Let T: ²->² be a linear transformation that maps u = Use the fact that T is linear to find the image of 3u + v. -16 0[-] 8 O-33 24 7 1 -55 32 -6 4
Let T: ²->² be a linear transformation that maps u = Use the fact that T is linear to find the image of 3u + v. -16 0[-] 8 O-33 24 7 1 -55 32 -6 4
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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![**Title: Understanding Linear Transformations in Vector Spaces**
**Introduction**
In this section, we will explore how to apply linear transformations to vectors in a given vector space. Specifically, we will understand how to use linearity properties to find the image of a combination of vectors given the transformation of other vectors.
**Linear Transformation Problem**
Consider a linear transformation \( T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) defined such that:
\[ u = \begin{bmatrix} -6 \\ 4 \end{bmatrix} \text{ maps to } \begin{bmatrix} -22 \\ 12 \end{bmatrix} \]
\[ v = \begin{bmatrix} -5 \\ 2 \end{bmatrix} \text{ maps to } \begin{bmatrix} -4 \\ 11 \end{bmatrix} \]
We are asked to find the image of the vector \( 3u + v \) under the linear transformation \( T \).
**Using the Linearity Property**
The linearity of the transformation \( T \) implies that for any vectors \( u \) and \( v \), and scalars \( a \) and \( b \), we have:
\[ T(au + bv) = aT(u) + bT(v) \]
Given:
\[ u = \begin{bmatrix} -6 \\ 4 \end{bmatrix}, \ T(u) = \begin{bmatrix} -22 \\ 12 \end{bmatrix} \]
\[ v = \begin{bmatrix} -5 \\ 2 \end{bmatrix}, \ T(v) = \begin{bmatrix} -4 \\ 11 \end{bmatrix} \]
We need to find:
\[ T(3u + v) \]
First, calculate the vector \( 3u + v \):
\[ 3u = 3 \begin{bmatrix} -6 \\ 4 \end{bmatrix} = \begin{bmatrix} -18 \\ 12 \end{bmatrix} \]
\[ 3u + v = \begin{bmatrix} -18 \\ 12 \end{bmatrix} + \begin{bmatrix} -5 \\ 2 \end{bmatrix} = \begin{bmatrix} -23 \\ 14 \end{bmatrix}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F157e66f5-e794-4af2-988e-1885b818a80a%2F567a509e-cacd-45ca-aeed-be6285282213%2Fse7wuao_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Linear Transformations in Vector Spaces**
**Introduction**
In this section, we will explore how to apply linear transformations to vectors in a given vector space. Specifically, we will understand how to use linearity properties to find the image of a combination of vectors given the transformation of other vectors.
**Linear Transformation Problem**
Consider a linear transformation \( T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) defined such that:
\[ u = \begin{bmatrix} -6 \\ 4 \end{bmatrix} \text{ maps to } \begin{bmatrix} -22 \\ 12 \end{bmatrix} \]
\[ v = \begin{bmatrix} -5 \\ 2 \end{bmatrix} \text{ maps to } \begin{bmatrix} -4 \\ 11 \end{bmatrix} \]
We are asked to find the image of the vector \( 3u + v \) under the linear transformation \( T \).
**Using the Linearity Property**
The linearity of the transformation \( T \) implies that for any vectors \( u \) and \( v \), and scalars \( a \) and \( b \), we have:
\[ T(au + bv) = aT(u) + bT(v) \]
Given:
\[ u = \begin{bmatrix} -6 \\ 4 \end{bmatrix}, \ T(u) = \begin{bmatrix} -22 \\ 12 \end{bmatrix} \]
\[ v = \begin{bmatrix} -5 \\ 2 \end{bmatrix}, \ T(v) = \begin{bmatrix} -4 \\ 11 \end{bmatrix} \]
We need to find:
\[ T(3u + v) \]
First, calculate the vector \( 3u + v \):
\[ 3u = 3 \begin{bmatrix} -6 \\ 4 \end{bmatrix} = \begin{bmatrix} -18 \\ 12 \end{bmatrix} \]
\[ 3u + v = \begin{bmatrix} -18 \\ 12 \end{bmatrix} + \begin{bmatrix} -5 \\ 2 \end{bmatrix} = \begin{bmatrix} -23 \\ 14 \end{bmatrix}
Transcribed Image Text:**Title: Non-Trivial Solutions for a Given Matrix**
**Problem Statement:**
Suppose \( A = \begin{bmatrix} 4 & 5 & -1 \\ 0 & 12 & -4 \\ -3 & 6 & 4 \end{bmatrix} \) that gives a non-trivial solution.
**Options:**
- \( \begin{bmatrix} 8 \\ 2 \\ 3 \end{bmatrix} \)
- \( \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} \)
- \( \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \)
- \( \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} \)
**Explanation:**
To determine which of the provided vectors represents a non-trivial solution for the matrix \( A \), one must solve the equation \( A\mathbf{x} = \mathbf{0} \).
A **non-trivial solution** means that \( \mathbf{x} \neq \mathbf{0} \), meaning the vector \( \mathbf{x} \) must not be the zero vector.
By analyzing the given options, we seek the vector that, when multiplied by matrix \( A \), results in the zero vector, which confirms that the matrix equation \( A\mathbf{x} = \mathbf{0} \) holds true.
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