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: 16

Answer

$A=\begin{bmatrix} -1 & 1\\ -2 & -2 \end{bmatrix}$ or $A=\begin{bmatrix} 2 & 1\\ -2 & -1 \end{bmatrix}$

Work Step by Step

Given: $A=\begin{bmatrix} x & 1\\ -2 & y \end{bmatrix}$ $A^2=AA=\begin{bmatrix} x & 1\\ -2 & y \end{bmatrix}.\begin{bmatrix} x & 1\\ -2 & y \end{bmatrix}=\begin{bmatrix} x^2-2 & x+y\\ -2x-2y & y^2-2 \end{bmatrix}$ $A^2=A$ $A^2-A=0$ $\begin{bmatrix} x^2-2 & x+y\\ -2x-2y & y^2-2 \end{bmatrix}-\begin{bmatrix} x & 1\\ -2 & y \end{bmatrix}=0$ $\begin{bmatrix} x^2-x-2 & x+y-1\\ -2x-2y+2 & y^2-2-y \end{bmatrix}=0$ $\rightarrow x^2-x-2=0$ $x=-1$ or $x=2$ and $y^2-2-y=0$ $y=-1$ or $y=2$ Substitute: $A=\begin{bmatrix} -1 & 1\\ -2 & -2 \end{bmatrix}$ or $A=\begin{bmatrix} 2 & 1\\ -2 & -1 \end{bmatrix}$
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