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: 7

Answer

$\begin{bmatrix} -3.6 & -0.75 \\ 6.3 & 2.25 \end{bmatrix}$

Work Step by Step

A matrix $A=\begin{bmatrix} a_{11} & a_{12} \\a_{21} &a_{22} \end{bmatrix} $ is said to be invertible when the determinant of a matrix $det=a_{11}a_{22}-a_{12}a_{21} \ne 0$ . Then, the inverse of a matrix $A$ can be computed as: $A^{-1}=\dfrac{1}{a_{11}a_{22}-a_{12}a_{21} }\begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} &a_{11} \end{bmatrix} $ and when $a_{11}a_{22}-a_{12}a_{21} =0$; then a matrix $A$ is not invertible. Now, we have: $det=(-1.5)(2.4)-(-4.2)(0.5)=-1.5$ This shows that matrix $A$ is invertible and its inverse can be calculated as: $A^{-1}=\dfrac{1}{-1.5} \begin{bmatrix} 2.4 & 0.5\\ -4.2 & -1.5 \end{bmatrix} =\begin{bmatrix} -3.6 & -0.75 \\ 6.3 & 2.25 \end{bmatrix}$
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