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Chapter 2: Q2.7-11Q (page 93)

A rotation on a computer screen is sometimes implemented as the product of two shear-and-scale transformations, which can speed up calculations that determine how a graphic image actually appears in terms of screen pixels. (The screen consists of rows and columns of small dots, called pixels.) The first transformation \({A_{\bf{1}}}\) shears vertically and then compresses each column of pixels; the second transformation \({A_2}\) shears

horizontally and then stretches each row of pixels. Let

\({A_1} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{sin\varphi }&{cos\varphi }&0\\0&0&1\end{array}} \right]\),

\[{A_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{sec\varphi }&{ - tan\varphi }&0\\0&1&0\\0&0&1\end{array}} \right]\]

Show that the composition of the two transformations is a rotation in \({\mathbb{R}^2}\).

Short Answer

Expert verified

The composition is \[{A_2}{A_1} = \left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\]. For a rotation in \({\mathbb{R}^2}\), the composition is the transformation matrix in coordinates.

Step by step solution

01

State the row-column rule

If the sum of the products of matching entries from row\(i\)of matrix A and column\(j\)of matrix B equals the item in row\(i\)and column\(j\)of AB, then it can be said that product AB is defined.

The product is \({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\).

02

Obtain the product of matrices

Consider the matrices\({A_1} = \left[ {\begin{array}{*{20}{c}}1&0&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\)and\[{A_2} = \left[ {\begin{array}{*{20}{c}}{\sec \varphi }&{ - \tan \varphi }&0\\0&1&0\\0&0&1\end{array}} \right]\].

Obtain the product of matrices\({A_2}{A_1}\), as shown below:

\[\begin{array}{c}{A_2}{A_1} = \left[ {\begin{array}{*{20}{c}}{\sec \varphi }&{ - \tan \varphi }&0\\0&1&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\sec \varphi \left( 1 \right) - \tan \varphi \left( {\sin \varphi } \right) + 0\left( 0 \right)}&{\sec \phi \left( 0 \right) - \tan \varphi \left( {\cos \varphi } \right) + 0\left( 0 \right)}&{\sec \phi \left( 0 \right) - \tan \varphi \left( 0 \right) + 0\left( 1 \right)}\\{0\left( 1 \right) + 1\left( {\sin \varphi } \right) + 0\left( 0 \right)}&{0\left( 0 \right) + 1\left( {\cos \varphi } \right) + 0\left( 0 \right)}&{0\left( 0 \right) + 1\left( 0 \right) + 0\left( 1 \right)}\\{0\left( 1 \right) + 0\left( {\sin \varphi } \right) + 1\left( 0 \right)}&{0\left( 0 \right) + 0\left( {\cos \varphi } \right) + 1\left( 0 \right)}&{0\left( 0 \right) + 0\left( 0 \right) + 1\left( 1 \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\sec \varphi - \tan \varphi \sin \varphi }&{ - \tan \varphi \cos \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\end{array}\]

Thus, \[{A_2}{A_1} = \left[ {\begin{array}{*{20}{c}}{\sec \varphi - \tan \varphi \sin \varphi }&{ - \tan \varphi \cos \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\].

03

Use the trigonometric identities

Simplify the trigonometric entry\[\sec \varphi - \tan \varphi \sin \varphi \]using a trigonometric identity, as shown below:

\[\begin{array}{c}\sec \varphi - \tan \varphi \sin \varphi = \frac{1}{{\cos \varphi }} - \frac{{{{\sin }^2}\varphi }}{{\cos \varphi }}\\ = \frac{{1 - {{\sin }^2}\varphi }}{{\cos \varphi }}\\ = \frac{{{{\cos }^2}\varphi }}{{\cos \varphi }}\\ = \cos \varphi \end{array}\]

Simplify \[ - \tan \varphi \cos \varphi \].

\[\begin{array}{c} - \tan \varphi \cos \varphi = - \frac{{\sin \varphi }}{{\cos \varphi }} \times \cos \varphi \\ = - \sin \varphi \end{array}\]

The product\({A_2}{A_1}\)becomes

\[{A_2}{A_1} = \left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\].

Therefore, for a rotation in \({\mathbb{R}^2}\), the composition is the transformation matrix in coordinates.

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Most popular questions from this chapter

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

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