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Classify the quadratic forms in Exercises $9-18 .$ Then make a change of variable, $\mathbf{x}=P \mathbf{y},$ that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct $P$ using the methods of Section $7.1 .$$$2 x_{1}^{2}+6 x_{1} x_{2}-6 x_{2}^{2}$$

Hence the required new quadratic form is $Q(y)=-7 y_{1}^{2}+3 y_{2}^{2}$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 2

Quadratic Forms

Introduction to Matrices

Campbell University

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Harvey Mudd College

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problem. 10 um, were given a quadratic for that is too sorry to ex Juan X one squared plus six x juan next to minus six x two squared. Okay, so the first thing we need to d'oh his problem is to write out the matrix a so based on this, or get a form. And because the coefficient of X one squared is too. So that means the first entry of diving off the diagonal is too. And since the coefficient off X two squared iss negative six. So that means the second entry for the diagonal is next six and the product of X y next to has the cooperation. Six. That means, um, the some off the first column, uh, some of first column, a second first colored woman said, Second row and the 1st 1st row, second column. We'll have six. So that means we can take the 1st 1st row second column to be three three, and we can take the second roll First column to be Greek so that these two entries have the some six. So this is our make 68 And the next thing is to find find the, uh, I got better. Uh, here we consider the determinant off a miners Lum di, and this will be determined and tough. Two minutes. Number three three negative six minus lambda and determined is zoo. Okay, so we expand this determinant. That will be two minor Lunda times Neck of six, minus Nanda and minus nine. He was zero. So the solution decision off this quadratic equation will be first time No one, We have three. And London, too, is negative. Seven. So since thes to London are not equal, I actually one of the wines positive, and number two is negative. So that means, uh, this it's called radical form is as the as across the vacation, which is not definite in a minute. Because in our textbook, the fear of five from our textbooks says, um, Symmetric matrix. If they could do that, if the quadratic form is positive, definite even on leave, the Eiken bellows are all positive. If it's an indefinite, that is the even happen. Only if the Matrix has both positive and negative, I get better. So in this case, we have both positive and negative values. So that means it's basics as the quadratic of warm that is indefinite and correspondingly we can write out the we're going to form with the under. Under the change of variables, I have three white, one squared, minus seven. Why two squared? Okay. And the last thing is to find out the corresponding Eigen Victor's car to act number one and number two. So that's first. So by our definition of Mike and Values and Aiken Victor's, we have this quality we take. So in this case, the yoga value has to be a two dimensional vector. So in this case, we let yeah, get ready to be x one x two. So, based on our first row of it takes a we will have two x one plus three x two ukuleles Well, on the right hand side is just a two off x one. Sorry, home. We have number three. So he's three up x one on this gives the relationship between X one and X two. So that is three dfx too is X one and also for lunga too, which is connective. Seven. Yeah, naked seven we have left inside. It's still saying, but the right hand side is connected. Seven next one so this gives. Say, um, this is three X two equals connective nine x one. So does x two. You causing neck too. Three x one. So we get right out. Uh, I can vector the first I can vector corresponding to loved a one will be one and three and we have to normalize it. So that gives one over square root 10 because three squared plus one square it is 10 and take the square root. So you have square with 10 and 31 So the second I can vector will be one neck of three and this is after the normalization. We have one over square root too. Times one negative street. So the matrix p in this problem will be the spin off these two vectors. So take out that one over square root. 10 inside will be 31 on one and three. Sorry. 31 So that's it for our problems. Problems. 10

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