91影视

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:

$A\stackrel{鈫�}{c}=\stackrel{鈫�}{0}$

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

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 $A\stackrel{鈫�}{c}=\stackrel{鈫�}{0}$. 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.

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

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

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

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 .

$c_{8}xy(x鈭�1)+c_{9}xy(y鈭�1)=0$

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

Now, for a point $(x,y)$on the cubic curve is either $x=0,y=0\text{鈥�}or\text{鈥�}y=1$. This set is graphed as follows:

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

Now, for a point $(x,y)$on the cubic curve is either $x=0,x=1\text{鈥�}or\text{鈥�}y=0$. This set is graphed as follows: