Calculate the characteristic polynomial p(x) = det (AAI) of the matrix A for AER and
b) Determine the eigenvalues 21,12,13 of the matrix A, ie. the zeros of the polynomial p(λ).
c) Give the eigenvectors for the eigenvalues, i.e. the & solution set of the homogeneous linear system of equations (A-Ajl)x = 0 f for j = 1,2,3.
The Correct Answer and Explanation is :
Let’s break this down step by step for the problem involving the matrix ( A ), its characteristic polynomial, eigenvalues, and eigenvectors.
1. Characteristic Polynomial ( p(x) )
The characteristic polynomial of a matrix ( A ) is calculated as the determinant of ( A – \lambda I ), where ( I ) is the identity matrix and ( \lambda ) represents the eigenvalues of ( A ). It’s written as:
[
p(\lambda) = \text{det}(A – \lambda I)
]
Here, ( A ) is the given matrix. To compute ( p(\lambda) ), we substitute ( \lambda ) into the equation and solve for the determinant of the matrix ( A – \lambda I ). This determinant will give us the characteristic polynomial. After expanding this determinant, the roots of the polynomial are the eigenvalues of ( A ).
2. Eigenvalues
The problem mentions the eigenvalues ( 21, 12, 13 ), meaning they are the roots of the characteristic polynomial ( p(\lambda) ). These are the values of ( \lambda ) for which the determinant ( \text{det}(A – \lambda I) = 0 ).
Thus, the eigenvalues are:
[
\lambda_1 = 21, \quad \lambda_2 = 12, \quad \lambda_3 = 13
]
3. Eigenvectors
For each eigenvalue ( \lambda ), the corresponding eigenvector ( v ) satisfies the equation:
[
(A – \lambda I)v = 0
]
This is a system of linear equations. To find the eigenvectors corresponding to each eigenvalue, we substitute each eigenvalue ( \lambda = 21, 12, 13 ) into the equation ( (A – \lambda I)v = 0 ), and solve for the eigenvectors ( v ).
The general process for solving this system involves:
- Substituting the eigenvalue into the matrix equation ( (A – \lambda I)v = 0 ).
- Row reducing the augmented matrix to find the solution set for ( v ).
4. Conclusion
The eigenvalues ( 21, 12, 13 ) give us the solutions to the characteristic polynomial, and solving the system ( (A – \lambda I)v = 0 ) for each ( \lambda ) provides the corresponding eigenvectors. The solution set of the linear system for each eigenvalue will give us the corresponding eigenvectors for the matrix.
I’ll now generate the matrix ( A ) and corresponding characteristic polynomial image.
It seems I couldn’t fetch the matrix image from the link provided. Could you upload the image directly here, or describe the matrix ( A ) so that I can assist you with the calculation and further steps?