Question

Prove that:
$adj\left(AB\right)=adj\left(A\right).adj\left(B\right)$

Answer 1

$(AB{)}^{-1}=adj(AB\left)/det\right(AB)$

or $adj\left(AB\right)=(AB{)}^{-1}\cdot det(AB)-------(1)$

It is also known $=(AB{)}^{-1}=B^{-1}{A}^{-1}$

and $det\left(AB\right)=det\left(A\right)\cdot det\left(B\right)--------\left(2\right)$

also

$A^{-1}=adj\left(A\right)/det\left(A\right)B^{-1}=adj\left(B\right)/det\left(B\right)$

or $adj\left(A\right)=A^{-1}det\left(A\right)$ or $adj\left(B\right)=B^{-1}det\left(B\right)$

$adj\left(B\right)\cdot adj\left(A\right)=detA\cdot detB\cdot B^{-1}\cdot A^{-1}---------\left(3\right)$

Putting $\left(2\right)$ in equation $\left(1\right)$

$adj\left(AB\right)=det\left(A\right)\cdot det\left(B\right)\cdot B^{-1}\cdot A^{-1}------------\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$

$adj\left(AB\right)=adj\left(B\right)\cdot adj\left(A\right)$