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Q47E

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Found in: Page 144

### 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.47.$\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{3}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{\right)}{\mathbf{,}}\mathbf{\left(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{\right)}$ .

The cubic that passes through the nine given points is of the form to ${c}_{9}xy\left(y-x\right)=0$.

See the step by step solution

## Step 1: Given Information

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:

,${c}_{1}+{x}_{i}{c}_{2}+{y}_{i}{c}_{3}+{x}_{i}^{2}{c}_{4}+{x}_{i}{y}_{i}{c}_{5}+{y}_{i}^{2}{c}_{6}+{x}_{i}^{3}{c}_{7}+{x}_{i}^{2}{y}_{i}{c}_{8}+{x}_{i}{y}_{i}^{2}{c}_{9}+{y}_{i}^{3}{c}_{10}=0$

There are nine points.

The system of five equations is written as follows:

$A\stackrel{\to }{c}=\stackrel{\to }{0}$

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 A

Plug in the nine points to derive the$A$ matrix.

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

Now, use gauss-Jordan elimination to solve the system $A\stackrel{\to }{c}=\stackrel{\to }{0}$. Note that the$A$ matrix is identical to the $A$matrix from Exercise 46l, with the addition of one row. Thus, the first eight rows are replaced with row echelon form in Exercise 46.

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

## Step 3: Showing that cubic through (0,0),(1,0),(2,0),(3,0),(0,1),(0,2), (0,3),(1,1),(2,2)

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

$\begin{array}{l}{c}_{1}=0\\ {c}_{2}=0\\ {c}_{3}=2{c}_{10}\\ {c}_{4}=0\\ {c}_{5}=2{c}_{10}\\ {c}_{6}=-3{c}_{10}\\ {c}_{7}=0\\ {c}_{8}=-{c}_{9}-{c}_{10}\\ {c}_{10}=0\end{array}$

While ${c}_{9}$ 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,

$\begin{array}{c}{c}_{9}{x}^{2}y+{c}_{9}x{y}^{2}=0\\ {c}_{9}xy\left(y-x\right)=0\end{array}$

## Step 3: Sketch of cubic

Now, for a point $\left(x,y\right)$ on the cubic curve is either $x=0,y=0\text{\hspace{0.17em}}or\text{\hspace{0.17em}}x=y$. This set is graphed as follows:

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