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SolveLinearSystem

# SolveLinearSystem - Math 306 – Lab 3 Systems of...

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Unformatted text preview: Math 306 – Lab 3 Systems of Differential Equations Printed Name: 1 Goal The goal of this lab is to solve the system of linear ordinary differential equations dx dt = ax + by dy dt = cx + dy. In order to accomplish this, we must first learn about eigenvalues, eigenvectors, and matrix exponentiation. 1.1 Eigenvalues and Eigenvectors Definition 1.1. Let A be an n × n matrix of numbers. A number λ ∈ C is an eigenvector of the matrix A if there exists a ::::::: non-zero vector x , called an eigenvector , such that Ax = λ x . Together, ( λ, x ) are called an eigenpair . Recall that I is the n × n identity matrix, that is a matrix with n 1’s on the diagonal and zeros everywhere else, that is I = 1 0 ··· 0 1 ··· . . . . . . . . . ··· . . . 0 0 ··· 1 0 0 ··· 1 Definition 1.2. Let A = a b c d . Then det( A ) = ad- bc. Note that determinants get a bit more complicated when larger matrices are used. To learn more about determinants, seek out a linear algebra book. Theorem 1.3. For a given n × n matrix A , the characteristic polynomial is defined to be p ( λ ) = det( A- λ I ) . The roots of the equation p ( λ ) = 0 are the eigenvalues of the matrix A . To find an eigenvector associated with a particular eigenvalue, you must solve the dependent system ( A- λ I ) x = . Example 1.1.1 Suppose that A = 6 3- 2- 1 . Find the eigenvalues and eigenvectors of A . Solution: First, find the characteristic polynomial, p ( λ ) = det( A- λ I ) = det 6 3- 2- 1- λ 1 0 0 1 = det 6 3- 2- 1- λ λ = det 6- λ 3- 2- 1- λ = (6- λ )(- 1- λ )- 3(- 2) = λ 2- 5 λ. So now, we solve the equation p ( λ ) = 0 to find all eigenvalues. This has solutions λ 2- 5 λ = 0 ⇐⇒ λ 1 = 0 or λ 2 = 5 . To find the eigenvectors, we return to the equation ( A- λ I ) x = . First, consider λ 1 = 0. Then, ( A- λ 1 I ) x = ⇐⇒ ( A- I ) x = ⇐⇒ Ax = ⇐⇒ 6 3- 2- 1 x 1 x 2 = . Dr. Emmert Spring 2011 I of X Math 306 – Lab 3 Systems of Differential Equations That is, what we really need to do is solve the system of equations 6 x 1 + 3 x 2 = 0- 2 x 1- x 2 = 0 . Since one row is a multiple of the other (this always happens), we quickly see that x 2 =- 2 x 1 and we are free to choose almost any value for x 1 (except zero! since that would force our eigenvector to be zero). Thus, we see that a family of eigenvectors is x = x 1- 2 x 1 = 1- 2 x 1 , x 1 ∈ R \{ } . (Note that x 1 ∈ R \{ } means use any real number except x 1 = 0.) What I normally do is pick a “nice” number...something that eliminates fractions (which we don’t have)...so I think x 1 = 1 will work nicely. Thus, an eigenvector associated with λ 1 = 0 is 1- 2 ....
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SolveLinearSystem - Math 306 – Lab 3 Systems of...

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