Math Problem Statement
1 −2 8 −1 −1 −5 =−7 0 0 0 1 −7 0 =7 0 0 0 0 0 1 =9 sorry its 3x6 matric
Solution
Ah, I see now! You're working with a matrix, which looks like this:
1 & -2 & 8 & -1 & -1 & -5 \\ 0 & 0 & 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ We will reduce this matrix to **reduced row echelon form (RREF)** using Gaussian and Gauss-Jordan elimination. Here's the step-by-step process. ### Step 1: Start with the given matrix $$A$$. $$A = \begin{bmatrix} 1 & -2 & 8 & -1 & -1 & -5 \\ 0 & 0 & 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ ### Step 2: Eliminate entries above the leading 1 in column 4. In the second row, we have a leading 1 in column 4. We want to eliminate the entry in column 4 of the first row by adding row 2 multiplied by $$1$$ to row 1: $$R_1 \to R_1 + R_2$$ This gives: $$A = \begin{bmatrix} 1 & -2 & 8 & 0 & -8 & -5 \\ 0 & 0 & 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ ### Step 3: Eliminate the entries above the leading 1 in column 6. In the third row, we have a leading 1 in column 6. To eliminate the entry in column 6 of row 1, add $$5$$ times row 3 to row 1: $$R_1 \to R_1 + 5R_3$$ This gives: $$A = \begin{bmatrix} 1 & -2 & 8 & 0 & -8 & 0 \\ 0 & 0 & 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ No other row operations are necessary since the matrix already satisfies the properties of **reduced row echelon form (RREF)**: - Each leading entry in a row is 1. - The leading entries are the only nonzero entries in their columns. - The leading 1s move from left to right as you move down the rows. ### Final RREF Matrix: \[ A = \begin{bmatrix} 1 & -2 & 8 & 0 & -8 & 0 \\Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Reduced Row Echelon Form (RREF)
Gaussian Elimination
Formulas
Row Operations: R1 → R1 + R2
Row Operations: R1 → R1 + 5R3
Theorems
Gauss-Jordan Elimination
Properties of RREF
Suitable Grade Level
College Level