# Thread: Compute p(A) where p(x) = ...?

1. ## Compute p(A) where p(x) = ...?

Hi! Can someone please explain this problem to me? I'm not sure what it's asking for or what I'm supposed to do...

Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)

and A =

[ a b
c d ]

^ That's a matrix.

I really don't understand the question.

2. ## Re: Compute p(A) where p(x) = ...?

Originally Posted by csgirl504
Hi! Can someone please explain this problem to me? I'm not sure what it's asking for or what I'm supposed to do...

Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)

and A =

[ a b
c d ]

^ That's a matrix.

I really don't understand the question.
This doesn't really make much sense. If you were to substitute the matrix, you would end up with a matrix plus a scalar, which obviously does not have matching dimensions...

Maybe you need to substitute $\displaystyle \displaystyle |\mathbf{A}|$ instead?

3. ## Re: Compute p(A) where p(x) = ...?

Originally Posted by csgirl504
Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)
and A =
[ a b
c d ]
If $\displaystyle p(x)=a_nx^n+\ldots +a_1x+a_0$ then, by definition $\displaystyle p(A)=a_nA^n+\ldots +a_1A+a_0I$ .

P.S. If you have covered the Cayley-Hamilton theorem you can immediately find that in our case the solution is the zero matrix

4. ## Re: Compute p(A) where p(x) = ...?

Originally Posted by Prove It
This doesn't really make much sense. If you were to substitute the matrix, you would end up with a matrix plus a scalar, which obviously does not have matching dimensions...
See my previous post.