Explanation Given: The data points are ( 0 , − 2 ) , ( 1 , − 1 ) , ( 2 , 1 ) , ( 3 , 2 ) , ( 4 , 2 ) . Approach: The least square line associated with the data points is given as, y = c 1 + x c 2 … ( 1 ) Here, ( x , y ) are the data points and c 1 and c 2 are constants. The least square solution x to the linear system A x = b is given as, x = ( A T A ) − 1 A T b ( A T A ) x = A T b … ( 2 ) Here, A is a matrix and A T is the transpose of the matrix. The transpose of a matrix is obtained by interchange of row and column of the matrix. Calculation: Substitute 0 for x and − 2 for y in equation ( 1 ) . − 2 = c 1 + 0 … ( 3 ) Substitute 1 for x and − 1 for y in equation ( 1 ) . − 1 = c 1 + c 2 … ( 4 ) Substitute 2 for x and 1 for y in equation ( 1 ) . 1 = c 1 + 2 c 2 … ( 5 ) Substitute 3 for x and 2 for y in equation ( 1 ) . 2 = c 1 + 3 c 2 … ( 6 ) Substitute 4 for x and 2 for y in equation ( 1 ) . 2 = c 1 + 4 c 2 … ( 7 ) Write the equation ( 3 ) , equation ( 4 ) , equation ( 5 ) equation ( 6 ) and equation ( 7 ) in the linear system form as A x = b . [ 1 0 1 1 1 2 1 3 1 4 ] [ c 1 c 2 ] = [ − 2 − 1 1 2 2 ] Compare the above system with linear system A x = b , this gives, A = [ 1 0 1 1 1 2 1 3 1 4 ] , x = [ c 1 c 2 ] and b = [ − 2 − 1 1 2 2 ] The transpose of the above matrix is, A T = [ 1 1 1 1 1 0 1 2 3 4 ] Multiply the matrix A T with A .

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Question

Chapter 5.5, Problem 15AP

To determine

The equation of the least square line associated with data points.

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Chapter 5.5, Problem 14AP

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Chapter 5 Solutions

Differential Equations and Linear Algebra (4th Edition)

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