Algebra 2 (1st Edition)

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

Chapter 3 Linear Systems and Matrices - Standardized Test Practice - Multiple Choice - Page 230: 5

Answer

Answer D

Work Step by Step

Given: $A=\begin{bmatrix} 2 & -3\\ -5 & 7 \end{bmatrix}$ $B=\begin{bmatrix} 4& 6\\ 0 &-1 \end{bmatrix}$ $AX=B$ $\begin{bmatrix} 2 & -3\\ -5 & 7 \end{bmatrix}.\begin{bmatrix} X\\ Y \end{bmatrix}=\begin{bmatrix} 4& 6\\ 0 &-1 \end{bmatrix}$ $\begin{bmatrix} X\\ Y \end{bmatrix}=\frac{1}{14-15}\begin{bmatrix} 7& 3\\ 5 &2 \end{bmatrix}\begin{bmatrix} 4& 6\\ 0 &-1 \end{bmatrix}$ $\begin{bmatrix} X\\ Y \end{bmatrix}=\frac{-1}{1}\begin{bmatrix} 28&39\\ 20 &28 \end{bmatrix}$ $\begin{bmatrix} X\\ Y \end{bmatrix}=\begin{bmatrix} -28 & -39\\ -20 &-28 \end{bmatrix}$ Hence, the correct answer is D.
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