{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M 340L HW 10 Solutions

M 340L HW 10 Solutions - cruz(fmc326 HW10 gilbert(56540...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
cruz (fmc326) – HW10 – gilbert – (56540) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points Using the fact that e x = 1 + x + 1 2! x 2 + . . . + 1 n ! x n + . . . , compute e tA as a matrix-valued function of t when A = b 5 4 2 1 B . 1. e tA = ± 2 e - t + e - t 2( e - 3 t e - t ) e - t e - 3 t 2 e - 3 t + e - t ² 2. e tA = ± 2 e - 3 t + e - t 2( e - t e - 3 t ) e - 3 t e - t 2 e - t + e - 3 t ² 3. e tA = ± 2 e - t e - 3 t 2( e - 3 t e - t ) e - t e - 3 t 2 e - 3 t e - t ² 4. e tA = ± 2 e - 3 t e - t 2( e - t e - 3 t ) e - 3 t e - t 2 e - t e - 3 t ² correct Explanation: If A can be diagonalized by A = PDP - 1 = P b d 1 0 0 d 2 B P - 1 , then e tA = Pe tD P - 1 = P b e td 1 0 0 e td 2 B P - 1 . Now A can be diagonalized if we can Fnd an eigenbasis of R 2 of eigenvectors v 1 , v 2 of A corresponding to eigenvalues λ 1 , λ 2 , for then: A = P b λ 1 0 0 λ 2 B P - 1 , P = [ v 1 v 2 ] . But det[ A λI ] = v v v v 5 λ 4 2 1 λ v v v v = λ 2 λ + 3 = ( λ + 3)( λ + 1) = 0 , i.e. , λ 1 = 3 and λ 2 = 1. Corresponding eigenvectors are v 1 = b 2 1 B , v 2 = b 1 1 B , so P = b 2 1 1 1 B , P - 1 = b 1 1 1 2 B . Consequently, e tA = b 2 1 1 1 Bb e - 3 t 0 0 e - t 1 1 1 2 B = ± 2 e - 3 t e - t 2( e - t e - 3 t ) e - 3 t e - t 2 e - t e - 3 t ² .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}