• November 15th 2008, 07:36 PM
Brokescholar
Prove that if A is an n x n matrix, then det(adj A) = [det(A)]^(n-1).

I'm having a little trouble figuring this one out. I tried plugging in various equations to see if I could make the two sides equal to each other, but I'm getting stuck.

Right now I have that adj A = A^-1 * det(A)
If I plug that in for adj A then I get:
det(A^-1 * det(A)) = [det(A)]^(n-1), but I'm getting stuck on the right side of the equation. I don't see how the (n-1) part comes in.

I tried plugging in equations for det(A)
I tried det(A) = adj(A) * A * In^-1 but I deaded ended with that equation too.

Any help about how to get started would be appreciated! Thanks!
• November 15th 2008, 11:14 PM
NonCommAlg
Quote:

Originally Posted by Brokescholar
Prove that if A is an n x n matrix, then det(adj A) = [det(A)]^(n-1).

be careful! $A$ might be non-invertible. first note that if $A=0,$ then from the definition of $\text{adj}(A)$ we have $\text{adj}(A)=0,$ and the claim is clearly true in this case.

so we'll assume that $A \neq 0.$ let $\det A =\alpha$ and $\text{adj}(A)=B.$ we know that: $AB= \alpha I.$ now consider two cases:

Case 1: $\alpha=0.$ in this case $AB = 0,$ and hence $B$ cannot be invertible, because $A \neq 0.$ so: $\det B=0=0^{n-1}=\alpha^{n-1}.$

Case 2: $\alpha \neq 0.$ we have: $\alpha(\det B)= (\det A)(\det B)=\det(AB)=\det(\alpha I)=\alpha^n,$ which gives us: $\det B = \alpha^{n-1}. \ \ \Box$