Matrix polynomials and inverses- Linear Algebra

lina29
Messages
84
Reaction score
0

Homework Statement


For p(x)=x4-2x3+3x2-3x+1 and

A= 1 1 1 -1
-1 0 -2 1
0 0 1 0
1 0 0 0

you can check that P(A)=0 using this find a polynomial q(x) so that q(A)=A-1. The point is A4-2A3+3A2-3A=A(-A3+2A2-3A+3I)=I

a) What is q(x)?

I don't really understand how to approach this problem. My initial though was that I had to solve the right side of the eqn(A4-2A3+3A2-3A) and that would be q(x). Am I on the right track? Also what's the purpose of p(x)?
 
Last edited:
Physics news on Phys.org
Well, you say, or were given, that "the point is A(-A^3+2A^2- 3A+ 3I)= I

Isn't the definition of "inverse" that AA^{-1}= I?
 
lina29 said:

Homework Statement


For p(x)=x4-2x3+3x2-3x+1 and

A= 1 1 1 -1
-1 0 -2 1
0 0 1 0
1 0 0 0

you can check that P(A)=0 using this find a polynomial q(x) so that q(A)=A-1. The point is A4-2A3+3A2-3A=A(-A3+2A2-3A+3I)=I

a) What is q(x)?

I don't really understand how to approach this problem. My initial though was that I had to solve the right side of the eqn(A4-2A3+3A2-3A) and that would be q(x). Am I on the right track? Also what's the purpose of p(x)?

Try to be more consistent with your letters. The functions you're working with are named p and q, so you shouldn't be using P in place of p. p(x) = x4-2x3+3x2-3x+1, so p(A) = A4-2A3+3A2-3A+I. Don't worry about the purpose of p - just take it as given for now.

You have A(-A3+2A2-3A+3I) = I. If A times whatever is the identity, then the whatever is the inverse of A. q(A) will be the inverse of A.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top