Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.4 Matrix Algebra - 10.4 Assess Your Understanding - Page 777: 34


$A^{-1}=\left[\begin{array}{rr} {-1}&{-1/2}\\ {-3}&{-2}\end{array}\right]$

Work Step by Step

In order to calculate the inverse of an $n$ by $n$ non-singular matrix $A$, we will proceed with the following steps: Step -1. Transform the given matrix $A$ into this form $\left[A|I_{n}\right]$ as follows: $\left[A|I_{n}\right]$ = $\left[\begin{array}{ll|ll} {-4}&{1}&{1}&{0}\\ {6}&{-2}&{0}&{1}\end{array}\right]$ Step-2. Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form by using Row operation: $R_{1}=r_{1}+r_{2}$ $\left[\begin{array}{rr|rr} {2}&{-1}&{1}&{1}\\ {6}&{-2}&{0}&{1}\end{array}\right]$ Now, use Row operations: $R_{2}=r_{2}-3r_{1}$ and $R_1=r_1+r_2$: $\left[\begin{array}{cc|cc}{2}&{0}&{-2}&{-1}\\{0}&{1}&{-3}&{-2}\end{array}\right]$ Finally divide Row-1 by $2$ to obtain: $\left[\begin{array}{cc|cc}{1}&{0}&{-1}&{-1/2}\\{0}&{1}&{-3}&{-2}\end{array}\right]$ Step-3: The reduced row echelon form of $\left[A|I_{n}\right]$ will be represented as the identity matrix $I_{n}$ on the left of the vertical line; and the $n$ by $n$ matrix on the right of the vertical line is the inverse of $A$. Thus, we have: $A^{-1}=\left[\begin{array}{rr} {-1}&{-1/2}\\ {-3}&{-2}\end{array}\right]$ (Note that if the identity matrix does not obtain on the left, then the matrix $A$ does have no inverse).
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