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Part 1: Let M solution to M = 2 0 5 5 -15 -5 48 -5 12 and let b= = 15 60 [24] 6 3 .There is a very obvious = b that you should be able to find by inspection. What is it?

Part 1: Let M solution to M = 2 0 5 5 -15 -5 48 -5 12 and let b= = 15 60 [24] 6 3 .There is a very obvious = b that you should be able to find by inspection. What is it?

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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**Part 1:** Let \( M = \begin{bmatrix} 4 & 5 & -5 & 48 \\ 2 & 5 & -5 & 12 \\ 0 & -15 & 15 & 60 \end{bmatrix} \) and let \(\vec{b} = \begin{bmatrix} 24 \\ 6 \\ 3 \end{bmatrix} \). There is a very obvious solution to \( M\vec{v} = \vec{b} \) that you should be able to find by inspection. What is it? **Part 2:** Use the fact that \(\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}\) is a basis for the null space of \( A \), together with your previous answer to write down all the solutions of \( M\vec{v} = \vec{b} \) in parametric vector form.
Transcribed Image Text:**Part 1:** Let \( M = \begin{bmatrix} 4 & 5 & -5 & 48 \\ 2 & 5 & -5 & 12 \\ 0 & -15 & 15 & 60 \end{bmatrix} \) and let \(\vec{b} = \begin{bmatrix} 24 \\ 6 \\ 3 \end{bmatrix} \). There is a very obvious solution to \( M\vec{v} = \vec{b} \) that you should be able to find by inspection. What is it? **Part 2:** Use the fact that \(\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}\) is a basis for the null space of \( A \), together with your previous answer to write down all the solutions of \( M\vec{v} = \vec{b} \) in parametric vector form.
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