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Q54E

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Found in: Page 145

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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.54. $\left(0,0\right)\left(1,0\right)\left(2,0\right)\left(0,1\right)\left(1,1\right)\left(2,1\right)\left(0,2\right)\left(1,2\right)\left(2,2\right)$.

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

${c}_{7}x\left(2-3x+{x}^{2}\right)+{c}_{10}y\left(2-3y+{y}^{2}\right)=0$.

See the step by step solution

Step 1: Given in the question.

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:

$\mathrm{A}\stackrel{⇀}{\mathrm{c}}=\stackrel{⇀}{0}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{A}=\left[\begin{array}{cccccccccc}1& {\mathrm{x}}_{1}& {\mathrm{y}}_{1}& {\mathrm{x}}_{1}^{2}& {\mathrm{x}}_{1}{\mathrm{y}}_{1}& {\mathrm{y}}_{1}^{2}& {\mathrm{x}}_{1}^{3}& {\mathrm{x}}_{1}^{2}{\mathrm{y}}_{1}& {\mathrm{x}}_{1}{\mathrm{y}}_{1}^{2}& {\mathrm{y}}_{1}^{3}\\ 1& {\mathrm{x}}_{2}& {\mathrm{y}}_{2}& {\mathrm{x}}_{2}^{2}& {\mathrm{x}}_{2}{\mathrm{y}}_{2}& {\mathrm{y}}_{2}^{2}& {\mathrm{x}}_{2}^{3}& {\mathrm{x}}_{2}^{2}{\mathrm{y}}_{2}& {\mathrm{x}}_{2}{\mathrm{y}}_{2}^{2}& {\mathrm{y}}_{2}^{3}\\ 1& {\mathrm{x}}_{3}& {\mathrm{y}}_{3}& {\mathrm{x}}_{3}^{2}& {\mathrm{x}}_{3}{\mathrm{y}}_{3}& {\mathrm{y}}_{3}^{2}& {\mathrm{x}}_{3}^{3}& {\mathrm{x}}_{3}^{2}{\mathrm{y}}_{3}& {\mathrm{x}}_{3}{\mathrm{y}}_{3}^{2}& {\mathrm{y}}_{3}^{3}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& {\mathrm{x}}_{9}& {\mathrm{y}}_{9}& {\mathrm{x}}_{9}^{2}& {\mathrm{x}}_{9}{\mathrm{y}}_{9}& {\mathrm{y}}_{9}^{2}& {\mathrm{x}}_{9}^{3}& {\mathrm{x}}_{9}^{2}{\mathrm{y}}_{9}& {\mathrm{x}}_{9}{\mathrm{y}}_{9}^{2}& {\mathrm{y}}_{9}^{3}\end{array}\right]$

Step 2: Apply gauss-Jordan elimination in the matrix .

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 $A\stackrel{⇀}{c}=\stackrel{⇀}{0}$. 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.

$\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 0& 8& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 2& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 1& 3& 2& 9& 6& 4& 27& 18& 12& 8\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\end{array}\right]$

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

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

${c}_{1}=0\phantom{\rule{0ex}{0ex}}{c}_{2}=2{c}_{7}\phantom{\rule{0ex}{0ex}}{c}_{3}=2{c}_{10}\phantom{\rule{0ex}{0ex}}{c}_{4}=0\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{4}=-3{\mathrm{c}}_{7}\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{5}=0\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{6}=-3{c}_{10}\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{8}=0\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{9}=0$

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

$2{c}_{7}x+2{c}_{10}y-3{c}_{7}{x}^{2}-3{c}_{10}{y}^{2}+{c}_{10}{y}^{3}=0\phantom{\rule{0ex}{0ex}}{c}_{7}\left(2x-3{x}^{2}+{x}^{3}\right)+{c}_{10}\left(2y-3{y}^{2}+{y}^{3}\right)=0\phantom{\rule{0ex}{0ex}}{c}_{7}x\left(2-3x+{x}^{2}\right)+{c}_{10}y\left(2-3y+{y}^{2}\right)=0$

Step 3: Sketch of cubics.

As the first example, substitute ${c}_{7}=1,{c}_{10}=0$. The cubic is

$2x-3{x}^{2}+{x}^{3}=0\phantom{\rule{0ex}{0ex}}x\left({x}^{2}-3x+2\right)=0\phantom{\rule{0ex}{0ex}}x\left(x-1\right)\left(x-2\right)=0$

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:

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