Question

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

50. $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}$.

Short answer

Thus, the cubic that passes through the nine given points is of the form $c_{9}xy(y-x)=0$.

Step-by-step solution

Given Information

Each point $P_i(x_i,y_i)$defines an equation in the 10 variables $c_1,c_2,\mathrm{.....},c_{10}$given by:

$c_1+x_i{c}_2+y_i{c}_3+x_i^2{c}_4+x_i{y}_i{c}_5+y_i^2{c}_6+x_i^3{c}_7+x_i^2{y}_i{c}_8+x_i{y}_i^2{c}_9+y_i^3{c}_{10}=0$

There are ten points.

The system of ten equations is written as follows:

$A\overrightarrow{c}=\overrightarrow{0}$

Where$A=\left[\begin{array}{cccccccccc}1& x_1& y_1& x_1^2& x_1{y}_1& y_1^2& x_1^3& x_1^2{y}_1& x_1{y}_1^2& y_1^3\\ 1& x_2& y_2& x_2^2& x_2{y}_2& y_2^2& x_2^3& x_2^2{y}_2& x_2{y}_2^2& y_2^3\\ 1& x_3& y_3& x_3^2& x_3{y}_3& y_3^2& x_3^3& x_3^2{y}_3& x_3{y}_3^2& y_3^3\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& x_{10}& y_{10}& x_{10}^2& x_{10}{y}_{10}& y_{10}^2& x_{10}^3& x_{10}^2{y}_{10}& x_{10}{y}_{10}^2& y_{10}^3\end{array}\right]$

Step 2:Apply gauss-Jordan elimination in the matrix

Plug in the ten points to derive the matrix.

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

Now, use gauss-Jordan elimination to solve the system $A\overrightarrow{c}=\overrightarrow{0}$. Note that the $A$matrix is identical to the $A$matrix from Exercise 47, with the addition of one row. Thus, the first nine rows is replaced with row echelon form in Exercise 47.

$\begin{array}{l}\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 3& 0& 9& 0& 0& 27& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 0& 3& 0& 0& 9& 0& 0& 0& 27\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 2& 4& 4& 4& 8& 8& 8& 8\\ 1& 3& 3& 9& 9& 9& 27& 27& 27& 27\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 1& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 1& 3& 3& 9& 9& 9& 27& 27& 27& 27\end{array}\right]\to \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 1& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& -3\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]\end{array}$

Showing that cubic passes through (0,0),(1,0),(2,0),(3,0),(0,1),(0,2), (0,3),(1,1),(2,2),(3,3)

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

$\begin{array}{c}{c}_1=0\\ c_2=0\\ c_3=0\\ c_4=0\\ c_5=0\\ c_6=0\\ c_7=0\\ c_8=-c_9\\ c_{10}=0\end{array}$

While are $c_9$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

$\begin{array}{c}-c_9{x}^{2}y+c_{9}x{y}^2=0\\ c_{9}xy(y-x)=0\end{array}$

Sketch of cubic

Now, for a point $(x,y)$on the cubic curve is either . This set is graphed as follows: