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Linear Algebra With Applications
Found in: Page 145
Linear Algebra With Applications

Linear Algebra With Applications

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 c72x-3x2+x3+c102y-3y2+y3=0 .

See the step by step solution

Step by Step Solution

Step 1: Given in the question.

Each point Pixi,yi defines an equation in the 10 variables c1,c2,.....,c10 given by:


There are eight points.

The system of eight equations is written as follows:


Where role="math" localid="1660376194262" A=1x1y1x12x1y1y12x13x12y1x1y12y131x2y2x22x2y2y22x23x22y2x2y22y231x3y3x32x3y3y32x33x32y3x3y32y331x8y8x82x8y8y82x83x82y8x8y82y83

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

Plug in the eight points to derive the A matrix.


Now, use gauss-Jordan elimination to solve the system Ac=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.



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 Ac=0 which satisfies:


While c7,c10 are free variables. Recall that the cubic equation is as follows:


Therefore, the cubic that passes through the eight given points is of the form role="math" localid="1660391605404" 2c7x+2c10y-3c7x2-3c10y2+c7x3+c10y3=0c72x-3x2+x3+c102y-3y2+y3=0 .

Step 3: Sketch of cubics.

As the first example, substitute c7=1,c10=0 . The cubic is


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

As the first example, substitute c7=0,c10=1 . The cubic is yy-1y-2=0 .

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

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