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Expert-verified 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.53. $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{\right)}$.

Thus, the cubic that passes through the nine given points is of the form $2{c}_{10}y-3{c}_{10}{y}^{2}+{c}_{10}{y}^{3}=0$.

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:

role="math" localid="1660365952521" ${{\mathbf{c}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{X}}}_{{\mathbf{i}}}{{\mathbf{c}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{i}}}{{\mathbf{c}}}_{{\mathbf{3}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{X}}}_{{\mathbf{i}}}^{{\mathbf{2}}}{{\mathbf{c}}}_{{\mathbf{4}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{X}}}_{{\mathbf{i}}}{{\mathbf{y}}}_{{\mathbf{i}}}{{\mathbf{c}}}_{{\mathbf{5}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{i}}}^{{\mathbf{3}}}{{\mathbf{c}}}_{{6}}{\mathbf{}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{i}}}^{{\mathbf{3}}}{{\mathbf{c}}}_{{\mathbf{7}}}{\mathbf{+}}{{\mathbf{x}}}_{{\mathbf{i}}}^{{\mathbf{2}}}{{\mathbf{y}}}_{{\mathbf{i}}}{{\mathbf{c}}}_{{\mathbf{8}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{x}}}_{{\mathbf{i}}}{{\mathbf{y}}}_{{\mathbf{i}}}^{{\mathbf{2}}}{{\mathbf{c}}}_{{\mathbf{9}}}{\mathbf{}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{i}}}^{{\mathbf{3}}}{{\mathbf{c}}}_{{\mathbf{10}}}{\mathbf{=}}{\mathbf{0}}$,

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 A.

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 A 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),(3,2).

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

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

While ${c}_{10}$ 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

$2{c}_{10}y-3{c}_{10}{y}^{2}+{c}_{10}{y}^{3}=0\phantom{\rule{0ex}{0ex}}{c}_{10}y\left({y}^{2}-3y+2\right)=0\phantom{\rule{0ex}{0ex}}y\left(y-1\right)\left(y-2\right)=0$

## Step 3: Sketch of cubics.

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

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

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$\left(x,y\right)$on the cubic curve is either$y=0,y=1ory=2$. This set is graphed as follows:  ### Want to see more solutions like these? 