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: 32

Answer

$A^2=\begin{bmatrix} 1 & -3 \\ 0 & 4 \end{bmatrix}$ $A^3=\begin{bmatrix} 1 & -7 \\ 0 & 8 \end{bmatrix}$

Work Step by Step

Given: $A=\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$ $A^2=AA=\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}.\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1-0 & -1-2 \\ 0+0 & 0+4 \end{bmatrix}=\begin{bmatrix} 1 & -3 \\ 0 & 4 \end{bmatrix}$ $A^3=A^2A=\begin{bmatrix} 1 & -3 \\ 0 & 4 \end{bmatrix}.\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1-0 & -1-6 \\ 0+0 & 0+8 \end{bmatrix}=\begin{bmatrix} 1 & -7 \\ 0 & 8 \end{bmatrix}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.