Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 12 - Matrices - 12-2 Matrix Multiplication - Practice and Problem-Solving Exercises - Page 778: 40

Answer

$D-2E=\begin{bmatrix} -3 & 12 & -1 \\ -2 & 3 & 5 \\ -4 & -3 & -4 \end{bmatrix}$

Work Step by Step

Multiplying each element of $E= \begin{bmatrix} 2 & -5 & 0 \\ 1 & 0 & -2 \\ 3 & 1 & 1 \end{bmatrix} $ by $2,$ and then subtracting the result from $D= \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & 1 \\ 2 & -1 & -2 \end{bmatrix} ,$ then \begin{align*} D-2E&= \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & 1 \\ 2 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 2(2) & -5(2) & 0(2) \\ 1(2) & 0(2) & -2(2) \\ 3(2) & 1(2) & 1(2) \end{bmatrix} \\\\&= \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & 1 \\ 2 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 4 & -10 & 0 \\ 2 & 0 & -4 \\ 6 & 2 & 2 \end{bmatrix} \\\\&= \begin{bmatrix} 1-4 & 2-(-10) & -1-0 \\ 0-2 & 3-0 & 1-(-4) \\ 2-6 & -1-2 & -2-2 \end{bmatrix} \\\\&= \begin{bmatrix} 1-4 & 2+10 & -1-0 \\ 0-2 & 3-0 & 1+4 \\ 2-6 & -1-2 & -2-2 \end{bmatrix} \\\\&= \begin{bmatrix} -3 & 12 & -1 \\ -2 & 3 & 5 \\ -4 & -3 & -4 \end{bmatrix} .\end{align*} Hence, $ D-2E=\begin{bmatrix} -3 & 12 & -1 \\ -2 & 3 & 5 \\ -4 & -3 & -4 \end{bmatrix} .$
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