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 - Guided Practice for Example 2 - Page 211: 4

Answer

$X=\begin{bmatrix} -1& -2\\ 4 &1 \end{bmatrix}$

Work Step by Step

Begin by finding the inverse of A $A^{-1}=\frac{1}{-24-0}\begin{bmatrix} 6 & -1\\ 0 & -4 \end{bmatrix}=\begin{bmatrix} \frac{-1}{4}& \frac{-1}{8}\\ 0 &\frac{1}{6} \end{bmatrix}$ To solve the equation for X: $A^{-1}AX=A^{-1}B$ $\begin{bmatrix} \frac{-1}{4}& \frac{-1}{8}\\ 0 &\frac{1}{6} \end{bmatrix}\begin{bmatrix} -4& 1\\ 0 &6 \end{bmatrix}X=\begin{bmatrix} \frac{-1}{4}& \frac{-1}{8}\\ 0 &\frac{1}{6} \end{bmatrix}\begin{bmatrix} 8& 9\\ 24 &6 \end{bmatrix}$ $\begin{bmatrix} 1& -1\\ 0 &1 \end{bmatrix}X=\begin{bmatrix} -5& -3\\ 4 &1 \end{bmatrix}$ $X=\begin{bmatrix} -1& -2\\ 4 &1 \end{bmatrix}$
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