College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Summary, Review, and Test - Review Exercises - Page 657: 39

Answer

$\begin{bmatrix}3&1\\2&1\end{bmatrix}$

Work Step by Step

We are given the matrix: $A=\begin{bmatrix} 1&-1\\-2&3\end{bmatrix}$ In order to find its inverse $A^{-1}$, first we compute its determinant $D$: $D=3(1)-(-1)(-2)=1$ Because $D\not=0$, matrix $A$ has inverse. We use the formula: $A^{-1}=\dfrac{1}{D}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$, where $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ $A^{-1}=\dfrac{1}{1}\begin{bmatrix}3&1\\2&1\end{bmatrix}$ $=\begin{bmatrix}3&1\\2&1\end{bmatrix}$
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