A change of variable, \( x = Py \), is made to transform the original quadratic form into one with no cross product. The new quadratic form is given below.\[ Q(Py) = 11y_1^2 + 6y_2^2 \]If the eigenvectors of the original matrix are as follows:\[ v_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad v_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]Find the matrices \( P \) and \( D \) such that \( A = PDP^T \).\[ D = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]\[ P = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]

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A change of variable, x = Py, is made to transform the original quadratic form into one with no cross product. The new quadratic form is given below. Q(Py) = 11y + 6y/2 If the eigenvectors of the original matrix are as follows: V1 = D= 1₁ Find the matrices P and D such that A P= v2 = = PDPT

A change of variable, x = Py, is made to transform the original quadratic form into one with no cross product. The new quadratic form is given below. Q(Py) = 11y + 6y/2 If the eigenvectors of the original matrix are as follows: V1 = D= 1₁ Find the matrices P and D such that A P= v2 = = PDPT

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A change of variable, \( x = Py \), is made to transform the original quadratic form into one with no cross product. The new quadratic form is given below.

\[ Q(Py) = 11y_1^2 + 6y_2^2 \]

If the eigenvectors of the original matrix are as follows:

\[ v_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad v_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]

Find the matrices \( P \) and \( D \) such that \( A = PDP^T \).

\[ D = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]

\[ P = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]

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P =

0	1
1	0

and D =

6	0
0	11

do not seem to be the correct answers, may there be a mistake somewhere? I went through by myself and got a similar answer, but it was also deemed incorrect. Just a bit confused.

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