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Linear Algebra With Applications
Found in: Page 144
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.

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 c9xy(yx)=0.

See the step by step solution

Step by Step Solution

Step 1: Given Information 

Each point Pi(xi,yi)defines an equation in the 10 variablesc1,c2,.....,c10 given by:


There are nine points.

The system of five equations is written as follows:



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

Plug in the nine points to derive theA matrix.


Now, use gauss-Jordan elimination to solve the system Ac=0. Note that theA matrix is identical to the Amatrix from Exercise 46l, with the addition of one row. Thus, the first eight rows are replaced with row echelon form in Exercise 46.


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 Ac=0 which satisfies:


While c9 are free variables. Recall that the cubic equation is as follows:


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


Step 3: Sketch of cubic

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

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