Algebra 2 (1st Edition)

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

Chapter 3 Linear Systems and Matrices - 3.8 Use Inverse Matrices to Solve Linear Systems - 3.8 Exercises - Quiz for Lessons 3.6-3.8 - Page 217: 4

Answer

$(B-A)C=\begin{bmatrix} -4 &3\\ 28 & 1 \end{bmatrix}$

Work Step by Step

$(B-A)C=(\begin{bmatrix} 2 &-3\\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1 & -4\\ 5 & 2 \end{bmatrix})\begin{bmatrix} -6 &-1\\ 2 & 4 \end{bmatrix}$ $=\begin{bmatrix} 2-1 &-3-(-4)\\ 0-5 & 1-2 \end{bmatrix}.\begin{bmatrix} -6 &-1\\ 2 & 4 \end{bmatrix}$ $=\begin{bmatrix} 1 &1\\ -5 & -1 \end{bmatrix}.\begin{bmatrix} -6 &-1\\ 2 & 4 \end{bmatrix}$ $=\begin{bmatrix} -6+2 &-1+4\\ 30-2 & 5-4 \end{bmatrix}$ $=\begin{bmatrix} -4 &3\\ 28 & 1 \end{bmatrix}$
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