Explanation Given: The given system is X ′ = ( 1 1 1 1 1 1 − 2 − 2 − 2 ) X and X ( 0 ) = ( 1 − 4 6 ) . Definition used: For any matrix A of order n × n , the expansion of e A t in power of the matrix A is e A t = 1 + A t + A 2 2 ! t 2 + A 3 3 ! t 3 + … + A k k ! t k + … . The general solution of the system X ′ = A X is given by X ( t ) = e A t C where C is the constant. Calculation: From the given system, the matrix is A = ( 1 1 1 1 1 1 − 2 − 2 − 2 ) and X ( 0 ) = ( 1 − 4 6 ) . Compute A 2 by multiplying matrix A with itself A 2 = A × A = ( 1 1 1 1 1 1 − 2 − 2 − 2 ) ⋅ ( 1 1 1 1 1 1 − 2 − 2 − 2 ) = ( 0 0 0 0 0 0 0 0 0 ) Multiply matrix A 2 with matrix A to obtain A 3 . A 3 = A 2 × A = ( 0 0 0 0 0 0 0 0 0 ) Multiply the matrix A 3 with the matrix A to obtain the value of the matrix A 4 . A 4 = A 3 × A = ( 0 0 0 0 0 0 0 0 0 ) Proceed in this way, the general value of a matrix raised to the power of a natural number is obtained as, A k = 0 , the zero matrix for k = 2 , 3 , 4 , …

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.4, Problem 13E

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To find: The solution of the given system subject to the initial condition X(0).

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Chapter 8.4, Problem 12E

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The solution of the given system subject to the initial condition X ( 0 ) .

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To find: The solution of the given system subject to the initial condition X(0).

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