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

Thus, the cubic that passes through the eight given points is of the form ${c}_{7}\left(2x-3{x}^{2}+{x}^{3}\right)+{c}_{10}\left(2y-3{y}^{2}+{y}^{3}\right)=0$ .

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:

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

There are eight points.

The system of eight equations is written as follows:

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

Where role="math" localid="1660376194262" $A=\left[\begin{array}{cccccccccc}1& {x}_{1}& {y}_{1}& {x}_{1}^{2}& {x}_{1}{y}_{1}& {y}_{1}^{2}& {x}_{1}^{3}& {x}_{1}^{2}{y}_{1}& {x}_{1}{y}_{1}^{2}& {y}_{1}^{3}\\ 1& {x}_{2}& {y}_{2}& {x}_{2}^{2}& {x}_{2}{y}_{2}& {y}_{2}^{2}& {x}_{2}^{3}& {x}_{2}^{2}{y}_{2}& {x}_{2}{y}_{2}^{2}& {y}_{2}^{3}\\ 1& {x}_{3}& {y}_{3}& {x}_{3}^{2}& {x}_{3}{y}_{3}& {y}_{3}^{2}& {x}_{3}^{3}& {x}_{3}^{2}{y}_{3}& {x}_{3}{y}_{3}^{2}& {y}_{3}^{3}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 1& {x}_{8}& {y}_{8}& {x}_{8}^{2}& {x}_{8}{y}_{8}& {y}_{8}^{2}& {x}_{8}^{3}& {x}_{8}^{2}{y}_{8}& {x}_{8}{y}_{8}^{2}& {y}_{8}^{3}\end{array}\right]$

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

Plug in the eight points to derive the A matrix.

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

Now, use gauss-Jordan elimination to solve the system $A\stackrel{\to }{c}=\stackrel{\to }{0}$ . Note that the matrix is identical to the A matrix from Exercise 44, with the addition of one row. Thus, the first five rows is replaced with row echelon form in Exercise 44.

$\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 8& 0& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 4& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 1& 0& 1& 0& 0& 1& 0& 0& 0\\ 1& 2& 0& 4& 0& 8& 0& 0& 0& 0\\ 1& 0& 1& 0& 0& 1& 0& 0& 0& 1\\ 1& 0& 2& 0& 0& 4& 0& 0& 0& 8\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 1& 2& 1& 4& 2& 1& 8& 4& 4& 1\\ 1& 1& 2& 1& 2& 4& 1& 2& 4& 8\end{array}\right]\to$

$\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 1& 1& 0& 1& 1& 3\\ 0& 0& 0& 0& 0& -1& 0& 2& 0& -3\\ 0& 0& 0& 0& 0& 2& 0& 0& 2& 6\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 1& 1& 0& 1& 1& 3\\ 0& 0& 0& 0& 0& -1& 0& 2& 0& -3\\ 0& 0& 0& 0& 0& 2& 0& 0& 2& 6\end{array}\right]\to \phantom{\rule{0ex}{0ex}}\left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 1& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 2& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 2& 0\end{array}\right]\to \left[\begin{array}{cccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& -2& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& -2\\ 0& 0& 0& 1& 0& 0& 3& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 1& 0\\ 0& 0& 0& 0& 1& 1& 0& 0& 0& 3\\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0\end{array}\right]\to$

## Step 3: Showing that cubics through(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(0,4),(1,1).(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(0,4),(1,1).

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

${c}_{1}=0\phantom{\rule{0ex}{0ex}}{c}_{2}=2{c}_{7}\phantom{\rule{0ex}{0ex}}{c}_{3}=2{c}_{10}\phantom{\rule{0ex}{0ex}}{c}_{4}=-3{c}_{7}\phantom{\rule{0ex}{0ex}}{c}_{5}=0\phantom{\rule{0ex}{0ex}}c6=-3{c}_{10}\phantom{\rule{0ex}{0ex}}{c}_{8}=0\phantom{\rule{0ex}{0ex}}{c}_{9}=0$

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 eight given points is of the form role="math" localid="1660391605404" $2{c}_{7}x+2{c}_{10}y-3{c}_{7}{x}^{2}-3{c}_{10}{y}^{2}+{c}_{7}{x}^{3}+{c}_{10}{y}^{3}=0\phantom{\rule{0ex}{0ex}}{c}_{7}\left(2x-3{x}^{2}+{x}^{3}\right)+{c}_{10}\left(2y-3{y}^{2}+{y}^{3}\right)=0$ .

## 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: