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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

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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

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

2. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{2}}\\{\bf{7}}&{\bf{4}}\end{aligned}} \right)\).

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

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