Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Question
**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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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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