MHB Adj ( adj A ) = ( det A )^(n-2) A (ARSLAN's question at Yahoo Answers)

[Fernando Revilla]

Here is a link to the question:

If A is an nxn matrix,prove that adj(adj(A))=Ax(det A)^(n-2)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello ARSLAN,

If $M$ is an invertible $n\times n$ matrix, then $M^{-1}=\dfrac{1}{\det M}\mbox{adj } M$ that is $\mbox{adj } M=(\det M)M^{-1}$.

Using well known properties ($\det (aM)=a^n\det M$, $(aM)^{-1}=a^{-1}M^{-1}$ etc):
$$\mbox{adj } \left(\mbox{adj }A \right)=\mbox{adj } \left((\det A)A^{-1}\right)=\det\left((\det A)A^{-1}\right)\cdot\left((\det A)A^{-1}\right)^{-1}\\\left((\det A)^n\cdot\frac{1}{\det A}\right)\cdot\left(\frac{1}{\det A}\cdot (A^{-1})^{-1}\right)=(\det A)^{n-2}A$$