Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 2 - Matrices and Systems of Linear Equations - 2.2 Matrix Algebra - Problems - Page 136: 14

Answer

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Work Step by Step

$A^2=A.A=\begin{bmatrix} -1 &0 &4\\ 1 & 1 & 2\\ -2 &3 & 0 \end{bmatrix}.\begin{bmatrix} -1 &0 &4\\ 1 & 1 & 2\\ -2 &3 & 0 \end{bmatrix}=\begin{bmatrix} -7 &12 &-4\\ -4 & 7 & 6\\ 5 &3 & -2 \end{bmatrix}$ $A^3=A^2.A=\begin{bmatrix} -7 &12 &-4\\ -4 & 7 & 6\\ 5 &3 & -2 \end{bmatrix}.\begin{bmatrix} -1 &0 &4\\ 1 & 1 & 2\\ -2 &3 & 0 \end{bmatrix}=\begin{bmatrix} 27 &0 &-4\\ -1 & 25 & -2\\ 2&-3 & 26 \end{bmatrix}$ Hence $A^3+A-26I_2=\begin{bmatrix} 27 &0 &-4\\ -1 & 25 & -2\\ 2&-3 & 26 \end{bmatrix}+\begin{bmatrix} -1 &0 &4\\ 1 & 1 & 2\\ -2 &3 & 0 \end{bmatrix}-\begin{bmatrix} 26&0 & 0\\ 0 &26&0\\ 0 &0 & 26 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0\\ 0&0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
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