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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![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 \( A = PDP^T \).
\[ D = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]
\[ P = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b609ed5-0879-4e23-8eee-4515e84ef2b9%2F469f3e4b-7bb0-4870-97e6-4f4b67f9813e%2F2zs0t1a_processed.png&w=3840&q=75)
Transcribed Image Text: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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