For Matrix A, find a basis for the eigenspace corresponding to λ = 3. 4 A = 3 0 0 2 2 - - 1 3 5
For Matrix A, find a basis for the eigenspace corresponding to λ = 3. 4 A = 3 0 0 2 2 - - 1 3 5
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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![**Problem:**
For Matrix \( A \), find a basis for the eigenspace corresponding to \( \lambda = 3 \).
\[ A = \begin{bmatrix} 4 & 0 & -1 \\ 3 & 0 & 3 \\ 2 & -2 & 5 \end{bmatrix} \]
**Solution:**
1. **Subtract \(\lambda I\) from \(A\):**
Here, \( \lambda = 3 \) and \( I \) is the identity matrix of the same dimension as \( A \).
\[ 3I = 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \]
\[ A - 3I = \begin{bmatrix} 4 & 0 & -1 \\ 3 & 0 & 3 \\ 2 & -2 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 3 & -3 & 3 \\ 2 & -2 & 2 \end{bmatrix} \]
2. **Set up the system of equations:**
The matrix above represents the system \((A - 3I)x = 0\).
\[ \begin{bmatrix} 1 & 0 & -1 \\ 3 & -3 & 3 \\ 2 & -2 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
3. **Solve the system:**
Reduce the matrix to row echelon form:
Step 1: Swap \( R2 \) and \( R1 \):
\[ \begin{bmatrix} 1 &](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9baea4c0-9d9e-491a-8271-bc2a2bbd8b45%2F09046f7e-65ed-4232-a103-552e9b4c9661%2F6x3taqn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
For Matrix \( A \), find a basis for the eigenspace corresponding to \( \lambda = 3 \).
\[ A = \begin{bmatrix} 4 & 0 & -1 \\ 3 & 0 & 3 \\ 2 & -2 & 5 \end{bmatrix} \]
**Solution:**
1. **Subtract \(\lambda I\) from \(A\):**
Here, \( \lambda = 3 \) and \( I \) is the identity matrix of the same dimension as \( A \).
\[ 3I = 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \]
\[ A - 3I = \begin{bmatrix} 4 & 0 & -1 \\ 3 & 0 & 3 \\ 2 & -2 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 3 & -3 & 3 \\ 2 & -2 & 2 \end{bmatrix} \]
2. **Set up the system of equations:**
The matrix above represents the system \((A - 3I)x = 0\).
\[ \begin{bmatrix} 1 & 0 & -1 \\ 3 & -3 & 3 \\ 2 & -2 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
3. **Solve the system:**
Reduce the matrix to row echelon form:
Step 1: Swap \( R2 \) and \( R1 \):
\[ \begin{bmatrix} 1 &
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