Answer
See below
Work Step by Step
Since $\lambda I_n=diag(\lambda, \lambda,...\lambda)$
Then we obtain:
$$e^{\lambda I_nt}=diag(e^{\lambda t},e^{\lambda t},...,e^{\lambda t})\\
e^{\lambda I_nt}=e^{\lambda t} I_n$$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 7 - Eigenvalues and Eigenvectors - 7.4 An Introduction to the Matrix Exponential Function - Problems - Page 465: 10
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