# Homework Help: Linear Algebra - polynomial functions of matrices

1. Sep 16, 2011

### Snoogx

1. The problem statement, all variables and given/known data
Given p(x) = x4+2x2+1 and
A =
[[1 1 -2 0]​
[0 1 0 2]​
[1 1 -1 1]​
[0 0 -2 -1]]​
p(A) = 0
Find a polynomial q(x) so that q(A) = A-1
a) What is q(x)?
b) Compute q(A) = A-1

2. Relevant equations
I found the Cayley-Hamilton theorem, which states: p(x) = det(A-xIn).

3. The attempt at a solution
I found the inverse of A, which is:
[[-1 -1 2 0]​
[4 1 -4 -2]​
[1 0 -1 -1]​
[-2 0 2 1]]​
From here I thought I could set q(x) = A-1.
[[-1-x -1 2 0]​
[4 1-x -4 -2]​
[1 0 -1-x -1]​
[-2 0 2 1-x]]​
Then I should just solve out to find the det(). I used the 2nd column as a reference point.
Doing this I end up with the equation, x4-6x2+1.
so q(x) = x4-6x2+1.

This seems to be the wrong answer. I'm not at all sure if I did this right, or if I made an error in my calculations. Would really appreciate if someone could look over and point out my error.

Thanks

2. Sep 16, 2011

### Staff: Mentor

You went at it the hard way.

Did you check that p(A) = 0? I did, and it turns out that this is so.

p(A) = 0 ==> A4 + 2A2 + I = 0
==> -A4 - 2A2 = I
If you can factor an A out of the left side, you'll be almost where you need to go.

3. Sep 16, 2011

### manifold1

I have a similar problem, and was wondering what part b was asking for.

4. Sep 16, 2011

### Staff: Mentor

5. Sep 16, 2011

### manifold1

That's what I thought, but it wants a polynomial. Thank you.

6. Sep 16, 2011

### Staff: Mentor

And A-1 will be that polynomial in A; i.e., q(A).

7. Sep 17, 2011

### HallsofIvy

As Mark44 said before, $-A^4 - 2A^2 = A(-A^3- 2A)= I$
Now what is $A^{-1}$?