.. For one of these matrices, V = -2 is an eigenvector. Find the matrix with this eigenvector. [1 2 3] 1 b. A₂ = 3 2 1 3] [1 a. A₁ = 3 L2 2 3 2 1 1 2 2 31 2 1 L2 1 0 [1 c. A3 = 3 2 [1 d. A4 = 3 L2 2 1] 2 1 1 1.
.. For one of these matrices, V = -2 is an eigenvector. Find the matrix with this eigenvector. [1 2 3] 1 b. A₂ = 3 2 1 3] [1 a. A₁ = 3 L2 2 3 2 1 1 2 2 31 2 1 L2 1 0 [1 c. A3 = 3 2 [1 d. A4 = 3 L2 2 1] 2 1 1 1.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Problem Description:**
1. For one of these matrices, \( \vec{v} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \) is an eigenvector. Find the matrix with this eigenvector.
Matrices:
a. \( A_1 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 2
\end{bmatrix} \)
b. \( A_2 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 0
\end{bmatrix} \)
c. \( A_3 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{bmatrix} \)
d. \( A_4 = \begin{bmatrix}
1 & 2 & 1 \\
3 & 2 & 1 \\
2 & 1 & 1
\end{bmatrix} \)
**Explanation:**
The given eigenvector \( \vec{v} \) is a vector that, when multiplied by the matrix \(A\), yields a scalar multiple of the same vector \( \vec{v} \). In mathematical terms, for a matrix \(A\) and an eigenvector \( \vec{v} \) with corresponding eigenvalue \(\lambda\),
\[ A \vec{v} = \lambda \vec{v} \]
The goal of this problem is to determine which of the provided matrices \(A_1\), \(A_2\), \(A_3\), or \(A_4\) has \( \vec{v} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \) as an eigenvector. This requires verifying that the multiplication of each matrix by the vector produces a result that is proportional to the original vector \( \vec{v} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F474861d3-aaf1-4b4c-afe2-e250091335e4%2F21533275-7307-4a2d-9e49-4f41a09468e6%2F5uchwz5_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
1. For one of these matrices, \( \vec{v} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \) is an eigenvector. Find the matrix with this eigenvector.
Matrices:
a. \( A_1 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 2
\end{bmatrix} \)
b. \( A_2 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 0
\end{bmatrix} \)
c. \( A_3 = \begin{bmatrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
2 & 1 & 3
\end{bmatrix} \)
d. \( A_4 = \begin{bmatrix}
1 & 2 & 1 \\
3 & 2 & 1 \\
2 & 1 & 1
\end{bmatrix} \)
**Explanation:**
The given eigenvector \( \vec{v} \) is a vector that, when multiplied by the matrix \(A\), yields a scalar multiple of the same vector \( \vec{v} \). In mathematical terms, for a matrix \(A\) and an eigenvector \( \vec{v} \) with corresponding eigenvalue \(\lambda\),
\[ A \vec{v} = \lambda \vec{v} \]
The goal of this problem is to determine which of the provided matrices \(A_1\), \(A_2\), \(A_3\), or \(A_4\) has \( \vec{v} = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \) as an eigenvector. This requires verifying that the multiplication of each matrix by the vector produces a result that is proportional to the original vector \( \vec{v} \).
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