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In Exercises 1 and $2,$ find the change of variable $\mathbf{x}=P \mathbf{y}$ that transforms the quadratic form $\mathbf{x}^{T} A \mathbf{x}$ into $\mathbf{y}^{T} D \mathbf{y}$ as shown.$$3 x_{1}^{2}+3 x_{2}^{2}+5 x_{3}^{2}+6 x_{1} x_{2}+2 x_{1} x_{3}+2 x_{2} x_{3}=7 y_{1}^{2}+4 y_{2}^{2}$$

$P=\left(\begin{array}{lll}{\mathbf{u}_{1}} & {\mathbf{u}_{2}} & {\mathbf{u}_{3}}\end{array}\right)$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 3

Constrained Optimization

Introduction to Matrices

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Okay, so for problem, too. Excuse me. Um, sorry. So we're keeping the quartet performs off x x transpose a Times X. And why transposed the times be at times Why? So we need to find a change of bearable And by the left inside off the give Any question we know our matrix matrix a should be three, three, five honey diagonal and 33 for 1st 1st Rule a versatile 2nd 2nd column because we have the coefficient of X one x two, which is six. So we have three and three for these two entries and same reason we have 111 on the other entries. And because work even were also given the the question on the right hand side, which is the quadratic poor mother off. Uh, why transpose times, times, t times. Why? So that that means our matrix d will be 740 and other terms will be zero. Excuse me. So with that information be given, we know our love. The one will be seven and number two we're before on London. Three will be zero. So the next thing is to find out the corresponding wagon vectors. So the idea is to apply our definition off organ vectors and I get values. So we just plugging different wagon values into this room is equality. And find out the vector. Be so I will not go through the process right now because he's already the previous married materials. So I'll just write down. Yeah, again, Vector's first Again, Victor is negative. 0.58 connective, their appointed 58. And that gives every 0.58 this after normalization. And second again, Victor will be negative. 0.41 elective 0.41 point 82. All right, the second. Sorry, the third, Duggan Victor will be going 71.71. So negative points him to want and zero. So our transformation matrix p will span. I'll be one b two b three, which is? It is a 0.58 naked 0.8 and 2.58. And the two point What do you want? 2.41 point 82 and 0.71. Negative. 0.71 0 So we're done

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