 Solution

Use a row of descending monomials $T(t)=[t^{2}t1]$ 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:The matrix multiplication isIn 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.