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Diagonalize the quadratic forms in Exercises 3540 by finding an orthogonal matrix Q such

Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) | 3rd Edition | ISBN: 9780538735452 | Authors: David Poole ISBN: 9780538735452 298

Solution for problem 5.5.31 Chapter 5

Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) | 3rd Edition

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Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) | 3rd Edition | ISBN: 9780538735452 | Authors: David Poole

Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) | 3rd Edition

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Problem 5.5.31

Diagonalize the quadratic forms in Exercises 3540 by finding an orthogonal matrix Q such that the change of variable x Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form.

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MAT 211­ Lecture 6­ 15.3 & 15.4 Section 15.3 Homework #8: ln(x,y)= x​ +y​2­y​x­2 lnx= 0 & lny= 0 1st Derivative:​ lnx= 2x­y​ lny= 2y­2yx I. 2x­y​=0 ­­­­­­­> x=1 2­y​=0 2=y​ y=­ 2 2​ 2y­2yx=0 ­­­­­­­> y=0 2x­y​ =0 2x­0=0 x=0 II. 2y(1­x)= 0 y=0 & x=1 Critical Points: (0,0); (1, ­ 2); (1, + 2) 2nd Derivative:​ lnxx= 2 lnxx (0,0)= 2 lnyy= 2­2x lnyy (0,0)= 2 lnxy= ­2y lnxy (0,0)= 0 H(0,0)= 2 * 2 ­ 0= 4 Since lnxx= 2>0, then we have a relative minimum at (0,0). ­Critical point at (1, ­ 2) : lnxx (1, ­ 2) =2 √ lnyy (1, ­ 2) =0 lnxy (1, ­ 2) =(­2)(­ 2) = 2 2 2 H (1, ­ 2)= 2 * 0 ­ (2 2)​ 0­8= ­8< 0 Since H is negative, we have a saddle point at (1, ­ 2). ­Critical point at (1, + 2) : lnxx (1, + 2)=2 lnyy (1, + 2)=0 lnxy (1, + 2)= ­2( 2) 2​ H(1, + 2) = 2* 0­ (­2 2)​= 0­8= ­8<0 Since H is negative, we have a saddle point at (1, + 2). Homework #11: M(c,f)= 11+8c+4c​ 2­40f­20cf+40f​2 Mc= 8+8c­20f­­­­­­­> 8+8c­20f= 0 divide by 4 2+2c­5f=0 Mf= ­40­20c+80f­­­­> ­40­20c+80f=0 divide by 20 ­2­c+4f=0

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Chapter 5, Problem 5.5.31 is Solved
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Textbook: Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign)
Edition: 3
Author: David Poole
ISBN: 9780538735452

The full step-by-step solution to problem: 5.5.31 from chapter: 5 was answered by , our top Math solution expert on 01/29/18, 04:03PM. The answer to “Diagonalize the quadratic forms in Exercises 3540 by finding an orthogonal matrix Q such that the change of variable x Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form.” is broken down into a number of easy to follow steps, and 38 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 7 chapters, and 1985 solutions. Since the solution to 5.5.31 from 5 chapter was answered, more than 240 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign), edition: 3. Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) was written by and is associated to the ISBN: 9780538735452.

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Diagonalize the quadratic forms in Exercises 3540 by finding an orthogonal matrix Q such