Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.6 Multiplying Matrices - 3.6 Exercises - Skill Practice - Page 200: 33

Answer

$A^2=\begin{bmatrix} 18 & -5 \\ -10 & 3 \end{bmatrix}$ $A^3=\begin{bmatrix} -82 &23 \\ 46& -13 \end{bmatrix}$

Work Step by Step

Given: $A=\begin{bmatrix} -4 & 1 \\ 2 & -1 \end{bmatrix}$ $A^2=AA=\begin{bmatrix} -4 & 1 \\ 2 & -1 \end{bmatrix}.\begin{bmatrix} -4 & 1 \\ 2 & -1 \end{bmatrix}=\begin{bmatrix} 16+2 & -4-1 \\ -8-2& 2+1 \end{bmatrix}=\begin{bmatrix} 18 & -5 \\ -10 & 3 \end{bmatrix}$ $A^3=A^2A=\begin{bmatrix} 18 & -5 \\ -10 & 3 \end{bmatrix}.\begin{bmatrix} -4 & 1 \\ 2 & -1 \end{bmatrix}=\begin{bmatrix} -72-10 & 18+5\\ 40+6 & -10-3 \end{bmatrix}=\begin{bmatrix} -82 &23 \\ 46& -13 \end{bmatrix}$
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