2 1 -1 -1 4 -1 1 2 = 1 -1 0 -1 1 -1 0 -1 1 300 0 -1 -1 020 -1 −1 −1 003 -1 -1 0
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
2. The matrix A is factored in the form PDP^−1. Use the Diagonalization Theorem to find the eigenvalues
of A and a basis for each eigenspace.
![On the left side, we have a matrix with the dimensions \( 3 \times 3 \):
\[
\begin{bmatrix}
2 & -1 & -1 \\
1 & 4 & 1 \\
-1 & -1 & 2
\end{bmatrix}
\]
This matrix is expressed as the product of three matrices on the right side of the equation. These matrices, each also with the dimensions \( 3 \times 3 \), are as follows:
The first matrix in the product is:
\[
\begin{bmatrix}
1 & -1 & 0 \\
-1 & 1 & -1 \\
0 & -1 & 1
\end{bmatrix}
\]
The second matrix is a diagonal matrix:
\[
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{bmatrix}
\]
The third matrix is:
\[
\begin{bmatrix}
0 & -1 & -1 \\
-1 & -1 & -1 \\
-1 & -1 & 0
\end{bmatrix}
\]
These matrices can be multiplied in sequence to yield the original matrix on the left. The multiplication follows standard matrix multiplication rules, which involve taking the dot product of the rows of the first matrix with the columns of the subsequent matrix.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d07e113-70c7-44ea-a54a-4dad627b44c8%2F99cb8c17-1427-4e10-8e13-feb8dfb94945%2Fyvo9zpi_processed.png&w=3840&q=75)
Transcribed Image Text:On the left side, we have a matrix with the dimensions \( 3 \times 3 \):
\[
\begin{bmatrix}
2 & -1 & -1 \\
1 & 4 & 1 \\
-1 & -1 & 2
\end{bmatrix}
\]
This matrix is expressed as the product of three matrices on the right side of the equation. These matrices, each also with the dimensions \( 3 \times 3 \), are as follows:
The first matrix in the product is:
\[
\begin{bmatrix}
1 & -1 & 0 \\
-1 & 1 & -1 \\
0 & -1 & 1
\end{bmatrix}
\]
The second matrix is a diagonal matrix:
\[
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{bmatrix}
\]
The third matrix is:
\[
\begin{bmatrix}
0 & -1 & -1 \\
-1 & -1 & -1 \\
-1 & -1 & 0
\end{bmatrix}
\]
These matrices can be multiplied in sequence to yield the original matrix on the left. The multiplication follows standard matrix multiplication rules, which involve taking the dot product of the rows of the first matrix with the columns of the subsequent matrix.
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