Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 5 - Eigenvalues and Eigenvectors - 5.3 Exercises - Page 288: 3

Answer

$A^k=\begin{bmatrix} a^k&0\\3a^k-3b^k&b^k \end{bmatrix}$

Work Step by Step

$A^k=PD^kP^{-1}$ $P=\begin{bmatrix} 1&0\\3&1 \end{bmatrix}$ $D^k=\begin{bmatrix} a^k&0\\0&b^k \end{bmatrix}$ $P^{-1}=\begin{bmatrix} 1&0\\-3&1 \end{bmatrix}$ $A^k=\begin{bmatrix} a^k&0\\3a^k-3b^k&b^k \end{bmatrix}$
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