= Solution
Use a row of descending monomials $T(t)=[t^2\ t\ 1]$ and a <matrix> with one <control point> per row. Expanding $(1-t)^2,2t(1-t),t^2$ gives their <coefficients> in the three columns of the basis <matrix>:
$$
M_B=\begin{pmatrix}1&-2&1\\-2&2&0\\1&0&0\end{pmatrix},\qquad
G_B=\begin{pmatrix}-2&0\\0&1\\0&0\end{pmatrix}.
$$
The <matrix multiplication> is
$$
\boxed{F(t)=T(t)M_BG_B
=[t^2\ t\ 1]\begin{pmatrix}-2&-2\\4&2\\-2&0\end{pmatrix}.}
$$
In particular, $T(t)M_B=[(1-t)^2\ 2t(1-t)\ t^2]$, so multiplying the geometric <matrix> takes exactly the weighted <control point> sum in part (a). The two output entries are $F_x(t),F_y(t)$. The basis <matrix> depends only on the degree and ordering convention; $G_B$ holds the particular geometry.
Back to article page
