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Consider the matrix B and the reduced row echelon form of B. 1 4 2 1 0 1 -4 1 B = 0 2 7 1 9. -7 2.

Consider the matrix B and the reduced row echelon form of B. 1 4 2 1 0 1 -4 1 B = 0 2 7 1 9. -7 2.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Linear algebra show all work and explain .
**Question:** Is the equation Bx = p consistent for all p in R^4? Explain. --- **Explanation:** To determine the consistency of the equation \( Bx = p \) for all \( p \) in \( \mathbb{R}^4 \): 1. **Understand the problem:** - We are given a matrix \( B \). - We need to determine if, for every vector \( p \) in \( \mathbb{R}^4 \), there exists a vector \( x \) such that \( Bx = p \). 2. **Matrix Condition:** - For the equation \( Bx = p \) to be consistent for all \( p \) in \( \mathbb{R}^4 \), matrix \( B \) must be able to span \( \mathbb{R}^4 \). - This implies that \( B \) must have full rank, meaning all rows (and columns) of \( B \) are independent. 3. **Matrix Dimensions:** - If \( B \) is a \( 4 \times 4 \) matrix: - \( B \) must be invertible. - An invertible matrix has a non-zero determinant. - It means \( \text{rank}(B) = 4 \). - Such a matrix can map any vector \( x \) to any vector \( p \) in \( \mathbb{R}^4 \). 4. **Conclusion:** - If \( B \) is a \( 4 \times 4 \) matrix with full rank (invertible), then \( Bx = p \) is consistent for all \( p \) in \( \mathbb{R}^4 \). - If \( B \) is not invertible (does not have full rank), there will be some vectors \( p \) in \( \mathbb{R}^4 \) for which \( Bx = p \) does not have a solution. We can thus state that for the equation \( Bx = p \) to be consistent for all \( p \) in \( \mathbb{R}^4 \), the matrix \( B \) must be a \( 4 \times 4 \) invertible matrix. Without information about the rank or specific properties of \( B \), we cannot conclus
Transcribed Image Text:**Question:** Is the equation Bx = p consistent for all p in R^4? Explain. --- **Explanation:** To determine the consistency of the equation \( Bx = p \) for all \( p \) in \( \mathbb{R}^4 \): 1. **Understand the problem:** - We are given a matrix \( B \). - We need to determine if, for every vector \( p \) in \( \mathbb{R}^4 \), there exists a vector \( x \) such that \( Bx = p \). 2. **Matrix Condition:** - For the equation \( Bx = p \) to be consistent for all \( p \) in \( \mathbb{R}^4 \), matrix \( B \) must be able to span \( \mathbb{R}^4 \). - This implies that \( B \) must have full rank, meaning all rows (and columns) of \( B \) are independent. 3. **Matrix Dimensions:** - If \( B \) is a \( 4 \times 4 \) matrix: - \( B \) must be invertible. - An invertible matrix has a non-zero determinant. - It means \( \text{rank}(B) = 4 \). - Such a matrix can map any vector \( x \) to any vector \( p \) in \( \mathbb{R}^4 \). 4. **Conclusion:** - If \( B \) is a \( 4 \times 4 \) matrix with full rank (invertible), then \( Bx = p \) is consistent for all \( p \) in \( \mathbb{R}^4 \). - If \( B \) is not invertible (does not have full rank), there will be some vectors \( p \) in \( \mathbb{R}^4 \) for which \( Bx = p \) does not have a solution. We can thus state that for the equation \( Bx = p \) to be consistent for all \( p \) in \( \mathbb{R}^4 \), the matrix \( B \) must be a \( 4 \times 4 \) invertible matrix. Without information about the rank or specific properties of \( B \), we cannot conclus
### Matrix and Reduced Row Echelon Form --- #### Consider the matrix \( B \) and the reduced row echelon form of \( B \). \[ B = \begin{bmatrix} 1 & 4 & 2 \\ 0 & 1 & -4 \\ 0 & 2 & 7 \\ 2 & 9 & -7 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] Here, the given matrix \( B \) is a 4x3 matrix: \[ \begin{bmatrix} 1 & 4 & 2 \\ 0 & 1 & -4 \\ 0 & 2 & 7 \\ 2 & 9 & -7 \end{bmatrix} \] And its reduced row echelon form (RREF) is: \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] ### Explanation of Matrices - **Matrix \( B \)**: This matrix is an initial matrix with 4 rows and 3 columns. - **Reduced Row Echelon Form of \( B \)**: This is the form of matrix \( B \) where every leading coefficient is 1, and all other entries in the column containing a leading coefficient are zero. Rows containing only zeros are at the bottom of the matrix. ### Steps to Achieve the RREF: 1. Perform row operations to substitute leading coefficients as 1. 2. Adjust the matrix so that all elements above and below these leading coefficients are 0. 3. Rearrange any rows with all zeros to the bottom of the matrix. The transformation shows how any non-zero matrix can be simplified to its RREF, preserving its row equivalence while making it simpler to derive solutions for equations involving that matrix.
Transcribed Image Text:### Matrix and Reduced Row Echelon Form --- #### Consider the matrix \( B \) and the reduced row echelon form of \( B \). \[ B = \begin{bmatrix} 1 & 4 & 2 \\ 0 & 1 & -4 \\ 0 & 2 & 7 \\ 2 & 9 & -7 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] Here, the given matrix \( B \) is a 4x3 matrix: \[ \begin{bmatrix} 1 & 4 & 2 \\ 0 & 1 & -4 \\ 0 & 2 & 7 \\ 2 & 9 & -7 \end{bmatrix} \] And its reduced row echelon form (RREF) is: \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] ### Explanation of Matrices - **Matrix \( B \)**: This matrix is an initial matrix with 4 rows and 3 columns. - **Reduced Row Echelon Form of \( B \)**: This is the form of matrix \( B \) where every leading coefficient is 1, and all other entries in the column containing a leading coefficient are zero. Rows containing only zeros are at the bottom of the matrix. ### Steps to Achieve the RREF: 1. Perform row operations to substitute leading coefficients as 1. 2. Adjust the matrix so that all elements above and below these leading coefficients are 0. 3. Rearrange any rows with all zeros to the bottom of the matrix. The transformation shows how any non-zero matrix can be simplified to its RREF, preserving its row equivalence while making it simpler to derive solutions for equations involving that matrix.
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