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