Row Reduction of Matrix to Row Echelon Form

Math Problem Statement

[ = \left[\begin{array}{cccc}1 & -1 & -2 & -6 \ 1 & -1 & -2 & -6 \ 2 & -2 & -4 & -12 \ 3 & -3 & -6 & -18\end{array}\right] ] row reduce this

Let's perform row reduction to simplify the given matrix step-by-step. The goal is to transform the matrix into row echelon form.

1 & -1 & -2 & -6 \\ 1 & -1 & -2 & -6 \\ 2 & -2 & -4 & -12 \\ 3 & -3 & -6 & -18 \end{bmatrix}$$ --- ### Step 1: Subtract row 1 from row 2, row 3, and row 4 to eliminate entries below the pivot in the first column. Perform the following operations: - $$R_2 \to R_2 - R_1$$ - $$R_3 \to R_3 - 2R_1$$ - $$R_4 \to R_4 - 3R_1$$ The resulting matrix is: $$\begin{bmatrix} 1 & -1 & -2 & -6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ --- ### Step 2: Notice that rows 2, 3, and 4 are all zero rows. The matrix is already in row echelon form. The final row-reduced echelon form is: $$\boxed{\begin{bmatrix} 1 & -1 & -2 & -6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}}$$ This indicates the system of equations represented by the matrix has no further independent information beyond the first row. --- Do you want to explore interpretation of this result or its implications? ### Follow-up questions: 1. What does this row reduction imply about the solutions to a system of equations represented by this matrix? 2. How do we interpret zero rows in the context of linear systems? 3. Would the solution be consistent if the augmented column changed? 4. How do we check if a matrix has dependent rows? 5. Can this matrix be converted to reduced row-echelon form? ### Tip: In row reduction, always check if rows are linearly dependent to minimize unnecessary operations.

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Linear Algebra
Row Reduction
Matrix Operations
Row Echelon Form

Row operations: R2 -> R2 - R1, R3 -> R3 - 2R1, R4 -> R4 - 3R1

Row Echelon Form
Linear Dependence
Rank of a Matrix

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