Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 2: Q2.7-14Q (page 93)

Show that the transformation in Exercise 7 is equivalent to a rotation about the origin followed by a translation by p. Find p.

Short Answer

Expert verified

The transformation is equivalent to a rotation about the origin. And the value of p is \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Obtain the translation and rotation matrices

As p is in\({\mathbb{R}^{\bf{2}}}\), thetranslation matrix to\({\bf{p}} = \left( { - 6, - 8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\).

As p is in\({\mathbb{R}^{\bf{2}}}\), thetranslation matrix to\({\bf{p}} = \left( {6,8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\).

Obtain therotation matrix through\(60^\circ \), as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \left( {60^\circ } \right)}&{ - \sin \left( {60^\circ } \right)}&0\\{\sin \left( {60^\circ } \right)}&{\cos \left( {60^\circ } \right)}&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\end{array}\)

02

Obtain the image of the homogeneous coordinate

Obtain the image of thehomogeneous coordinate rotating about\(60^\circ \)and translated by\({\bf{p}} = \left( { - 6, - 8} \right)\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\)

By rotating about the origin, it becomes

\(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\).

Translation at\({\bf{p}} = \left( {6,8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\).

03

Obtain the product of matrices

Compute the product of\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\), as shown below:

\[\]

Thus, the translation matrix is \(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&{3 + 4\sqrt 3 }\\{\sqrt 3 /2}&{1/2}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\).

04

Write the translation matrix

Recall that the transformation onhomogeneous coordinates for graphics has the matrix of the form\(\left[ {\begin{array}{*{20}{c}}A&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

Compare\(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&{3 + 4\sqrt 3 }\\{\sqrt 3 /2}&{1/2}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\)with \(\left[ {\begin{array}{*{20}{c}}A&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

This gives\(A = \left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}\\{\sqrt 3 /2}&{1/2}\end{array}} \right]\)and\({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).

Thetranslation matrix is represented as

\(\left[ {\begin{array}{*{20}{c}}1&0&{3 + 4\sqrt 3 }\\0&1&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

The matrix can also be represented as\(\left[ {\begin{array}{*{20}{c}}I&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&0\\{{0^T}}&1\end{array}} \right]\).

Thus, the transformation is equivalent to rotation about the origin.

And \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free