Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

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. $(0, 0), (1, 0), (2, 0), (3, 0), (0, 1), (0, 2), (0, 3), (1, 1), (2, 2)$ .

The cubic that passes through the nine given points is of the form to $c_{9} x y (y - x) = 0$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

Each point $P_{i} (x_{i}, y_{i})$ 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:

$A \vec{c} = \vec{0}$

Where $A = [\begin{matrix} 1 & x_{1} & y_{1} & x_{1}^{2} & x_{1} y_{1} & y_{1}^{2} & x_{1}^{3} & x_{1}^{2} y_{1} & x_{1} y_{1}^{2} & y_{1}^{3} \\ 1 & x_{2} & y_{2} & x_{2}^{2} & x_{2} y_{2} & y_{2}^{2} & x_{2}^{3} & x_{2}^{2} y_{2} & x_{2} y_{2}^{2} & y_{2}^{3} \\ 1 & x_{3} & y_{3} & x_{3}^{2} & x_{3} y_{3} & y_{3}^{2} & x_{3}^{3} & x_{3}^{2} y_{3} & x_{3} y_{3}^{2} & y_{3}^{3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & x_{9} & y_{9} & x_{9}^{2} & x_{9} y_{9} & y_{9}^{2} & x_{9}^{3} & x_{9}^{2} y_{9} & x_{9} y_{9}^{2} & y_{9}^{3} \end{matrix}]$ .

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

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

$A = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 2 & 0 & 4 & 0 & 0 & 8 & 0 & 0 & 0 \\ 1 & 3 & 0 & 9 & 0 & 0 & 27 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 2 & 0 & 0 & 4 & 0 & 0 & 0 & 8 \\ 1 & 0 & 3 & 0 & 0 & 9 & 0 & 0 & 0 & 27 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 4 & 4 & 4 & 8 & 8 & 8 & 8 \end{matrix}]$

Now, use gauss-Jordan elimination to solve the system $A \vec{c} = \vec{0}$ . 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.

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

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 $A \vec{c} = \vec{0}$ 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:

$c_{1} + x c_{2} + y c_{3} + x^{2} c_{4} + x y c_{5} + y^{2} c_{6} + x^{3} c_{7} + x^{2} y c_{8} + x y^{2} c_{9} + y^{3} c_{10} = 0$

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

$\begin{matrix} c_{9} x^{2} y + c_{9} x y^{2} = 0 \\ c_{9} x y (y - x) = 0 \end{matrix}$

Step 3: Sketch of cubic

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

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Given Information

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

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

Step 3: Sketch of cubic

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Given Information

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

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

Step 3: Sketch of cubic

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