Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 27E from Chapter 3.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 27E
Chapter:
Problem:
Let S be the parallelogram determined by the vectorsb1 = and b2 = , and let A = Compute the area of the image of S under the mapping × ↦ Ax.
Step-by-Step Solution
Given Information
We are given with two vectors that form a parallelogram S \[ \mathbf { b } _ { 1 } = \left[ \begin{array} { r } { - 2 } \\ { 3 } \end{array} \right] \text { and } \mathbf { b } _ { 2 } = \left[ \begin{array} { r } { - 2 } \\ { 5 } \end{array} \right] \] We have to compute the area of the image of S under the mapping $\mathbf { x } \mapsto A \mathbf { x } $
Step-1: The Area
Area of the parallelogram: \[\begin{array}{l} \left| { det \left[ {{{\bf{b}}_1}\quad {{\bf{b}}_2}} \right]} \right| = | det \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 2}\\ 3&5 \end{array}} \right]\\ = |( - 2)(5) - ( - 2)(3)|\\ = | - 10 + 6|\\ = 4 \end{array}\]
Step-2: The Area of Image
\[\begin{array}{l} | det A| \times \{ {\rm{ Area of }}S\} = 4\cdot\left| {\begin{array}{*{20}{c}} 6&{ - 3}\\ { - 3}&2 \end{array}} \right|\\ = 4(6 \times 2 - 3 \times 3)\\ = 4(3)\\ = 12 \end{array}\] Therefore,
The area of image of S under the mapping is 12
We are given with two vectors that form a parallelogram S \[ \mathbf { b } _ { 1 } = \left[ \begin{array} { r } { - 2 } \\ { 3 } \end{array} \right] \text { and } \mathbf { b } _ { 2 } = \left[ \begin{array} { r } { - 2 } \\ { 5 } \end{array} \right] \] We have to compute the area of the image of S under the mapping $\mathbf { x } \mapsto A \mathbf { x } $
Step-1: The Area
Area of the parallelogram: \[\begin{array}{l} \left| { det \left[ {{{\bf{b}}_1}\quad {{\bf{b}}_2}} \right]} \right| = | det \left[ {\begin{array}{*{20}{c}} { - 2}&{ - 2}\\ 3&5 \end{array}} \right]\\ = |( - 2)(5) - ( - 2)(3)|\\ = | - 10 + 6|\\ = 4 \end{array}\]
Step-2: The Area of Image
\[\begin{array}{l} | det A| \times \{ {\rm{ Area of }}S\} = 4\cdot\left| {\begin{array}{*{20}{c}} 6&{ - 3}\\ { - 3}&2 \end{array}} \right|\\ = 4(6 \times 2 - 3 \times 3)\\ = 4(3)\\ = 12 \end{array}\] Therefore,
The area of image of S under the mapping is 12