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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.

Rotate points through \({\bf{45}}^\circ \) about the point \(\left( {{\bf{3}},{\bf{7}}} \right)\).

Short Answer

Expert verified

\(\left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}&{3 + 2\sqrt 2 }\\{\frac{{\sqrt 2 }}{2}}&{\frac{{\sqrt 2 }}{2}}&{7 - 5\sqrt 2 }\\0&0&1\end{array}} \right]\)

Step by step solution

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01

Find the matrix for translation

The point \(\left( {3,7} \right)\) must be shifted back to the origin.

The translation matrix is

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}\\0&1&{ - 7}\\0&0&1\end{array}} \right]\).

02

Find the matrix for rotation

The matrix for rotation by \(45^\circ \) is

\(\left[ {\begin{array}{*{20}{c}}{\cos 45^\circ }&{ - \sin 45^\circ }&0\\{\sin 45^\circ }&{\cos 45^\circ }&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&0\\0&0&1\end{array}} \right]\).

03

Find the matrix for translation

The point must again be shifted back to \(\left( {3,7} \right)\). The translation matrix is

\(\left[ {\begin{array}{*{20}{c}}1&0&3\\0&1&7\\0&0&1\end{array}} \right]\).

04

Find the combined matrix of transformation

The combined matrix for transformation can be expressed as

\(\left[ {\begin{array}{*{20}{c}}1&0&3\\0&1&7\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 3}\\0&1&{ - 7}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}&{3 + 2\sqrt 2 }\\{\frac{{\sqrt 2 }}{2}}&{\frac{{\sqrt 2 }}{2}}&{7 - 5\sqrt 2 }\\0&0&1\end{array}} \right]\).

So, the transformed matrix is \(\left[ {\begin{array}{*{20}{c}}{\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}&{3 + 2\sqrt 2 }\\{\frac{{\sqrt 2 }}{2}}&{\frac{{\sqrt 2 }}{2}}&{7 - 5\sqrt 2 }\\0&0&1\end{array}} \right]\).

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Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) …(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

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