PLEASE do both. I will give thumbs up!!! PLEASE. # Linear Algebra: Vector Spaces and Span## Given Problem:Let \[ A = \begin{bmatrix} 1 & 0 & -6 \\0 & 3 & -5 \\-5 & 9 & 4 \end{bmatrix} \]and \[ b = \begin{bmatrix}9 \\-2 \\-29 \end{bmatrix} \]Denote the columns of A by \(\mathbf{a_1}\), \(\mathbf{a_2}\), \(\mathbf{a_3}\), and let \(W = \text{Span}\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\).### Questions:1. **Is \(\mathbf{b}\) in \(\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\)? How many vectors are in \(\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\)?**2. **Is \(\mathbf{b}\) in \(W\)? How many vectors are in \(W\)?**3. **Show that \(\mathbf{a_2}\) is in \(W\). [Hint: Row operations are unnecessary.]**---### Solution:**a. Is \(\mathbf{b}\) in \(\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\)? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.**- **A.** No, \(\mathbf{b}\) is not in \(\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\) since \(\mathbf{b}\) is not equal to \(\mathbf{a_1}\), \(\mathbf{a_2}\), or \(\mathbf{a_3}\). ✔- **B.** Yes, \(\mathbf{b}\) is in \(\{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}\}\) since \(\mathbf{b} = a\). \[\text{(Type a whole number.)}\]- **C. ### Educational Resource on Linear Combinations of Vectors### Problem StatementUse the accompanying figure to write each vector listed as a linear combination of vectors **u** and **v**.Vectors **b**, **w**, **x**, and **y**![Vector Diagram](image_url)**Write **b** as a linear combination of **u** and **v**.**\[ \mathbf{b} = (\_\_\_\_) \mathbf{u} + (\_\_\_\_) \mathbf{v} \](Type integers or decimals.)### Explanation of the DiagramThe diagram provided in the figure consists of vectors with their linear combinations represented on a grid. Each vector is positioned relative to vectors **u** and **v** along their respective axes. We see a central origin point **0** with the following vectors and positions in relation to **u** and **v**:- **u**: A unit vector along the positive x-axis.- **v**: A unit vector along the positive y-axis.- **b**: Positioned at the intersection of the grid lines representing 0.5 units of **u** and 1 unit of **v**.Some other vectors shown for reference are:- **d**, **c**, **a**: Various other vector positions.- **w**, **x**, **y**, **z**: Other vectors to be expressed as combinations of **u** and **v**.### Example Calculation#### Vector **b**To express vector **b** as a linear combination of **u** and **v**:1. **b** is located at a point which, in terms of vector addition, can be represented as: - **0.5** units in the direction of **u** (along the x-axis). - **1** unit in the direction of **v** (along the y-axis).Thus, vector **b** can be written as:\[ \mathbf{b} = 0.5 \mathbf{u} + 1 \mathbf{v} \]### Interactive ComponentYou can now type the values into the provided fields to check your understanding:\[ \mathbf{b} = (\_\_\_\_) \mathbf{u} + (\_\_\_\_) \mathbf{v} \]Remember to type integers or decimals.---This exercise explains how to break down vectors into linear combinations of basis vectors, enhancing your understanding

Since you have posted a question with multiple subparts we will solve the first one. As per the…

-59 4 - 29 a. Is b in (a₁, a2, a3}? How many vectors are in {a₁, a2, a3}? b. Is b in W? How many vectors are in W? c. Show that a2 is in W. [Hint: Row operations are unnecessary.] a. Is b in (a₁, a2, a3}? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. No, b is not in (a₁, a2, a3} since b is not equal to a₁, a2, or a3. B. Yes, b is in (a₁, a2, a3} since b = a (Type a whole number.) OC. Yes, b is in (a₁, a2, a3} since, although b is not equal to a₁, a2, or a3, it can be expressed as a linear combination of them. In particular, b = (a₁ + a₂ + ( )a3. (Simplify your answers.) OD. No, b is not in (a₁, a2, a3} since it cannot be generated by a linear combination of a₁, a2, and a3. How many vectors are in (a₁, a2, a3}? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is(are) 3 vector(s) in (a₁, a2, a3}. (Type a whole number.) OB. There are infinitely many vectors in (a₁, a2, a3). b. Is b in W? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. OA. No, b is not in W since b is not equal to a₁, a2, or a3. OB. Yes, b is in W since, although b is not equal to a₁, a2, or a3, it can be expressed as a linear combination of them. In particular, b = (a₁ + a₂ + ( )a3. (Simplify your answers.) OC. Yes, b is in W since b=a (Type a whole number.) OD. No, b is not in W since it cannot be generated by a linear combination of a₁, a2, and a3.

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter7: Triangles

Section: Chapter Questions

Problem 1RP: We mentioned in Section 7.5 that our algebraic treatment of vectors could be attributed, in part, to...