Question

How many cubics can you fit through nine distinct points${\mathit{P}}_{\mathbf{1}}\left(x_1,y_1\right)\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{P}}_{\mathbf{9}}\left(x_9,y_9\right)$?. Describe all possible scenarios, and give an example in each case.

Short answer

Thus, the dimension of the kernel is at least 1.

Step-by-step solution

Given in the question.

If nine points to fit to the cubic as follows:

$\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{xc}}_{\mathbf{2}}\mathbf{+}{\mathbf{yc}}_{\mathbf{3}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{4}}\mathbf{+}{\mathbf{xyc}}_{\mathbf{5}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{6}}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{7}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{yc}}_{\mathbf{8}}\mathbf{+}{\mathbf{xy}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{9}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{10}}\mathbf{=}\mathbf{0}$

A system of 9 equations with 10 unknowns. This equal to the kernel of a $9\times 10$matrix.

Thus, the dimension of the kernel is at least 1.

All possible scenarios for nine distinct points.

If in fact the kernel is equal one there is only one solution because all the other variables is in terms of the others and is reduced to same solution, any multiple of a solution is equivalent to th solution itself.

Example for the case.

If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

$\left(\begin{array}{c}0,0\end{array}\right),\left(\begin{array}{c}1,0\end{array}\right),\left(\begin{array}{c}3,0\end{array}\right),\left(\begin{array}{c}4,0\end{array}\right),\left(\begin{array}{c}0,1\end{array}\right),\left(\begin{array}{c}0,2\end{array}\right),\left(\begin{array}{c}0,3\end{array}\right),\left(\begin{array}{c}0,4\end{array}\right),\left(\begin{array}{c}1,1\end{array}\right).$

Plug in the ten points to derive the A matrix.

$A=\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 3& 0& 9& 0& 0& 27& 0& 0& 0\\ 1& 4& 0& 16& 0& 0& 64& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 0& 3& 0& 0& 9& 0& 0& 0& 27\\ 1& 0& 4& 0& 0& 16& 0& 0& 0& 64\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\end{array}\right]$

Now, use gauss-Jordan elimination to solve the system role="math" localid="1660376531377" $A\stackrel{\rightharpoonup }{c}=\stackrel{\rightharpoonup }{o}$Note that the A matrix is identical to the A matrix from Exercise 48, with the addition of one row. Thus, the first nine rows is replaced with row echelon form in Exercise 48 and the new row is moved to the bottom.

$$\begin{gathered} \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 3& 0& 9& 0& 0& 27& 0& 0& 0\\ 1& 4& 0& 16& 0& 0& 64& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 0& 3& 0& 0& 9& 0& 0& 0& 27\\ 1& 0& 4& 0& 0& 16& 0& 0& 0& 64\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 1& 1& -1\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 4& 0& 0& 16& 0& 0& 0& 64\end{array}\right]\to \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 1& 1& -1\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]\to \end{gathered}$$uncaught exception: Invalid chunk

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Showing that cubics through (0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(0,4),(1,1).

The solution of the equation$A\stackrel{\rightharpoonup }{c}=\stackrel{\rightharpoonup }{0}$which satisfies:

$$\begin{gathered} c_1=0 \\ {c}_2=0 \\ {c}_3=0 \\ {c}_4=0 \\ {c}_5=-c_s-c_9 \\ {c}_6=0 \\ {c}_7=0 \\ {c}_{10}=0 \end{gathered}$$

While $c_s-c_9$are free variables. Recall that the cubic equation is as follows:

$c_1+x{c}_2+y{c}_3+x_{}^2{c}_4+xy{c}_5+y_{}^2{c}_6+x^3{c}_7+x^{2}y{c}_8+x{y}^2{c}_9+y^3{c}_{10}=0$

Therefore, the cubic that passes through the nine given points is of the form .

$$\begin{gathered} \left(-c_8-c_9\right)xy+c_8{x}^{2}y+c_{9}x{y}^2=0 \\ {c}_8\left(x^{2}y-xy\right)+c_9\left(x{y}^2-xy\right)=0 \\ {c}_{8}xy\left(x-1\right)+c_{9}xy\left(y-1\right)=0 \\ xy\left(c_8\left(x-1\right)+c_9\left(y-1\right)\right)=0 \end{gathered}$$