Let $\mathbf{u} = \begin{bmatrix} -21 \\ 2 \\ -8 \end{bmatrix}$ and $A = \begin{bmatrix} 5 & 9 & -17 \\ 2 & 1 & 1 \\ 1 & 3 & -7 \end{bmatrix}$. Is $\mathbf{u}$ in the subset of $\mathbb{R}^3$ spanned by the columns of $A$? Why or why not?---Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or decimal for each matrix element.)- A. Yes, multiplying $A$ by the vector $\begin{bmatrix} \text{ } \end{bmatrix}$ writes $\mathbf{u}$ as a linear combination of the columns of $A$.- B. No, the reduced row echelon form of the augmented matrix is $\begin{bmatrix} \text{ } \end{bmatrix}$, which is an inconsistent system.

Given u=−212−8 and A=59−1721113−7. We have to check whether vector u is the subset of R3 spanned by…

Answered: - 21 5 9 - 17 Let u = 2 and A = 2 1 Is…

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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Related questions

Question

Let $\mathbf{u} = \begin{bmatrix} -21 \\ 2 \\ -8 \end{bmatrix}$ and $A = \begin{bmatrix} 5 & 9 & -17 \\ 2 & 1 & 1 \\ 1 & 3 & -7 \end{bmatrix}$. Is $\mathbf{u}$ in the subset of $\mathbb{R}^3$ spanned by the columns of $A$? Why or why not?

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Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or decimal for each matrix element.)

- A. Yes, multiplying $A$ by the vector $\begin{bmatrix} \text{ } \end{bmatrix}$ writes $\mathbf{u}$ as a linear combination of the columns of $A$.

- B. No, the reduced row echelon form of the augmented matrix is $\begin{bmatrix} \text{ } \end{bmatrix}$, which is an inconsistent system.

Expert Solution

Step 1

Given $u = [\begin{matrix} - 21 \\ 2 \\ - 8 \end{matrix}]$ and $A = [\begin{matrix} 5 & 9 & - 17 \\ 2 & 1 & 1 \\ 1 & 3 & - 7 \end{matrix}]$ .

We have to check whether vector $u$ is the subset of $R^{3}$ spanned by the columns of $A$ .

The column vectors of $A$ are $u_{1} = (5, 2, 1)$ , $u_{2} = (9, 1, 3)$ and $u_{3} = (- 17, 1, - 7)$ . If vector $u$ is the span of the column of $A$ then there must exist scalars $a, b a n d c$ such that $a u_{1} + b u_{2} + c u_{3} = u$ . This reduces to a system of $3$ linear equations in $3$ variables. We will use Cramer rule here.

$a u_{1} + b u_{2} + c u_{3} = u$

$\begin{array}{l} \Rightarrow a [5, 2, 1] + b [9, 1, 3] + c [- 17, 1, - 7] = [- 21, 2, - 8] \\ \Rightarrow 5 a + 9 b - 17 c = - 21 \\ 2 a + b + c = 2 \\ a + 3 b - 7 c = - 8 \end{array}$