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Show from first principles, Le., by using the definition of linear inde- pendence, that if µ = z + iy, y # 0 is an eigenvalue of a real matrix A with associated eigenvector e=u+ iw, then the two real solutions Y() = "(u cos dt – w sin be) and Z(1) = e“(usin de + w cos be) are linearly independent solutions of X = AX. (b) Use (a) to solve the system x - (; )x. NB: Real solutions are required. Hint: Recall that if the roots of C(A) = 0 occur in complex conjugate parts, then any one of the two eigenvalues yields two linearly indepen- dent solutions. The second will yleld solutions which are identical (up to a constant) to the solutions already found. Check Theorem 2.19 (Equation 2.2) and Example 2.20 on page 32 of the study guide in this regard.

Show from first principles, Le., by using the definition of linear inde- pendence, that if µ = z + iy, y # 0 is an eigenvalue of a real matrix A with associated eigenvector e=u+ iw, then the two real solutions Y() = "(u cos dt – w sin be) and Z(1) = e“(usin de + w cos be) are linearly independent solutions of X = AX. (b) Use (a) to solve the system x - (; )x. NB: Real solutions are required. Hint: Recall that if the roots of C(A) = 0 occur in complex conjugate parts, then any one of the two eigenvalues yields two linearly indepen- dent solutions. The second will yleld solutions which are identical (up to a constant) to the solutions already found. Check Theorem 2.19 (Equation 2.2) and Example 2.20 on page 32 of the study guide in this regard.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.6: The Algebra Of Matrices
Problem 37E
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This is an ordinary differential equation question. Please help with both (a) and (b)

Document1 - Microsoft Word (Product Activation Failed) W File Home Insert Page Layout References Mailings Review View % Cut A Find - - A A =,三,“ 请 T Aal AaBbCcDc AaBbCcD AaBbC AaBbCc AaBI AqBbCcl 1 No Spaci. Heading 1 Heading 2 Calibri (Body) - 11 Aa E Copy a Replace Paste IU- abe x, x* ab A I Normal Change Styles - Select - Title Subtitle Format Painter 6. Clipboard Font Paragraph Styles Editing 2:1:1:1 1::2:1·3:1:4:15.16.17:18.1 9.1 10. 1 11. | 12.1 13.1 14:1 15: 1 17: 1 18: (a) Show from first principles, i.e., by using the definition of linear inde- pendence, that if u = x+ iy, y +0 is an eigenvalue of a real matrix A with associated eigenvector v = u+ iw, then the two real solutions Y(t) = ea" (u cos bt – w sin bt) and Z(t) = e" (usin bt + w cos bt) are linearly independent solutions of X = AX. (b) Use (a) to solve the system X = ( :) 3 X. -8 W NB: Real solutions are required. Hint: Recall that if the roots of C(A) = 0 occur in complex conjugate parts, then any one of the two eigenvalues yields two linearly indepen- dent solutions. The second will yield solutions which are identical (up to a constant) to the solutions already found. Check Theorem 2.19 (Equation 2.2) and Example 2.20 on page 32 of the study guide in this regard. ll 06:53 AM 2022-03-24 Page: 1 of 1 Words: 3 B A e E E = 709% +
Transcribed Image Text:Document1 - Microsoft Word (Product Activation Failed) W File Home Insert Page Layout References Mailings Review View % Cut A Find - - A A =,三,“ 请 T Aal AaBbCcDc AaBbCcD AaBbC AaBbCc AaBI AqBbCcl 1 No Spaci. Heading 1 Heading 2 Calibri (Body) - 11 Aa E Copy a Replace Paste IU- abe x, x* ab A I Normal Change Styles - Select - Title Subtitle Format Painter 6. Clipboard Font Paragraph Styles Editing 2:1:1:1 1::2:1·3:1:4:15.16.17:18.1 9.1 10. 1 11. | 12.1 13.1 14:1 15: 1 17: 1 18: (a) Show from first principles, i.e., by using the definition of linear inde- pendence, that if u = x+ iy, y +0 is an eigenvalue of a real matrix A with associated eigenvector v = u+ iw, then the two real solutions Y(t) = ea" (u cos bt – w sin bt) and Z(t) = e" (usin bt + w cos bt) are linearly independent solutions of X = AX. (b) Use (a) to solve the system X = ( :) 3 X. -8 W NB: Real solutions are required. Hint: Recall that if the roots of C(A) = 0 occur in complex conjugate parts, then any one of the two eigenvalues yields two linearly indepen- dent solutions. The second will yield solutions which are identical (up to a constant) to the solutions already found. Check Theorem 2.19 (Equation 2.2) and Example 2.20 on page 32 of the study guide in this regard. ll 06:53 AM 2022-03-24 Page: 1 of 1 Words: 3 B A e E E = 709% +
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