Program Explanation Given information: The system of differential equations is shown below. x ' = [ 2 5 0 2 ] x The initial condition for the above differential system is shown below. x ( 0 ) = [ 4 7 ] Explanation: The coefficient matrix for the system of differential equations is written as, A = [ 2 5 0 2 ] Express the matrix coefficient A as the sum of a nilpotent matrix and a multiple of the identity matrix as, A = [ 2 5 0 2 ] = [ 2 0 0 2 ] + [ 0 5 0 0 ] = D + B Here, D is a diagonal matrix and diagonal matrix is commutative with any same order square matrix. It is known that for any commutative matrices D and B the following formula holds. e D + B = e D e B Therefore, the value of e A t is calculated as shown below. e A t = e ( D + B ) t = e D t + B t = e D t e B t Further diagonal matrix D , the exponent of D is a diagonal matrix that has an exponential of each diagonal element of D . e A t = [ e 2 t 0 0 e 2 t ] e B t = [ e 2 t 0 0 e 2 t ] ( I + B t + B 2 t 2 2 ! + B 3 t 3 3 ! + ⋯ + B n t n n ! + ⋯ ) As the square of matrix B is zero matrix, therefore matrix B is nilpotent B n = 0 for n ≥ 2

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

See similar textbooks

Question

Chapter 5.6, Problem 25P

Program Plan Intro

Program Description: Purpose of the problem is to solve the initial value problem by the use of sum of a nilpotent matrix and a multiple of the identity matrix.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.6, Problem 24P

Chapter 5.6, Problem 26P

Students have asked these similar questions

Please solve.

Computer Science matlab

Consider the system of linear equations X2 + 2x3 X4 -1 X1 X2 X3 -X1 X2 + X3 + 3.x4 1 - Xị + 2x2 X4 9 Use Matlab to find the LU decomposition of the coefficient matrix A and then solve the resulting triangular system using forward and backward substitutions programs. (You need to provide the codes)

