Question

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

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

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

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

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.