(a) Show matrix A+In is invertible and find (A+In)-1 in terms of A.
(b) exp(A) =In +A +(1/2!)A2+ (1/3!)A3+..+(1/2010!)A2010
Show exp(A) is invertible and find (exp(A))-1 in terms of A
A is square matrix with size n*n such that A2011=0
The Attempt at a Solution
I made it to A*(A2010)=0
it's in the same form as Ax=0, so can I say that solution, x=A2010
(well, I don't know whether this will help me so I just wrote it down)
(a) To show matrix (A+In) is invertible, I tried to let (A+In)x=0
and I don't know how to continue anymore T^T
I just know if I can find out that solution for (A+In)x=0 is trival then it is invertible.
But, am I approaching it in a correct path?
(b) I guess we use the same approach as a?