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 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\(s\):

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

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