like to continue solving this step-by-step or ask for clarifications? Here are some related questions: 1. What is the significance of a matrix being in RREF? 2. Can RREF of a matrix be unique? 3. How does RREF help in solving systems of linear equations? 4. What conditions make a matrix invertible? 5. Can every matrix be reduced to RREF? **Tip:** Always try to work with smaller entries in a matrix to simplify your row operations!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Row Operations
Reduced Row Echelon Form (RREF)
Formulas
Row operations: Row swapping, Row scaling, Row addition/subtraction
Theorems
Reduced Row Echelon Form
Uniqueness of RREF
Suitable Grade Level
Grades 11-12 or College Level