Explanation Given: The given matrix is, A = [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] Calculation: The given matrix is, A = [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] The characteristic equation of A is, det ( λ I − A ) = 0 … ( 1 ) The value of λ I − A is, λ I − A = λ [ 1 0 0 0 1 0 0 0 1 ] − [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] = [ λ 0 0 0 λ 0 0 0 λ ] − [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] = [ λ − 3 2 − 2 2 λ 1 − 2 1 λ ] Substitute [ λ − 3 2 − 2 2 λ 1 − 2 1 λ ] for λ I − A in equation (1). det ( [ λ − 3 2 − 2 2 λ 1 − 2 1 λ ] ) = 0 ( λ − 3 ) ( λ 2 − 1 ) − 2 ( 2 λ + 2 ) − 2 ( 2 + 2 λ ) = 0 λ 3 − 3 λ 2 − 9 λ − 5 = 0 ( λ + 1 ) ( λ + 1 ) ( λ − 5 ) = 0 The eigenvalues of A by characteristic equation are λ = − 1 , 5 Thus, the eigenvalues of A are λ 1 = − 1 , and λ 2 = 5 . For eigenvalue λ 1 = − 1 , Substitute − 1 for λ in λ I − A . ( − 1 ) I − A = ( − 1 ) [ 1 0 0 0 1 0 0 0 1 ] − [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] ( − 1 ) I − A = [ − 1 0 0 0 − 1 0 0 0 − 1 ] − [ 3 − 2 2 − 2 0 − 1 2 − 1 0 ] ( − 1 ) I − A = [ − 4 2 − 2 2 − 1 1 − 2 1 − 1 ] ... ( 2 ) Equation (2) can be reduced by R 2 → 2 R 2 + R 1 and R 3 → 2 R 3 − R 1 as, [ − 4 2 − 2 0 0 0 0 0 0 ] The reduction shows that, − 4 x 1 + 2 x 2 − 2 x 3 = 0 4 x 1 = 2 x 2 − 2 x 3 x 1 = 1 2 x 2 − 1 2 x 3 Let x 2 = t and x 3 = s , then every eigenvector of λ 1 will be of the form,

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN: 9781305658004

Author: Ron Larson

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 7.CR, Problem 12CR

To determine

To find:

A is diagonalizable and a nonsingular matrix P.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.CR, Problem 11CR

Chapter 7.CR, Problem 13CR

Students have asked these similar questions

Assume {u1, U2, us} spans R³. Select the best statement. A. {U1, U2, us, u4} spans R³ unless u is the zero vector. B. {U1, U2, us, u4} always spans R³. C. {U1, U2, us, u4} spans R³ unless u is a scalar multiple of another vector in the set. D. We do not have sufficient information to determine if {u₁, u2, 43, 114} spans R³. OE. {U1, U2, 3, 4} never spans R³. F. none of the above

Assume {u1, U2, 13, 14} spans R³. Select the best statement. A. {U1, U2, u3} never spans R³ since it is a proper subset of a spanning set. B. {U1, U2, u3} spans R³ unless one of the vectors is the zero vector. C. {u1, U2, us} spans R³ unless one of the vectors is a scalar multiple of another vector in the set. D. {U1, U2, us} always spans R³. E. {U1, U2, u3} may, but does not have to, span R³. F. none of the above

Let H = span {u, v}. For each of the following sets of vectors determine whether H is a line or a plane. Select an Answer u = 3 1. -10 8-8 -2 ,v= 5 Select an Answer -2 u = 3 4 2. + 9 ,v= 6

Chapter 7 Solutions

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,...

Knowledge Booster

$Algebra$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.

A is diagonalizable and a nonsingular matrix P.

Solution Summary: The author explains that A is diagonalizable and a nonsingular matrix P.

Videos

To find: A is diagonalizable and a nonsingular matrix P.

Want to see the full answer?

Chapter 7 Solutions

To find:

A is diagonalizable and a nonsingular matrix P.