Question

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

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

Short answer

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

Step-by-step solution

Matrix equation for Bezier curve

The matrix whose columns are the four control points is called ageometry 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 tis replaced by a parameter\(s\):

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

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