Explanation Result used: The characteristic equation of the matrix A is det ( A − λ I ) = 0 . “The values of λ satisfy the characteristic equation may be either real or complex valued are called eigenvalues of A and the nonzero solution of equation ( A − λ I ) x = 0 are obtained by using such a value of λ are called eigenvectors corresponding to that eigenvalue”. Calculation: The given matrix is A = ( 3 2 2 1 4 1 − 2 − 4 − 1 ) . By the result stated above, the eigenvalues λ and the eigenvector X satisfy the equation ( A − λ I ) x = 0 . That is, ( 3 − λ 2 2 1 4 − λ 1 − 2 − 4 − 1 − λ ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) . Find the eigenvalue of the equation. det ( A − λ I ) = 0 | 3 − λ 2 2 1 4 − λ 1 − 2 − 4 − 1 − λ | = 0 ( 3 − λ ) ( ( 4 − λ ) ( − 1 − λ ) + 4 ) − 2 ( − 1 − λ + 2 ) + 2 ( − 4 − 8 + 2 λ ) = 0 ( 3 − λ ) ( − 4 − λ + λ + λ 2 + 4 ) − 2 + 2 λ − 24 + 4 λ = 0 ( λ − 1 ) ( λ 2 − 5 λ + 6 ) = 0 On further simplification. ( λ − 1 ) ( λ 2 − 5 λ + 6 ) = 0 ( λ − 1 ) ( λ − 2 ) ( λ − 3 ) = 0 λ − 1 = 0 or λ − 2 = 0 or λ − 3 = 0 Find the eigenvector for λ = 1 . ( 2 2 2 1 3 1 − 2 − 4 − 2 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) ( 2 2 2 0 2 0 0 − 2 0 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) ( subtract row 1 multiplied by 1 2 from row 2 ) ( 2 2 2 0 2 0 0 0 0 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) ( Add second row and third row ) Since first and last columns are equal, it can be concluded that the eigenvector ( 1 , 0 , − 1 ) T corresponds to λ = 1 . The corresponding system of equation of above matrix is 2 x 1 + 2 x 2 + 2 x 3 = 0 2 x 2 = 0 . That is, x 1 = − x 3 and x 2 = 0 . If x 3 = 1 , then x 1 = − 1 . The eigenvector for λ = 1 is X ( 1 ) = ( 1 0 − 1 ) _ . Find the eigenvector for λ = 2

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Question

Chapter 7.3, Problem 23P

To determine

The eigenvalues and eigenvectors of the matrix (322141−2−4−1).

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.3, Problem 22P

Chapter 7.3, Problem 24P

Students have asked these similar questions

Find the first four nonzero terms in a power series expansion about x=0 for a general solution to the given differential equation w''-14x^2w'+w=0

In this exercise, we will investigate a technique to prove that a language is notregular. This tool is called the pumping lemma.The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the wordw is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may bedivided into three pieces, w = xyz, satisfying the following conditions:1. For each i ∈ N, xy^i z ∈ L(M ).2. |y| > 0 (i.e., y contains at least one character).3. |xy| ≤ p (i.e., the string xy has at most p characters). Use the pumping lemma to show the following language is not regular (HINT: Use proof by contradictionto assume the language is regular and apply the pumping lemma to the language):L = {0^k1^k | k ∈ N}

A prefix of length ℓ of some word w are the first ℓ characters (in order) of w.1. Construct a context-free grammar for the language: L = {w ∈ {a, b}∗ | every prefix of w has at least as many a’s as b’s}2. Explain why every word generated by your context-free grammar (in Part 1) is contained in L. Then,prove via induction that every w ∈ L is produced by your context-free grammar.

Chapter 7 Solutions

Elementary Differential Equations

Ch. 7.1 - In each of Problems 1 through 4, transform the...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - In each of Problems 1 through 4, transform the...Ch. 7.1 - Prob. 5P Ch. 7.1 - In each of Problems 5 and 6, transform the given...Ch. 7.1 - Systems of first order equations can sometimes be...Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - In each of Problems 8 through 12, proceed as in...

Knowledge Booster

The eigenvalues and eigenvectors of the matrix ( 3 2 2 1 4 1 − 2 − 4 − 1 ) .

The eigenvalues and eigenvectors of the matrix (322141−2−4−1).

Want to see the full answer?

Chapter 7 Solutions