Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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 \( A = PDP^T \).
\[ D = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]
\[ P = \begin{bmatrix} \phantom{-} \quad & \phantom{-} \quad \\ \phantom{-} \quad & \phantom{-} \quad \end{bmatrix} \]](https://content.bartleby.com/qna-images/question/8b609ed5-0879-4e23-8eee-4515e84ef2b9/469f3e4b-7bb0-4870-97e6-4f4b67f9813e/2zs0t1a_processed.png)
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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Follow-up Question
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.
Solution
by Bartleby Expert
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
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.
Solution
by Bartleby Expert
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