91影视

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

${\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{2}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{3}}\mathbf{}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{4}}\mathbf{}\mathbf{+}{\mathbf{X}}_{\mathbf{i}}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{5}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_6\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{7}}\mathbf{+}{\mathbf{x}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{y}}_{\mathbf{i}}{\mathbf{c}}_{\mathbf{8}}\mathbf{}\mathbf{+}{\mathbf{x}}_{\mathbf{i}}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{2}}{\mathbf{c}}_{\mathbf{9}}\mathbf{}\mathbf{+}{\mathbf{y}}_{\mathbf{i}}^{\mathbf{3}}{\mathbf{c}}_{\mathbf{10}}\mathbf{=}\mathbf{0}$

There are nine points.

The system of nine equations is written as follows:

$$\begin{gathered} \mathrm{A}\stackrel{鈬赌}{\mathrm{c}}=\stackrel{鈬赌}{0} \\ \mathrm{Here},\mathrm{A}=\left[\begin{array}{cccccccccc}1& {\mathrm{x}}_1& {\mathrm{y}}_1& {\mathrm{x}}_1^2& {\mathrm{x}}_1{\mathrm{y}}_1& {\mathrm{y}}_1^2& {\mathrm{x}}_1^3& {\mathrm{x}}_1^2{\mathrm{y}}_1& {\mathrm{x}}_1{\mathrm{y}}_1^2& {\mathrm{y}}_1^3\\ 1& {\mathrm{x}}_2& {\mathrm{y}}_2& {\mathrm{x}}_2^2& {\mathrm{x}}_2{\mathrm{y}}_2& {\mathrm{y}}_2^2& {\mathrm{x}}_2^3& {\mathrm{x}}_2^2{\mathrm{y}}_2& {\mathrm{x}}_2{\mathrm{y}}_2^2& {\mathrm{y}}_2^3\\ 1& {\mathrm{x}}_3& {\mathrm{y}}_3& {\mathrm{x}}_3^2& {\mathrm{x}}_3{\mathrm{y}}_3& {\mathrm{y}}_3^2& {\mathrm{x}}_3^3& {\mathrm{x}}_3^2{\mathrm{y}}_3& {\mathrm{x}}_3{\mathrm{y}}_3^2& {\mathrm{y}}_3^3\\ 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�& 鈰�\\ 1& {\mathrm{x}}_9& {\mathrm{y}}_9& {\mathrm{x}}_9^2& {\mathrm{x}}_9{\mathrm{y}}_9& {\mathrm{y}}_9^2& {\mathrm{x}}_9^3& {\mathrm{x}}_9^2{\mathrm{y}}_9& {\mathrm{x}}_9{\mathrm{y}}_9^2& {\mathrm{y}}_9^3\end{array}\right] \end{gathered}$$

Plug in the nine points to derive the A matrix.

$A=\left[\begin{array}{cccccccccc}1& 2& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 2& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right]$

Now, use gauss-Jordan elimination to solve the system $A\stackrel{鈬赌}{c}=\stackrel{鈬赌}{0}$. Note that the matrix is identical to the A matrix from Exercise 52, with the addition of one row. Thus, the first eight rows is replaced with row echelon form in Exercise 52.

$$\begin{gathered} \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 2& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right]鈫�\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right] \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right]鈫�\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right] \\ \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right] \end{gathered}$$

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

$$\begin{gathered} c_1=0 \\ {c}_2=2c_7 \\ {c}_3=2c_{10} \\ {c}_4=0 \\ {\mathrm{c}}_4=-3{\mathrm{c}}_7 \\ {\mathrm{c}}_5=0 \\ {\mathrm{c}}_6=-3c_{10} \\ {\mathrm{c}}_8=0 \\ {\mathrm{c}}_9=0 \end{gathered}$$

While $c_7,c_{10}$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

$$\begin{gathered} 2c_{7}x+2c_{10}y-3c_7{x}^2-3c_{10}{y}^2+c_{10}{y}^3=0 \\ {c}_7\left(2x-3x^2+x^3\right)+c_{10}\left(2y-3y^2+y^3\right)=0 \\ {c}_{7}x\left(2-3x+x^2\right)+c_{10}y\left(2-3y+y^2\right)=0 \end{gathered}$$

As the first example, substitute$c_7=1,c_{10}=0$. The cubic is

$$\begin{gathered} 2x-3x^2+x^3=0 \\ x\left(x^2-3x+2\right)=0 \\ x\left(x-1\right)\left(x-2\right)=0 \end{gathered}$$

Now, for a point ( x, y ) on the cubic curve is either $x=0,x=1orx=2$. This set is graphed as follows:

As the first example, substitute $c_7=0,c_{10}=1$. The cubic is $y\left(y-1\right)\left(y-2\right)=0$.

Now, for a point (x,y) on the cubic curve is either $y=0,y=1ory=2$. This set is graphed as follows: