Linear Algebra and Its Applications (5th Edition)

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

Chapter 3 - Determinants - 3.3 Exercises - Page 187: 20

Answer

6 square units

Work Step by Step

We take the parallelogram as; $\textbf{ABCD with A=(0,0); B=(-2,4); C=(4,-5); D=(2,-1)}$ A parallelogram has 2 vectors with 4 vertices. The two vectors are $\textbf{u}$ and $\textbf{v}$ $\textbf{u} =\begin{bmatrix}2\\-1\end{bmatrix}$ $\textbf{v} =\begin{bmatrix}-2\\4\end{bmatrix}$ By forming a matrix $\textbf{Z}$ from the vectors $\textbf{u}$ and $\textbf{v}$ $\textbf{Z} =\begin{bmatrix}2&-2\\-1&4\end{bmatrix}$ The area of a parallelogram is the absolute value of the determinant of $\textbf{Z}$ $Area\,of\,the\,parallelogram\,ABCD=|det\begin{bmatrix}2&-2\\-1&4\end{bmatrix}|=|det(8-2)|=6$ sq units
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