# Linear Algebra - polynomial functions of matrices

## Homework Statement

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

## Homework Equations

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

## 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

Mark44
Mentor

## Homework Statement

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

## Homework Equations

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

## 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

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.

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

Mark44
Mentor
I have a similar problem, and was wondering what part b was asking for.

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

Mark44
Mentor
And A-1 will be that polynomial in A; i.e., q(A).

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