Adjoint of the Inverse: Proving [adj(A)]^{-1} = adj(A^{-1})

[Rijad Hadzic]

Homework Statement

Prove that $[adj(A)]^{-1} = adj(A^{-1})$

Homework Equations

The Attempt at a Solution

Ok. So if

$1/det(a) * adj(a) = A^{-1}$ is true,

then $adj(A) = A^{-1} det(A)$

then

$[adj(A)]^{-1} = 1/det(A^{-1}) * adj(A^{-1}) * det(A)$

now the statement would be proved if det(A) and det(A^{-1}) were both = 1, but it doesn't say that's the case in the problem. How would I go about completing this?

 

[Rijad Hadzic]

Ok hold on a sec guys. I've been relying on the help of others a bit too much. I'm going to mark the thread as solved and then open it again if I truly can't figure it out.

 

[fresh_42]

If we take $\operatorname{adj}(B) = B^{-1} \cdot \det(B)$ as defining equation for the adjoint matrix, what does this mean for $B=A^{-1}$ and how does it relate to the case $B=A\,$?

 

[Mark44]

Rijad, for future reference, in TeX expressions if the exponent is more than one character, you have to have braces -- { } -- around the expression. I have fixed your first post.