Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 3 - Matrices - 3.3 The Inverse of a Matrix - Exercises 3.3 - Page 178: 12

Answer

\[x_1=1\;,\; x_2=0\]

Work Step by Step

\[x_1-x_2=1\\ 2x_1+x_2=2\] Here $A=\left[\begin{array}{cc} 1&-1\\ 2&1\end{array}\right]\;,\;b=\left[\begin{array}{c} 1\\ 2\end{array}\right]$ and $X=\left[\begin{array}{c} x_1\\ x_2\end{array}\right]$ Consider $\;A=\left[\begin{array}{cc} 1&-1\\ 2&1\end{array}\right]$ Here $|A|=1(1)-(-1)2=3\neq 0$ $\Rightarrow A$ is invertible Using Theorem 3.8 $\;\Rightarrow A^{-1}=\frac{1}{1-(-2)}\left[\begin{array}{cc} 1&1\\ -2&1\end{array}\right]$ $\;\Rightarrow A^{-1}=\frac{1}{3}\left[\begin{array}{cc} 1&1\\ -2&1\end{array}\right]$ $X=A^{-1}b$ $\Rightarrow X=\frac{1}{3}\left[\begin{array}{cc} 1&1\\ -2&1\end{array}\right]\left[\begin{array}{c} 1\\ 2\end{array}\right]$ $\Rightarrow X=\frac{1}{3}\left[\begin{array}{c} 1+2\\ -2+2\end{array}\right]$ $\Rightarrow X=\frac{1}{3}\left[\begin{array}{c} 3\\ 0\end{array}\right]$ $\Rightarrow X=\left[\begin{array}{c} 1\\ 0\end{array}\right]$ $\Rightarrow \left[\begin{array}{c} x_1\\ x_2\end{array}\right] =\left[\begin{array}{c} 1\\ 0\end{array}\right]$ $\Rightarrow x_1=1\;,\; x_2=0$ Hence $x_1=1\;,\; x_2=0$.
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