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Question 10 152 0 0 1 Describe all solutions of Ax = 0 in parametric vector form, where A = 0 0 0 0 7 x = 4₂ O x = 2₂ x = 2₂ 5 0 0 0 51 1 0 0 0 5 1 0 0 + IL + 14 +14 201 7 1 0 8 0 7 1 0 0 7 1 + as + Is + 15 + IS ToTomo -4 1 0 -4 0 0 4 0 0 0 -6 9 0 -74 -8 0 0 1 0 0 0

Question 10 152 0 0 1 Describe all solutions of Ax = 0 in parametric vector form, where A = 0 0 0 0 7 x = 4₂ O x = 2₂ x = 2₂ 5 0 0 0 51 1 0 0 0 5 1 0 0 + IL + 14 +14 201 7 1 0 8 0 7 1 0 0 7 1 + as + Is + 15 + IS ToTomo -4 1 0 -4 0 0 4 0 0 0 -6 9 0 -74 -8 0 0 1 0 0 0

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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### Question 10 **Problem Statement:** Describe all solutions of \( \mathbf{A}\mathbf{x} = \mathbf{0} \) in parametric vector form, where: \[ \mathbf{A} = \begin{bmatrix} 1 & 5 & 2 & -6 & 9 & 0 \\ 0 & 1 & -7 & 4 & -8 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \] **Choices:** 1. \[ \mathbf{x} = x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 2. \[ \mathbf{x} = x_2 \begin{bmatrix} 5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 3. \[ \mathbf{x} = x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 4. \[ \mathbf{x} =
Transcribed Image Text:### Question 10 **Problem Statement:** Describe all solutions of \( \mathbf{A}\mathbf{x} = \mathbf{0} \) in parametric vector form, where: \[ \mathbf{A} = \begin{bmatrix} 1 & 5 & 2 & -6 & 9 & 0 \\ 0 & 1 & -7 & 4 & -8 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \] **Choices:** 1. \[ \mathbf{x} = x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 2. \[ \mathbf{x} = x_2 \begin{bmatrix} 5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 3. \[ \mathbf{x} = x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix} \] 4. \[ \mathbf{x} =
**Linear Algebra: Finding Non-Trivial Solutions** **Problem Statement:** Find a non-trivial solution of the equation \( A \mathbf{x} = \mathbf{0} \), where \( A = \begin{bmatrix} -15 & -3 \\ 30 & 6 \\ -10 & -2 \end{bmatrix} \). **Options:** 1. \(\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) 2. \(\mathbf{x} = \begin{bmatrix} -5 \\ 1 \end{bmatrix}\) 3. \(\mathbf{x} = \begin{bmatrix} 5 \\ -1 \end{bmatrix}\) 4. \(\mathbf{x} = \begin{bmatrix} 1 \\ -5 \end{bmatrix}\) 5. \(\mathbf{x} = \begin{bmatrix} -1 \\ 5 \end{bmatrix}\) **Explanation:** In this problem, we are given a matrix \( A \) and we need to find a non-trivial solution \(\mathbf{x}\) such that \( A \mathbf{x} = \mathbf{0} \). A non-trivial solution is a solution where \(\mathbf{x} \neq \mathbf{0} \). You are required to find the correct \(\mathbf{x}\) vector from the given options that satisfies this equation.
Transcribed Image Text:**Linear Algebra: Finding Non-Trivial Solutions** **Problem Statement:** Find a non-trivial solution of the equation \( A \mathbf{x} = \mathbf{0} \), where \( A = \begin{bmatrix} -15 & -3 \\ 30 & 6 \\ -10 & -2 \end{bmatrix} \). **Options:** 1. \(\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) 2. \(\mathbf{x} = \begin{bmatrix} -5 \\ 1 \end{bmatrix}\) 3. \(\mathbf{x} = \begin{bmatrix} 5 \\ -1 \end{bmatrix}\) 4. \(\mathbf{x} = \begin{bmatrix} 1 \\ -5 \end{bmatrix}\) 5. \(\mathbf{x} = \begin{bmatrix} -1 \\ 5 \end{bmatrix}\) **Explanation:** In this problem, we are given a matrix \( A \) and we need to find a non-trivial solution \(\mathbf{x}\) such that \( A \mathbf{x} = \mathbf{0} \). A non-trivial solution is a solution where \(\mathbf{x} \neq \mathbf{0} \). You are required to find the correct \(\mathbf{x}\) vector from the given options that satisfies this equation.
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