prove that $\det(\operatorname{adj}(kA))=k^{n(n-1)}\det(A)^{n-1}$

Question

If $A$ is a non singular square matrix of order $n$ and $k$ is non zero scalar prove that $\det(\text{adj}(kA))=k^{n(n-1)}\det(A)^{n-1}$. Any help will be appreciated. thanks

Answer 1

We have $ \det (kA) =k^n \det (A)$ . We also have Laplace's formula (see https://en.wikipedia.org/wiki/Adjugate_matrix#Inverses) \begin{eqnarray*} A \operatorname{Adj}(A) = (\det(A)^n 1_n \\ \end{eqnarray*} \begin{eqnarray*} kA \operatorname{Adj}(kA) = (\det(kA)^n 1_n \\ \end{eqnarray*} Take $\det$ of this \begin{eqnarray*} \det (kA \operatorname{Adj}(kA)) =\det( \det (kA) 1_n) \\ \end{eqnarray*} Use $\det(AB)=\det(A) \det(B)$ \begin{eqnarray*} \det (kA ) \det ((\operatorname{Adj}(kA))) = k^n \det (A ) \det ((\operatorname{Adj}(kA))) =\det(k^n \det (A)1_n) =(k^n \det (A)) ^n\\\\ \end{eqnarray*} Rearrange a bit & you will have the desired result.