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Q52E

Expert-verified

Linear Algebra With Applications

Found in: Page 145

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.

52. $(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2)$ .

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given in the question.

Each point $P_{i} (x_{i}, y_{i})$ defines an equation in the 10 variables $c_{1}, c_{2}, . . . . ., c_{10}$ given by:

$c_{1} + x_{i} c_{2} + y_{i} c_{2} + x_{i}^{2} c_{4} + x_{i} y_{i} c_{5} + y_{i}^{2} c_{6} + x_{i}^{3} c_{7} + x_{i}^{3} y_{i} c_{8} + x_{i} y_{i}^{2} c_{9} + y_{i}^{3} c_{10} = 0$ ,

There are eight points.

The system of eight equations is written as follows:

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

Where role="math" localid="1660376194262" $A = [\begin{matrix} 1 & x_{1} & y_{1} & x_{1}^{2} & x_{1} y_{1} & y_{1}^{2} & x_{1}^{3} & x_{1}^{2} y_{1} & x_{1} y_{1}^{2} & y_{1}^{3} \\ 1 & x_{2} & y_{2} & x_{2}^{2} & x_{2} y_{2} & y_{2}^{2} & x_{2}^{3} & x_{2}^{2} y_{2} & x_{2} y_{2}^{2} & y_{2}^{3} \\ 1 & x_{3} & y_{3} & x_{3}^{2} & x_{3} y_{3} & y_{3}^{2} & x_{3}^{3} & x_{3}^{2} y_{3} & x_{3} y_{3}^{2} & y_{3}^{3} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & x_{8} & y_{8} & x_{8}^{2} & x_{8} y_{8} & y_{8}^{2} & x_{8}^{3} & x_{8}^{2} y_{8} & x_{8} y_{8}^{2} & y_{8}^{3} \end{matrix}]$

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

Plug in the eight points to derive the A matrix.

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

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

$[\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 2 & 0 & 4 & 0 & 8 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 2 & 0 & 0 & 4 & 0 & 0 & 0 & 8 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 4 & 2 & 1 & 8 & 4 & 4 & 1 \\ 1 & 1 & 2 & 1 & 2 & 4 & 1 & 2 & 4 & 8 \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 2 & 0 & 4 & 0 & 8 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 2 & 0 & 0 & 4 & 0 & 0 & 0 & 8 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 4 & 2 & 1 & 8 & 4 & 4 & 1 \\ 1 & 1 & 2 & 1 & 2 & 4 & 1 & 2 & 4 & 8 \end{matrix}] \to$

$[\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 2 & 0 & - 3 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 2 & 6 \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 2 & 0 & - 3 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 2 & 6 \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{matrix}] \to$

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

$c_{1} = 0 c_{2} = 2 c_{7} c_{3} = 2 c_{10} c_{4} = - 3 c_{7} c_{5} = 0 c 6 = - 3 c_{10} c_{8} = 0 c_{9} = 0$

While $c_{7}, c_{10}$ are free variables. Recall that the cubic equation is as follows:

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

Therefore, the cubic that passes through the eight given points is of the form role="math" localid="1660391605404" $2 c_{7} x + 2 c_{10} y - 3 c_{7} x^{2} - 3 c_{10} y^{2} + c_{7} x^{3} + c_{10} y^{3} = 0 c_{7} (2 x - 3 x^{2} + x^{3}) + c_{10} (2 y - 3 y^{2} + y^{3}) = 0$ .

Step 3: Sketch of cubics.

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

$2 x - 3 x^{2} + x^{3} = 0 x (x^{2} - 3 x + 2) = 0 x (x - 1) (x - 2) = 0$ .

Now, for a point (x,y) on the cubic curve is either $x = 0, x = 1 o r x = 2$ . This set is graphed as follows:

As the first example, substitute $c_{7} = 0, c_{10} = 1$ . The cubic is $y (y - 1) (y - 2) = 0$ .

Now, for a point (x,y) on the cubic curve is either $y = 0, y = 1 o r y = 2$ . This set is graphed as follows:

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