i need help working it out - 25 87Let u=-2 and A = 1 2 1 Is u in the subset of R' spanned by the columns of A? Why or why not?133 1 8Select the correct choice below and fillin the answer box to complete your choice.(Type an integer or decimal for each matrix element.)O A. No, the reduced echelon form of the augmented matrix iswhich is an inconsistent system.B. Yes, multiplying A by the vector writes u as a linear combination of the columns of A.

Name: i need help working it out - 25 87Let u=-2 and A = 1 2 1 Is u in the s
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Description: i need help working it out - 25 87Let u=-2 and A = 1 2 1 Is u in the subset of R' spanned by the columns of A? Why or why not?133 1 8Select the correct choice below and fillin the answer box to complete your choice.(Type an integer or decimal for eac

Given A = 587121318 Consider the augmented matrix…

Answered: -2 5 8 7 Let u = -2 and A =12 1 Is u in…

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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i need help working it out

- 2
5 87
Let u=
-2 and A = 1 2 1 Is u in the subset of R' spanned by the columns of A? Why or why not?
13
3 1 8
Select the correct choice below and fillin the answer box to complete your choice.
(Type an integer or decimal for each matrix element.)
O A. No, the reduced echelon form of the augmented matrix is
which is an inconsistent system.
B. Yes, multiplying A by the vector writes u as a linear combination of the columns of A.

Expert Solution

Step 1

Given $A = [\begin{matrix} 5 & 8 & 7 \\ 1 & 2 & 1 \\ 3 & 1 & 8 \end{matrix}]$

Consider the augmented matrix

$[\begin{matrix} 5 & 8 & 7 & - 2 \\ 1 & 2 & 1 & - 2 \\ 3 & 1 & 8 & 13 \end{matrix}] R_{1} \leftrightarrow \frac{R_{1}}{5} [\begin{matrix} 1 & 1.6 & 1.4 & - 0.4 \\ 1 & 2 & 1 & - 2 \\ 3 & 1 & 8 & 13 \end{matrix}] R_{2} \to R_{2} - R_{1} [\begin{matrix} 1 & 1.6 & 1.4 & - 0.4 \\ 0 & 0.4 & - 0.4 & - 1.6 \\ 3 & 1 & 8 & 13 \end{matrix}] R_{3} \to R_{3} - 3 R_{1} [\begin{matrix} 1 & 1.6 & 1.4 & - 0.4 \\ 0 & 0.4 & - 0.4 & - 1.6 \\ 0 & - 3.8 & 3.8 & 14.2 \end{matrix}] R_{2} \leftrightarrow \frac{R_{2}}{0.4} [\begin{matrix} 1 & 1.6 & 1.4 & - 0.4 \\ 0 & 1 & - 1 & - 4 \\ 0 & - 3.8 & 3.8 & 14.2 \end{matrix}] R_{3} \to R_{3} + 3.8 R_{2} [\begin{matrix} 1 & 1.6 & 1.4 & - 0.4 \\ 0 & 1 & - 1 & - 4 \\ 0 & 0 & 0 & - 1 \end{matrix}] R_{1} \to R_{1} - 1.6 R_{2} [\begin{matrix} 1 & 0 & 3 & 6 \\ 0 & 1 & - 1 & - 4 \\ 0 & 0 & 0 & - 1 \end{matrix}]$

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