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5 The matrix A= -2 7 0 has eigenvalues A₁ = 2 and A₂ = 3. An eigenvector for 2 is = 2 1 -2 3 [-2 10 07 ܒ -147 [] The vectors 2 and 3 are linearly indepedent and the defect of A is 0 Using 1, 2 and 3, write the general solution z(t) to the system - At : and 3 has eigenvectors ₂ = -10] -5 and 3 = 3 (Use c_1, c_2, and c_3 for c₁, c₂ and C3.)

5 The matrix A= -2 7 0 has eigenvalues A₁ = 2 and A₂ = 3. An eigenvector for 2 is = 2 1 -2 3 [-2 10 07 ܒ -147 [] The vectors 2 and 3 are linearly indepedent and the defect of A is 0 Using 1, 2 and 3, write the general solution z(t) to the system - At : and 3 has eigenvectors ₂ = -10] -5 and 3 = 3 (Use c_1, c_2, and c_3 for c₁, c₂ and C3.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The matrix A = -2 10 07 -2 7 0 has eigenvalues A₁ = 2 and A₂ = 3. An eigenvector for 2 is v₁ = -2 3 and 3 has eigenvectors ₂ : -107 -5 3 and U3 7 (Use c_1, c_2, and c_3 for C₁, C₂ and C3.) 4 The vectors ₂ and 3 are linearly indepedent and the defect of A is 0 Using 71, 72 and 3, write the general solution z(t) to the system z= Aż : 5 2
Transcribed Image Text:The matrix A = -2 10 07 -2 7 0 has eigenvalues A₁ = 2 and A₂ = 3. An eigenvector for 2 is v₁ = -2 3 and 3 has eigenvectors ₂ : -107 -5 3 and U3 7 (Use c_1, c_2, and c_3 for C₁, C₂ and C3.) 4 The vectors ₂ and 3 are linearly indepedent and the defect of A is 0 Using 71, 72 and 3, write the general solution z(t) to the system z= Aż : 5 2
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