Linear Algebra and Its Applications (5th Edition)

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

Chapter 3 - Determinants - 3.1 Exercises - Page 170: 38

Answer

$\det kA=k^{2}\det A$

Work Step by Step

$\det k\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{vmatrix}ka&kb\\kc&kd\end{vmatrix}=(ka\cdot kd)-(kb\cdot kc)=k^{2}ad-k^{2}bc=k^{2}(ad-bc)=k^{2}\det A$ Note that this formula applies only to $2\times 2$ matrices, but the analog for the $n\times n$ case is readily derived by similar calculations.
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