Explanation Given: The given matrix is: A = [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] Procedure used: The formula is: A k = P B k P − 1 Finding eigenvalues and eigenvectors: Let A be an n × n matrix. 1. Form the characteristics equation | λ I − A | = 0 . 2. Find the real roots of the characteristics equation. These are the eigenvalues of A . 3. For each eigenvalue λ i , find the eigenvectors corresponding to λ i by solving the homogeneous system ( λ i I − A ) X = 0 . This can be required row reducing an n × n matrix. Calculation: Consider the given matrix: A = [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] The roots of the characteristic equation are: | [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | [ 2 − λ 3 − 2 − 2 − 5 − λ 0 − 2 − 1 4 − λ ] | = 0 ( 2 − λ ) ⋅ | − 5 − λ 0 − 1 4 − λ | − 3 ⋅ | − 2 0 − 2 4 − λ | − 2 ⋅ | − 2 − 5 − λ − 2 − 1 | = 0 Further simplify as: ( 2 − λ ) [ ( − 5 − λ ) ( 4 − λ ) ] − 3 [ − 2 ( 4 − λ ) ] − 2 [ 2 + 2 ( − 5 − λ ) ] = 0 − λ ( λ + 4 ) ( λ − 5 ) = 0 λ = 0 , − 4 , 5 Find the eigenvector corresponding to eigenvalue λ = 0 by solving the homogeneous system ( A − 0 I ) X = 0 : Let X = [ x 1 x 2 x 3 ] . ( [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] − 0 [ 1 0 0 0 1 0 0 0 1 ] ) [ x 1 x 2 x 3 ] = 0 ( [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] − [ 0 0 0 0 0 0 0 0 0 ] ) [ x 1 x 2 x 3 ] = 0 [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] [ x 1 x 2 x 3 ] = 0 Using reduced row-echelon form: [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] → [ 1 0 − 5 2 0 1 1 0 0 0 ] It indicate as, x 1 − 5 2 x 3 = 0 and x 2 + x 3 = 0 . Choose the arbitrary value x 3 = 1 , then x 1 = 5 2 x 2 = − 1 . The eigenvector is: P 1 = [ 5 2 − 1 1 ] Find the eigenvector corresponding to eigenvalue λ = − 4 by solving the homogeneous system ( A + 4 I ) X = 0 : Let X = [ x 1 x 2 x 3 ] .

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN: 9781305658004

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 7.2, Problem 36E

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The matrix A8 for the given matrix A=[23−2−2−50−2−14].

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Chapter 7.2, Problem 35E

Chapter 7.2, Problem 37E

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Elementary Linear Algebra (MindTap Course List)

Ch. 7.1 - Verifying Eigenvalues and Eigenvectors in...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and Eigenvectors in...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Determining Eigenvectors In Exercise 9-12,...Ch. 7.1 - Determining Eigenvectors In Exercise 9-12,...

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The matrix A 8 for the given matrix A = [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] .

Solution Summary: The author explains how to find the matrix A8 for the given matrix.

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To find:

The matrix A8 for the given matrix A=[23−2−2−50−2−14].

Want to see the full answer?

Chapter 7 Solutions

The matrix A 8 for the given matrix A = [ 2 3 − 2 − 2 − 5 0 − 2 − 1 4 ] .

Solution Summary: The author explains how to find the matrix A8 for the given matrix.

Videos

To find: The matrix A8 for the given matrix A=[23−2−2−50−2−14].

Want to see the full answer?

Chapter 7 Solutions

To find:

The matrix A8 for the given matrix A=[23−2−2−50−2−14].