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Q18E

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Found in: Page 437

### 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

# 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 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).