Finding Characteristic Polynomial of Matrix B

In summary, the problem is asking for the characteristic polynomial of an invertible matrix and the roots of that polynomial are the eigenvalues of the inverse matrix.
  • #1
Boom101
16
0

Homework Statement


B =
|a 1 -5 |
|-2 b -8 |
|2 3 c |

Find the characteristic polynomial of the following matrix.


Homework Equations


None


The Attempt at a Solution


So basically I have to find the det(B-λI). No matter what I do to the matrix I can't make the result simple. I can't find a simple answer using any cofactor expansion of the matrix. I get to (a-λ)[(b-λ)(c-λ)+24)] - [(-2c+2λ)+16] -5[-6-(2b-2λ)] which comes out to about 20 different characters. I tried rearranging the matrix to add a 0 to help with cofactor expansion but that doesn't help cause I'm left with like b+3-λ which doesn't help. Is there just no simple answer or am I doing something wrong?
 
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  • #2
Your characteristic polynomial seems correct to me. You only need to simplify it further...
 
  • #3
Ok thanks. I just have another question. Let M be an nxn matrix this is invertible and let A be an nxn matrix. show that the eigenvalues of (inverse of M) are the roots of the polynomial p(λ) = det(A-λM). I just have no idea where to go on this one.
 
  • #4
Boom101 said:
Ok thanks. I just have another question. Let M be an nxn matrix this is invertible and let A be an nxn matrix. show that the eigenvalues of (inverse of M) are the roots of the polynomial p(λ) = det(A-λM). I just have no idea where to go on this one.
What's the exact problem statement? I haven't done any work on this, but there don't seem to be enough conditions on A and M for this to be true.

However, for this to be true, if λ1 is an eigenvalue of M-1, then for some nonzero vector x, M-1x = λ1x.
 
  • #5
That was the exact problem statement. It was b) from which a) was the question above. Regarding my first question, I'm left with about 20 different combinations of variables which I can't simplify. If I can't simplify, should I just leave it the way it is?
 
  • #6
The problem is just asking for the characteristic polynomial. You could multiply things out and give it in terms of λ3, λ2, λ, and a constant.

For the second question you asked, it seems to me that you're supposed to conclude that matrix A is actually the identity matrix I.
 
  • #7
Thanks again mark.
 
  • #8
Thanks again mark.
 

1. How do you find the characteristic polynomial of a matrix?

To find the characteristic polynomial of a matrix, you first need to find the determinant of the matrix. Then, you need to subtract the identity matrix multiplied by the eigenvalue from the original matrix. Finally, set the determinant equal to zero and solve for the eigenvalue, which will be the coefficient of the resulting polynomial.

2. What is the purpose of finding the characteristic polynomial of a matrix?

The characteristic polynomial of a matrix helps us to find the eigenvalues and eigenvectors of the matrix. This information is useful in many areas of mathematics, including linear algebra, differential equations, and geometry.

3. Can any type of matrix have a characteristic polynomial?

Yes, any square matrix can have a characteristic polynomial. However, only square matrices have eigenvalues and eigenvectors, so only they can have a characteristic polynomial.

4. How does the degree of the characteristic polynomial relate to the size of the matrix?

The degree of the characteristic polynomial is equal to the size of the matrix. For example, a 3x3 matrix will have a characteristic polynomial of degree 3.

5. Are there any special techniques for finding the characteristic polynomial of a complex matrix?

No, the same process can be used to find the characteristic polynomial of a complex matrix as for a real matrix. However, the calculations may be more complex due to the inclusion of complex numbers. It is important to follow the same steps and be careful with the calculations in order to find the correct characteristic polynomial.

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