The vectors v.= span R but do not form a basis. Find two different ways to express -3 - 3 as a linear combination of V V V 24 v = Write as a linear combination of v. v. V when the coefficient of V is 0. 24 [2అ-ం·
The vectors v.= span R but do not form a basis. Find two different ways to express -3 - 3 as a linear combination of V V V 24 v = Write as a linear combination of v. v. V when the coefficient of V is 0. 24 [2అ-ం·
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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Question
![### Linear Combinations and Vector Spaces
**Learning Objective:** Understand how to express a vector as a linear combination of a set of vectors.
---
#### Problem Statement
The vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ -4 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 2 \\ -11 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 0 \\ -3 \end{bmatrix} \) span \(\mathbb{R}^2\) but do not form a basis.
**Task:** Find two different ways to express \(\begin{bmatrix} -3 \\ 24 \end{bmatrix}\) as a linear combination of \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \).
#### Approach
Write \(\begin{bmatrix} -3 \\ 24 \end{bmatrix}\) as a linear combination of \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \) when the coefficient of \( \mathbf{v}_3 \) is 0.
\[
\begin{bmatrix} -3 \\ 24 \end{bmatrix} = \_ \cdot \mathbf{v}_1 + \_ \cdot \mathbf{v}_2
\]
This leads to the following equation:
\[
\begin{bmatrix} -3 \\ 24 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -4 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ -11 \end{bmatrix}
\]
To solve for \(c_1\) and \(c_2\), set up the following system of linear equations:
\[
-3 = c_1 + 2c_2
\]
\[
24 = -4c_1 - 11c_2
\]
Solving the above system will provide the coefficients \(c_1\) and \(c_2\).
**Note:** There might be multiple correct solutions due to the vectors spanning \(\mathbb{R}^2\) but not forming a basis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9225fa68-a539-4743-ab4b-10a3968e5a82%2F7b82b88e-dbe4-43a7-a798-d687186158c6%2Fswlf05n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Linear Combinations and Vector Spaces
**Learning Objective:** Understand how to express a vector as a linear combination of a set of vectors.
---
#### Problem Statement
The vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ -4 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 2 \\ -11 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 0 \\ -3 \end{bmatrix} \) span \(\mathbb{R}^2\) but do not form a basis.
**Task:** Find two different ways to express \(\begin{bmatrix} -3 \\ 24 \end{bmatrix}\) as a linear combination of \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \).
#### Approach
Write \(\begin{bmatrix} -3 \\ 24 \end{bmatrix}\) as a linear combination of \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \) when the coefficient of \( \mathbf{v}_3 \) is 0.
\[
\begin{bmatrix} -3 \\ 24 \end{bmatrix} = \_ \cdot \mathbf{v}_1 + \_ \cdot \mathbf{v}_2
\]
This leads to the following equation:
\[
\begin{bmatrix} -3 \\ 24 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -4 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ -11 \end{bmatrix}
\]
To solve for \(c_1\) and \(c_2\), set up the following system of linear equations:
\[
-3 = c_1 + 2c_2
\]
\[
24 = -4c_1 - 11c_2
\]
Solving the above system will provide the coefficients \(c_1\) and \(c_2\).
**Note:** There might be multiple correct solutions due to the vectors spanning \(\mathbb{R}^2\) but not forming a basis.
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