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To find the eigenvalues and eigenvectors of the matrix [tex]\( A \)[/tex], we need to follow these steps:
### Step 1: Compute the Characteristic Polynomial
The characteristic polynomial is determined by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( \lambda \)[/tex] is an eigenvalue and [tex]\( I \)[/tex] is the identity matrix of the same size as [tex]\( A \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 3 & 20 \end{pmatrix} \][/tex]
The identity matrix [tex]\( I \)[/tex] is:
[tex]\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]
So, [tex]\( A - \lambda I \)[/tex] is:
[tex]\[ A - \lambda I = \begin{pmatrix} 2-\lambda & 1 & 3 \\ 1 & 2-\lambda & 3 \\ 3 & 3 & 20-\lambda \end{pmatrix} \][/tex]
Next, we compute the determinant:
[tex]\[ \text{det}(A - \lambda I) = \text{det}\begin{pmatrix} 2-\lambda & 1 & 3 \\ 1 & 2-\lambda & 3 \\ 3 & 3 & 20-\lambda \end{pmatrix} \][/tex]
### Step 2: Solve the Characteristic Polynomial
Finding the roots of this determinant gives us the eigenvalues:
After solving the eigenvalue equation (characteristic polynomial), we find that the eigenvalues are:
[tex]\[ \lambda_1 = 21, \quad \lambda_2 = 1, \quad \lambda_3 = 2 \][/tex]
### Step 3: Find the Eigenvectors
Next, we need to find the eigenvectors corresponding to each eigenvalue. For each eigenvalue [tex]\( \lambda \)[/tex], we solve the equation [tex]\( (A - \lambda I)x = 0 \)[/tex] for the eigenvector [tex]\( x \)[/tex].
For [tex]\( \lambda_1 = 21 \)[/tex]:
[tex]\[ A - 21I = \begin{pmatrix} 2-21 & 1 & 3 \\ 1 & 2-21 & 3 \\ 3 & 3 & 20-21 \end{pmatrix} = \begin{pmatrix} -19 & 1 & 3 \\ 1 & -19 & 3 \\ 3 & 3 & -1 \end{pmatrix} \][/tex]
Solving [tex]\( (A - 21I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_1 = \begin{pmatrix} -0.162 \\\ -0.162 \\\ -0.973 \end{pmatrix} \][/tex]
For [tex]\( \lambda_2 = 1 \)[/tex]:
[tex]\[ A - I = \begin{pmatrix} 2-1 & 1 & 3 \\ 1 & 2-1 & 3 \\ 3 & 3 & 20-1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 1 & 3 \\ 3 & 3 & 19 \end{pmatrix} \][/tex]
Solving [tex]\( (A - I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_2 = \begin{pmatrix} -0.707 \\\ 0.707 \\\ -1.677 \times 10^{-16} \end{pmatrix} \][/tex]
For [tex]\( \lambda_3 = 2 \)[/tex]:
[tex]\[ A - 2I = \begin{pmatrix} 2-2 & 1 & 3 \\ 1 & 2-2 & 3 \\ 3 & 3 & 20-2 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 3 \\ 1 & 0 & 3 \\ 3 & 3 & 18 \end{pmatrix} \][/tex]
Solving [tex]\( (A - 2I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_3 = \begin{pmatrix} -0.688 \\\ -0.688 \\\ 0.229 \end{pmatrix} \][/tex]
### Summary
The eigenvalues of matrix [tex]\( A \)[/tex] are:
[tex]\[ \lambda_1 = 21, \quad \lambda_2 = 1, \quad \lambda_3 = 2 \][/tex]
The corresponding eigenvectors are:
[tex]\[ x_1 = \begin{pmatrix} -0.162 \\\ -0.162 \\\ -0.973 \end{pmatrix}, \quad x_2 = \begin{pmatrix} -0.707 \\\ 0.707 \\\ -1.677 \times 10^{-16} \end{pmatrix}, \quad x_3 = \begin{pmatrix} -0.688 \\\ -0.688 \\\ 0.229 \end{pmatrix} \][/tex]
### Step 1: Compute the Characteristic Polynomial
The characteristic polynomial is determined by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( \lambda \)[/tex] is an eigenvalue and [tex]\( I \)[/tex] is the identity matrix of the same size as [tex]\( A \)[/tex].
For matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 3 & 20 \end{pmatrix} \][/tex]
The identity matrix [tex]\( I \)[/tex] is:
[tex]\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]
So, [tex]\( A - \lambda I \)[/tex] is:
[tex]\[ A - \lambda I = \begin{pmatrix} 2-\lambda & 1 & 3 \\ 1 & 2-\lambda & 3 \\ 3 & 3 & 20-\lambda \end{pmatrix} \][/tex]
Next, we compute the determinant:
[tex]\[ \text{det}(A - \lambda I) = \text{det}\begin{pmatrix} 2-\lambda & 1 & 3 \\ 1 & 2-\lambda & 3 \\ 3 & 3 & 20-\lambda \end{pmatrix} \][/tex]
### Step 2: Solve the Characteristic Polynomial
Finding the roots of this determinant gives us the eigenvalues:
After solving the eigenvalue equation (characteristic polynomial), we find that the eigenvalues are:
[tex]\[ \lambda_1 = 21, \quad \lambda_2 = 1, \quad \lambda_3 = 2 \][/tex]
### Step 3: Find the Eigenvectors
Next, we need to find the eigenvectors corresponding to each eigenvalue. For each eigenvalue [tex]\( \lambda \)[/tex], we solve the equation [tex]\( (A - \lambda I)x = 0 \)[/tex] for the eigenvector [tex]\( x \)[/tex].
For [tex]\( \lambda_1 = 21 \)[/tex]:
[tex]\[ A - 21I = \begin{pmatrix} 2-21 & 1 & 3 \\ 1 & 2-21 & 3 \\ 3 & 3 & 20-21 \end{pmatrix} = \begin{pmatrix} -19 & 1 & 3 \\ 1 & -19 & 3 \\ 3 & 3 & -1 \end{pmatrix} \][/tex]
Solving [tex]\( (A - 21I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_1 = \begin{pmatrix} -0.162 \\\ -0.162 \\\ -0.973 \end{pmatrix} \][/tex]
For [tex]\( \lambda_2 = 1 \)[/tex]:
[tex]\[ A - I = \begin{pmatrix} 2-1 & 1 & 3 \\ 1 & 2-1 & 3 \\ 3 & 3 & 20-1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 1 & 3 \\ 3 & 3 & 19 \end{pmatrix} \][/tex]
Solving [tex]\( (A - I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_2 = \begin{pmatrix} -0.707 \\\ 0.707 \\\ -1.677 \times 10^{-16} \end{pmatrix} \][/tex]
For [tex]\( \lambda_3 = 2 \)[/tex]:
[tex]\[ A - 2I = \begin{pmatrix} 2-2 & 1 & 3 \\ 1 & 2-2 & 3 \\ 3 & 3 & 20-2 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 3 \\ 1 & 0 & 3 \\ 3 & 3 & 18 \end{pmatrix} \][/tex]
Solving [tex]\( (A - 2I)x = 0 \)[/tex] will give us the eigenvector:
[tex]\[ x_3 = \begin{pmatrix} -0.688 \\\ -0.688 \\\ 0.229 \end{pmatrix} \][/tex]
### Summary
The eigenvalues of matrix [tex]\( A \)[/tex] are:
[tex]\[ \lambda_1 = 21, \quad \lambda_2 = 1, \quad \lambda_3 = 2 \][/tex]
The corresponding eigenvectors are:
[tex]\[ x_1 = \begin{pmatrix} -0.162 \\\ -0.162 \\\ -0.973 \end{pmatrix}, \quad x_2 = \begin{pmatrix} -0.707 \\\ 0.707 \\\ -1.677 \times 10^{-16} \end{pmatrix}, \quad x_3 = \begin{pmatrix} -0.688 \\\ -0.688 \\\ 0.229 \end{pmatrix} \][/tex]
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