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Chapter 2: Q2.7-5Q (page 93)

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

Reflect points through the x-axis, and then rotate \({\bf{30}}^\circ \) about the origin.

Short Answer

Expert verified

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

Step by step solution

01

Find the matrix for reflection

The reflection matrix can be written as

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

02

Find the matrix for rotation

Thematrix for rotation is

\(\left[ {\begin{array}{*{20}{c}}{\cos 30^\circ }&{ - \sin 30^\circ }&0\\{\sin 30^\circ }&{\cos 30^\circ }&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{2}}&{ - \frac{1}{2}}&0\\{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}&0\\0&0&1\end{array}} \right]\).

03

Find the combined matrix of transformation

The combined matrix for transformation can be expressed as

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

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

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Most popular questions from this chapter

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.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Suppose the first two columns, \({{\bf{b}}_1}\) and \({{\bf{b}}_2}\), of Bare equal. What can you say about the columns of AB(if ABis defined)? Why?

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

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].
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