Math Problem Statement
Reduced row elecon fom convert
Reduced row echelon form conversion of matrix
Solution
The image you provided shows a matrix, and it looks like you're performing row reduction (Gaussian elimination) to convert the matrix into reduced row echelon form (RREF). I will walk you through the process to complete the reduction and verify the steps.
The matrix given is:
0 & 0 & 7 \\ 2 & -10 & 0 \\ 4 & 12 & 1 \\ 5 & 6 & 7 \end{bmatrix}$$ Let's now proceed with the row reduction step-by-step: ### Step 1: Make the leading entry in the first row a 1. Since the first entry in the first column is a zero, we swap rows to bring a non-zero entry to the top. We can swap row 1 with row 2 or row 4 (whichever you prefer). Let's swap row 1 with row 2: $$\begin{bmatrix} 2 & -10 & 0 \\ 0 & 0 & 7 \\ 4 & 12 & 1 \\ 5 & 6 & 7 \end{bmatrix}$$ ### Step 2: Make the leading 2 in row 1 a 1. We can divide the entire first row by 2: $$\begin{bmatrix} 1 & -5 & 0 \\ 0 & 0 & 7 \\ 4 & 12 & 1 \\ 5 & 6 & 7 \end{bmatrix}$$ ### Step 3: Eliminate the first column entries below the pivot (the 1 in row 1). To eliminate the 4 and 5 below the pivot, we perform the following row operations: - $$R_3 = R_3 - 4 \cdot R_1$$ - $$R_4 = R_4 - 5 \cdot R_1$$ After applying these operations, the matrix becomes: $$\begin{bmatrix} 1 & -5 & 0 \\ 0 & 0 & 7 \\ 0 & 32 & 1 \\ 0 & 31 & 7 \end{bmatrix}$$ ### Step 4: Make the leading entry in the second row a 1. To do this, divide the entire second row by 7: $$\begin{bmatrix} 1 & -5 & 0 \\ 0 & 0 & 1 \\ 0 & 32 & 1 \\ 0 & 31 & 7 \end{bmatrix}$$ ### Step 5: Eliminate the second column entries above and below the pivot in the second row. To eliminate the second column entries, we use the following row operations: - $$R_1 = R_1 - 0 \cdot R_2$$ (This doesn't change row 1 since it's already zero in the third column.) - $$R_3 = R_3 - 32 \cdot R_2$$ - $$R_4 = R_4 - 31 \cdot R_2$$ After performing these operations, the matrix becomes: $$\begin{bmatrix} 1 & -5 & 0 \\ 0 & 0 & 1 \\ 0 & 32 & 0 \\ 0 & 31 & 0 \end{bmatrix}$$ ### Step 6: Eliminate the 32 and 31 entries in the third and fourth rows. We can subtract the third row from the fourth to finish the elimination process: $$\begin{bmatrix} 1 & -5 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ This is the reduced row echelon form (RREF) of the matrix. Do you want more details on this or have any questions? ### Five Related Questions: 1. How do you determine the pivot positions when performing row reduction? 2. What are the advantages of using RREF in solving systems of linear equations? 3. Can every matrix be transformed into reduced row echelon form? 4. How can row reduction be used to find the inverse of a matrix? 5. What are the differences between Gaussian elimination and Gauss-Jordan elimination? ### Tip: When performing row reduction, always focus on clearing columns below and above the pivots as early as possible to streamline the process of reaching RREF.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Row Reduction
Gaussian Elimination
Gauss-Jordan Elimination
Formulas
Row reduction formulas for Gaussian elimination
RREF transformation formulas
Theorems
Gaussian Elimination Theorem
Gauss-Jordan Elimination Method
Suitable Grade Level
Grades 10-12 (Undergraduate Level)