6 1 -3 0; λ = 2,3,5 6-1 6-6 5
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.**
\[
\begin{bmatrix}
6 & 1 & -3 \\
6 & -1 & 0 \\
6 & -6 & 5
\end{bmatrix}; \quad \lambda = 2, 3, 5
\]
---
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
- **A.**
For \( P = \, \boxed{\,}\), \( D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \).
(Simplify your answer.)
- **B.**
The matrix cannot be diagonalized.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb73e221-fc24-4f22-8ff5-c1a81fef495f%2F01c6f5d9-7df1-42a6-85eb-9475b0b056d3%2Fyru24l5_processed.png&w=3840&q=75)
Transcribed Image Text:**Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.**
\[
\begin{bmatrix}
6 & 1 & -3 \\
6 & -1 & 0 \\
6 & -6 & 5
\end{bmatrix}; \quad \lambda = 2, 3, 5
\]
---
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
- **A.**
For \( P = \, \boxed{\,}\), \( D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \).
(Simplify your answer.)
- **B.**
The matrix cannot be diagonalized.
Expert Solution
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Step 1: To find the eigenvector:
Step by step
Solved in 4 steps with 4 images
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