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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 .$$$\begin{array}{l}{2 x_{1}^{2}+2 x_{2}^{2}-6 x_{1} x_{2}-6 x_{1} x_{3}-6 x_{1} x_{4}-6 x_{2} x_{3}-} \\ {6 x_{2} x_{4}-2 x_{3} x_{4}}\end{array}$$

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Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 2

Quadratic Forms

Introduction to Matrices

McMaster University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

09:57

Classify the quadratic for…

04:07

03:15

07:50

02:33

05:22

02:35

03:01

05:52

03:42

Make a change of variable,…

a quick radical form and his two ex. Once we're plus two extra square, try to run you down. It's two x one squared plus two x two square, thanks to sorry you ask us where minus six ex wives too. Minus six x one x tree and minus six ex wives for and minus six x two X tree. That's two next to x three and minus six x to export X two x four and the last term minus two extra x four two x three x four. All right, So the corresponding matrix a might ISS even order form will be. We first have to 200 on the diagonal do zero Cerro zero. All right, so ex wives, too has coefficient inactive. Six. That means we have naked three year elective three here and Xbox three has negative six. So again they give three here Negative three and the same for X two at three. So we have a connective three here, connective three here and for extra s for we have negative three here and active three here and for the last term x three x four. So that is connected. One here and negative one, and the rest terms will be zero. Oh, wait. Um I'm sorry. Are we still have it? Ex wife's for with the with coefficient. Negative six. So here should be a negative three. And here's B negative three. So this is our matrix eight. All right. And by using the programming, we can find the Aiken values. I'm the one negative seven. And London too. ISS five lum, which is the same as London three. And number four is one. So we have connective seven. But we also have to positive numbers so that, beings, the organic form is indefinite and definite. Okay? And we can just write down the corresponding forgetting form and the change of bearable that will be effective seven by once. Weird. Plus white. Why two squared us by my pre squared and class. Why four squared? All right. And corresponding. Aiken Vector. I'll just write down directly to save us more time. This will be first. I can vector will be all ones. The second again vector will be connective 2011 The third eye can vector is And I took 1100 The last I can vector. We'll be 00 negative. One and one. All right. And remember, we need to normalize these four vectors for anyone. We have skater on half. I guess we have one squared plus one squared plus one squared, plus one squared. That's four. And you take it Square roots. So that's two 1111 Okay. Or be too active. Two squared were before and plus one plus one Take a square root, so that would be one over square root. Six. We're six and times the naked too. 014 and three. Again. We also have, um, negative 11 for the president. 1st 2 entries. And that means you have to take one over square root to asthma scaler for me. Three. So that is negative. 1100 And before. So that is There was a rule. Say, there was a row next 11 So that's the same. We just take where to when I was going to. And 00 negative one. And the one hence, our Matrix p will be the span off these four vectors. So that is First column will be 1/2 one. Have that one, huh? And one hat second column will be naked to over a square of six zero and one over square six and the one over square with six. And the third column will be one over square to next one over square root, too. And the one over is going to and zero. So the last column will be suicidal next. One hopeless player, too, and one over square root. So here's our Matrix Peak.

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