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.

$(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(1,1)$

Short answer

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

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 nine points.

The system of five equations is written as follows:

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

Where$A=\left[\begin{array}{cccccccccc}1& x_1& y_1& x_1^2& x_1{y}_1& y_1^2& x_1^3& x_1^2{y}_1& x_1{y}_1^2& y_1^3\\ 1& x_2& y_2& x_2^2& x_2{y}_2& y_2^2& x_2^3& x_2^2{y}_2& x_2{y}_2^2& y_2^3\\ 1& x_3& y_3& x_3^2& x_3{y}_3& y_3^2& x_3^3& x_3^2{y}_3& x_3{y}_3^2& y_3^3\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& x_9& y_9& x_9^2& x_9{y}_9& y_9^2& x_9^3& x_9^2{y}_9& x_9{y}_9^2& y_9^3\end{array}\right]$

Step 2:Apply gauss-Jordan elimination in the matrix

Plug in the nine points to derive the matrix.

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

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 46, with the addition of one row. Thus, the first eight rows are replaced with row echelon form in Exercise 46 and the new row is moved to the bottom.

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

Showing that cubic passes through(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2), (0,3),(1,1) 

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

$\begin{array}{c}{c}_1=0\\ c_2=0\\ c_3=2c_{10}\\ c_4=0\\ c_5=-c_8-c_9+c_{10}\\ c_6=-3c_{10}\\ c_7=0\\ c_{10}=0\end{array}$

While $c_8,c_9$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 .

$c_{8}xy(x-1)+c_{9}xy(y-1)=0$

Sketch of cubic

As the first example, substitute $c_8=0,c_9=1$. The cubic is $xy(x-1)=0$.

Now, for a point $(x,y)$on the cubic curve is either $x=0,y=0ory=1$. This set is graphed as follows:

As the first example, substitute $c_8=1,c_9=0$. The cubic is $xy(y-1)=0$.

Now, for a point $(x,y)$on the cubic curve is either $x=0,x=1ory=0$. This set is graphed as follows: