# Math Help - Proving adj(a^trace)=[adj(a)]^trace?

1. ## Proving adj(a^trace)=[adj(a)]^trace?

Hello, I had 3 homework problems that I am having trouble with. I really have problems when it comes to proving things. Here goes

Assume A is an n x n invertible matrix. Prove the following identities:

a) adj( A^trace) = [adj(A)]^trace

b) adj( A^-1) = [adj(A)]^-1

c) adj(k*A) = k^(n-1) * adj(A), k E Real Number.

So I've tried a bit, starting out with

A^-1 = 1/det(a) * adj(a) then
A^(-1) * det(a) = adj(a) then I do the trace of both sides and don't really know where to go from there...

[det(a)*A^(-1)]^trace = [adj(A)]^trace

CAn anyone help me out?

2. Originally Posted by forensic91
Hello, I had 3 homework problems that I am having trouble with. I really have problems when it comes to proving things. Here goes

Assume A is an n x n invertible matrix. Prove the following identities:

a) adj( A^trace) = [adj(A)]^trace

b) adj( A^-1) = [adj(A)]^-1

c) adj(k*A) = k^(n-1) * adj(A), k E Real Number.

So I've tried a bit, starting out with

A^-1 = 1/det(a) * adj(a) then
A^(-1) * det(a) = adj(a) then I do the trace of both sides and don't really know where to go from there...

[det(a)*A^(-1)]^trace = [adj(A)]^trace

CAn anyone help me out?
for a) and c) use the definition of adjoint matrix (cofactors!). part b) is trivial because: $I=\text{adj}(AA^{-1})=\text{adj}(A^{-1}) \text{adj}(A).$