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

In Exercises 3-8, find the \({\bf{3}} \times {\bf{3}}\) matrices that produce the described composite 2D transformations, using homogenous coordinates.

Rotate points through \({\bf{60}}^\circ \) about the point \(\left( {{\bf{6}},{\bf{8}}} \right)\).

Short Answer

Expert verified

\(\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}&{3 + 4\sqrt 3 }\\{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\)

Step by step solution

01

Find the matrix for translation

The point \(\left( {6,8} \right)\) must be shifted back to the origin.

The translation matrix is

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

02

Find the matrix for rotation

The matrix for rotation by \(60^\circ \) is

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

03

Find the matrix for translation

The point must again be shifted back to \(\left( {6,8} \right)\). The translation matrix is:

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

04

Find the combined matrix of transformation

The combined matrix for transformation can be expressed as

\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}&0\\{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}&{3 + 4\sqrt 3 }\\{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\).

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

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Most popular questions from this chapter

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

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\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

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In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

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Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).
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