Chapter 5 Solutions

EBK DIFFERENTIAL EQUATIONS

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. (53) by substituting the expressions...Ch. 5.3 - Prob. 35P Ch. 5.3 - Prob. 36P Ch. 5.3 - Prob. 37P Ch. 5.3 - Prob. 38P Ch. 5.3 - Prob. 39P Ch. 5.3 - Prob. 40P Ch. 5.4 - Prob. 1P Ch. 5.4 - Prob. 2P Ch. 5.4 - Prob. 3P Ch. 5.4 - Prob. 4P Ch. 5.4 - Prob. 5P Ch. 5.4 - Prob. 6P Ch. 5.4 - Prob. 7P Ch. 5.4 - Prob. 8P Ch. 5.4 - Prob. 9P Ch. 5.4 - Prob. 10P Ch. 5.4 - Prob. 11P Ch. 5.4 - Prob. 12P Ch. 5.4 - Prob. 13P Ch. 5.4 - Prob. 14P Ch. 5.4 - Prob. 15P Ch. 5.4 - Prob. 16P Ch. 5.4 - Prob. 17P Ch. 5.4 - Prob. 18P Ch. 5.4 - Prob. 19P Ch. 5.4 - Prob. 20P Ch. 5.4 - Prob. 21P Ch. 5.4 - Prob. 22P Ch. 5.4 - Prob. 23P Ch. 5.4 - Prob. 24P Ch. 5.4 - Prob. 25P Ch. 5.4 - Prob. 26P Ch. 5.4 - Prob. 27P Ch. 5.4 - Prob. 28P Ch. 5.4 - Prob. 29P Ch. 5.5 - Prob. 1P Ch. 5.5 - Prob. 2P Ch. 5.5 - Prob. 3P Ch. 5.5 - Prob. 4P Ch. 5.5 - Prob. 5P Ch. 5.5 - Prob. 6P Ch. 5.5 - Prob. 7P Ch. 5.5 - Prob. 8P Ch. 5.5 - Prob. 9P Ch. 5.5 - Prob. 10P Ch. 5.5 - Prob. 11P Ch. 5.5 - Prob. 12P Ch. 5.5 - Prob. 13P Ch. 5.5 - Prob. 14P Ch. 5.5 - Prob. 15P Ch. 5.5 - Prob. 16P Ch. 5.5 - Prob. 17P Ch. 5.5 - Prob. 18P Ch. 5.5 - Prob. 19P Ch. 5.5 - Prob. 20P Ch. 5.5 - Prob. 21P Ch. 5.5 - Prob. 22P Ch. 5.5 - Prob. 23P Ch. 5.5 - Prob. 24P Ch. 5.5 - Prob. 25P Ch. 5.5 - Prob. 26P Ch. 5.5 - Prob. 27P Ch. 5.5 - Prob. 28P Ch. 5.5 - Prob. 29P Ch. 5.5 - Prob. 30P Ch. 5.5 - Prob. 31P Ch. 5.5 - Prob. 32P Ch. 5.5 - Prob. 33P Ch. 5.5 - Prob. 34P Ch. 5.5 - Prob. 35P Ch. 5.5 - Prob. 36P Ch. 5.6 - Prob. 1P Ch. 5.6 - Prob. 2P Ch. 5.6 - Prob. 3P Ch. 5.6 - Prob. 4P Ch. 5.6 - Prob. 5P Ch. 5.6 - Prob. 6P Ch. 5.6 - Prob. 7P Ch. 5.6 - Prob. 8P Ch. 5.6 - Prob. 9P Ch. 5.6 - Prob. 10P Ch. 5.6 - Prob. 11P Ch. 5.6 - Prob. 12P Ch. 5.6 - Prob. 13P Ch. 5.6 - Prob. 14P Ch. 5.6 - Prob. 15P Ch. 5.6 - Prob. 16P Ch. 5.6 - Prob. 17P Ch. 5.6 - Prob. 18P Ch. 5.6 - Prob. 19P Ch. 5.6 - Prob. 20P Ch. 5.6 - Prob. 21P Ch. 5.6 - Prob. 22P Ch. 5.6 - Prob. 23P Ch. 5.6 - Prob. 24P Ch. 5.6 - Prob. 25P Ch. 5.6 - Prob. 26P Ch. 5.6 - Prob. 27P Ch. 5.6 - Prob. 28P Ch. 5.6 - Prob. 29P Ch. 5.6 - Prob. 30P Ch. 5.6 - Prob. 31P Ch. 5.6 - Prob. 32P Ch. 5.6 - Prob. 33P Ch. 5.6 - Prob. 34P Ch. 5.6 - Prob. 35P Ch. 5.6 - Prob. 36P Ch. 5.6 - Prob. 37P Ch. 5.6 - Prob. 38P Ch. 5.6 - Prob. 39P Ch. 5.6 - Prob. 40P Ch. 5.7 - Prob. 1P Ch. 5.7 - Prob. 2P Ch. 5.7 - Prob. 3P Ch. 5.7 - Prob. 4P Ch. 5.7 - Prob. 5P Ch. 5.7 - Prob. 6P Ch. 5.7 - Prob. 7P Ch. 5.7 - Prob. 8P Ch. 5.7 - Prob. 9P Ch. 5.7 - Prob. 10P Ch. 5.7 - Prob. 11P Ch. 5.7 - Prob. 12P Ch. 5.7 - Prob. 13P Ch. 5.7 - Prob. 14P Ch. 5.7 - Prob. 15P Ch. 5.7 - Prob. 16P Ch. 5.7 - Prob. 17P Ch. 5.7 - Prob. 18P Ch. 5.7 - Prob. 19P Ch. 5.7 - Prob. 20P Ch. 5.7 - Prob. 21P Ch. 5.7 - Prob. 22P Ch. 5.7 - Prob. 23P Ch. 5.7 - Prob. 24P Ch. 5.7 - Prob. 25P Ch. 5.7 - Prob. 26P Ch. 5.7 - Prob. 27P Ch. 5.7 - Prob. 28P Ch. 5.7 - Prob. 29P Ch. 5.7 - Prob. 30P Ch. 5.7 - Prob. 31P Ch. 5.7 - Prob. 32P Ch. 5.7 - Prob. 33P Ch. 5.7 - Prob. 34P

Knowledge Booster

Program Description: Purpose of the problem is to solve the initial value problem by the use of sum of a nilpotent matrix and a multiple of the identity matrix.

Solution Summary: The author explains how to solve the initial value problem by the use of sum of a nilpotent matrix and multiple of the identity matrix.

Program Description: Purpose of the problem is to solve the initial value problem by the use of sum of a nilpotent matrix and a multiple of the identity matrix.

Want to see the full answer?

Chapter 5 Solutions