Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 3 - Determinants - 3.2 Exercises - Page 178: 31

Answer

\begin{align*} & AA^{-1}=I \\ & \Rightarrow \det{AA^{-1}}=\det I=1 \\ & =(\det A) (\det{A^{-1}}) \\ & \Rightarrow\det A^{-1}=\frac{1}{\det A} \end{align*}

Work Step by Step

\begin{align*} & AA^{-1}=I \\ & \Rightarrow \det{AA^{-1}}=\det I=1 \\ & =(\det A) (\det{A^{-1}}) \\ & \Rightarrow\det A^{-1}=\frac{1}{\det A} \end{align*}
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