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Q18E

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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Use partitioned matrix multiplication to compute the following matrix product, which appears in the alternative formula (5) for a Bezier curve.

\(\left( {\begin{aligned}{{}}1&0&0&0\\{ - 3}&3&0&0\\3&{ - 6}&3&0\\{ - 1}&3&{ - 3}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\)

The matrix product is \({\mathop{\rm x}\nolimits} \left( s \right) = \left( {\begin{aligned}{{}}1&{ - 3\left( {1 - s} \right)}&{3{{\left( {1 - s} \right)}^2}}&{ - {{\left( {1 - s} \right)}^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Matrix equation for Bezier curve

The matrix whose columns are the four control points is called a geometry matrix \(G\). The \(4 \times 4\) matrix of polynomial coefficients is the Bezier basis matrix \({M_B}\).

If \({\mathop{\rm u}\nolimits} \left( t \right)\) is the column vector of power of t, then the Bezier curve is given by \({\mathop{\rm x}\nolimits} \left( t \right) = G{M_B}{\mathop{\rm u}\nolimits} \left( t \right)\).

The parameter t is replaced by a parameter \(s\):

\({\mathop{\rm x}\nolimits} \left( s \right) = {\mathop{\rm u}\nolimits} {\left( s \right)^T}M_B^T\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\)

Step 2: Use partitioned matrix multiplication to compute the matrix product

Recall the alternative formula for a Bezier curve as shown below:

\({\mathop{\rm x}\nolimits} \left( s \right) = \left( {\begin{aligned}{{}}{{{\left( {1 - s} \right)}^3}}&{3s{{\left( {1 - s} \right)}^2}}&{3{s^2}\left( {1 - s} \right)}&{{s^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\)…(5)

Consider the matrix as \({M_B} = \left( {\begin{aligned}{{}}1&0&0&0\\{ - 3}&3&0&0\\3&{ - 6}&3&0\\{ - 1}&3&{ - 3}&1\end{aligned}} \right)\).

Consider the matrix \({\mathop{\rm u}\nolimits} {\left( s \right)^T} = \left( {\begin{aligned}{{}}1&s&{{s^2}}&{{s^3}}\end{aligned}} \right)\).

\(\begin{aligned}{}{\mathop{\rm x}\nolimits} \left( s \right) &= {\mathop{\rm u}\nolimits} {\left( s \right)^T}M_B^T\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1&s&{{s^2}}&{{s^3}}\end{aligned}} \right){\left( {\begin{aligned}{{}}1&0&0&0\\{ - 3}&3&0&0\\3&{ - 6}&3&0\\{ - 1}&3&{ - 3}&1\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1&s&{{s^2}}&{{s^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}1&{ - 3}&3&{ - 3}\\0&3&{ - 6}&{ - 3}\\0&0&3&1\\0&0&0&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1&{ - 3 + 3s}&{3 - 6s + 3{s^2}}&{ - 1 + 3s - 3{s^2} + {s^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1&{ - 3\left( {1 - s} \right)}&{3{{\left( {1 - s} \right)}^2}}&{ - {{\left( {1 - s} \right)}^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\end{aligned}\)

Thus, the matrix product is \({\mathop{\rm x}\nolimits} \left( s \right) = \left( {\begin{aligned}{{}}1&{ - 3\left( {1 - s} \right)}&{3{{\left( {1 - s} \right)}^2}}&{ - {{\left( {1 - s} \right)}^3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{{\mathop{\rm p}\nolimits} _0}}\\{{{\mathop{\rm p}\nolimits} _1}}\\{{{\mathop{\rm p}\nolimits} _2}}\\{{{\mathop{\rm p}\nolimits} _3}}\end{aligned}} \right)\).

Most popular questions for Math Textbooks

In Exercises 21–24, a, b, and c are noncollinear points in \({\mathbb{R}^{\bf{2}}}\) and p is any other point in \({\mathbb{R}^{\bf{2}}}\). Let \(\Delta {\bf{abc}}\) denote the closed triangular region determined by a, b, and c, and let \(\Delta {\bf{pbc}}\) be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that \(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\) is positive, where \(\overrightarrow {\bf{a}} \), \(\overrightarrow {\bf{b}} \) and \(\overrightarrow {\bf{c}} \) are the standard homogeneous forms for the points.

23. Let p be any point in the interior of \(\Delta {\bf{abc}}\), with barycentric coordinates \(\left( {r,s,t} \right)\), so that

\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}r\\s\\t\end{array}} \right] = \widetilde {\bf{p}}\)

Use Exercise 21 and a fact about determinants (Chapter 3) to show that

\(r = \left( {area of \Delta pbc} \right)/\left( {area of \Delta abc} \right)\)

\(s = \left( {area of \Delta apc} \right)/\left( {area of \Delta abc} \right)\)

\(t = \left( {area of \Delta abp} \right)/\left( {area of \Delta abc} \right)\)

Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)

Question: In Exercises 21 and 22, mark each statement True or False. Justify each answer.

22 a. If \(d\)is a real number and \(f\) is a nonzero linear functional defined on \({\mathbb{R}^n}\) , then \(f:d\)is a hyperplane in \({\mathbb{R}^n}\) .

b. Given any vector n and any real number \(d\), the set \(\left\{ {x:n \cdot x = d} \right\}\) is a hyperplane.

c. If \(A\) and \(B\) are nonempty disjoint sets such that \(A\) is compact and \(B\) is closed, then there exists a hyperplane that strictly separates \(A\) and \(B\).

d. If there exists a hyperplane \(H\) such that \(H\) does not strictly separate two sets \(A\) and \(B\), then \(\left( {{\rm{conv}}\,A} \right) \cap \left( {{\rm{conv}}\,B} \right) \ne \emptyset \) and \(B\).

Question: 12. In Exercises 11 and 12, mark each statement True or False. Justify each answer.

a. If \(S = \left\{ {\bf{x}} \right\}\), then \({\rm{aff}}\,S\) is the empty set.

b. A set is affine if and only if it contains its affine hull.

c. A flat of dimension 1 is called a line.

d. A flat of dimension 2 is called a hyper plane.

e. A flat through the origin is a subspace.

In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

1.\(\left( {\begin{aligned}{{}}3\\{ - 3}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\6\end{aligned}} \right),\left( {\begin{aligned}{{}}2\\0\end{aligned}} \right)\)
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