Let V₁ = -3 and V₂ = Av= පස ද.. [8] 3 be eigenvectors of the matrix A which correspond to the eigenvalues A₁ = -3 and 2 = 2, respectively. Let v = -4v₁ + 4v₂. Find Av. 0
Let V₁ = -3 and V₂ = Av= පස ද.. [8] 3 be eigenvectors of the matrix A which correspond to the eigenvalues A₁ = -3 and 2 = 2, respectively. Let v = -4v₁ + 4v₂. Find Av. 0
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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![### Eigenvalues and Eigenvectors
**Problem Statement:**
Let \(\mathbf{v}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 3 \\ 0 \end{bmatrix}\) be eigenvectors of the matrix \( A \) which correspond to the eigenvalues \(\lambda_1 = -3 \) and \(\lambda_2 = 2 \), respectively.
Let \(\mathbf{v} = -4\mathbf{v}_1 + 4\mathbf{v}_2\). Find \( A\mathbf{v} \).
**Solution:**
Given:
\[
\mathbf{v}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 3 \\ 0 \end{bmatrix}
\]
Eigenvalues:
\[
\lambda_1 = -3, \quad \lambda_2 = 2
\]
Vector \(\mathbf{v}\):
\[
\mathbf{v} = -4\mathbf{v}_1 + 4\mathbf{v}_2
\]
\[
\mathbf{v} = -4 \begin{bmatrix} 0 \\ 1 \end{bmatrix} + 4 \begin{bmatrix} 3 \\ 0 \end{bmatrix}
\]
To find \( A\mathbf{v} \):
The image displays an equation with placeholders indicating each element of the resulting vector \(A \mathbf{v}\):
\[
A\mathbf{v} = \begin{bmatrix} \boxed{\phantom{0}} \\ \boxed{\phantom{0}} \\ \boxed{\phantom{0}} \end{bmatrix}
\]
Where the boxed areas represent the elements of the resulting vector after calculating \(A\mathbf{v}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb2133c9-e1e5-4d56-9c72-044227328930%2F91e394fa-7c40-4201-a7ab-2bd5d362ce6a%2Fdghyc5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Eigenvalues and Eigenvectors
**Problem Statement:**
Let \(\mathbf{v}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 3 \\ 0 \end{bmatrix}\) be eigenvectors of the matrix \( A \) which correspond to the eigenvalues \(\lambda_1 = -3 \) and \(\lambda_2 = 2 \), respectively.
Let \(\mathbf{v} = -4\mathbf{v}_1 + 4\mathbf{v}_2\). Find \( A\mathbf{v} \).
**Solution:**
Given:
\[
\mathbf{v}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 3 \\ 0 \end{bmatrix}
\]
Eigenvalues:
\[
\lambda_1 = -3, \quad \lambda_2 = 2
\]
Vector \(\mathbf{v}\):
\[
\mathbf{v} = -4\mathbf{v}_1 + 4\mathbf{v}_2
\]
\[
\mathbf{v} = -4 \begin{bmatrix} 0 \\ 1 \end{bmatrix} + 4 \begin{bmatrix} 3 \\ 0 \end{bmatrix}
\]
To find \( A\mathbf{v} \):
The image displays an equation with placeholders indicating each element of the resulting vector \(A \mathbf{v}\):
\[
A\mathbf{v} = \begin{bmatrix} \boxed{\phantom{0}} \\ \boxed{\phantom{0}} \\ \boxed{\phantom{0}} \end{bmatrix}
\]
Where the boxed areas represent the elements of the resulting vector after calculating \(A\mathbf{v}\).
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