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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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):
$$\mbox{adj } \left(\mbox{adj }A \right)=\mbox{adj } \left((\det A)A^{-1}\right)=\det\left((\det A)A^{-1}\right)\cdot\left((\det A)A^{-1}\right)^{-1}\\\left((\det A)^n\cdot\frac{1}{\det A}\right)\cdot\left(\frac{1}{\det A}\cdot (A^{-1})^{-1}\right)=(\det A)^{n-2}A$$