Prove that if $A$ is ANY $n\times n$ matrix, then $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$.
first of all, since I did it about 3 or 4 times before, I started off by proving the case where $A$ is an INVERTIBLE $n\times n$ matrix from $A^{-1} = \operatorname{adj}(A)/\det(A)$.
but how do I prove it when $A$ isn't invertible? I wanna do something like plug $0$ in for $\det(A)$ but how can I use $A^{-1}$ in my proof if $A$ is not invertible? do I have to just substitute $A$ for $B$ and continue using $B^{-1}$ where needed? If so why is that even allowed?
Best Answer
Use the Laplace expansion to prove the formula $$ A\times\text{adj}A=\det(A)I. $$ If $\det(A)\neq 0$, then your conclusion follows. Now, consider $\det(A)=0$. Then, you have $A\times\text{adj}(A)=0$. If $A$ is the $0$ matrix, then $\text{adj}(A)$ is also the $0$ matrix and thus has determinant $0$. If $A$ has a nonzero row, then this row together with $A\times\text{adj}(A)=0$ provides a linear dependence between the rows of $\text{adj}(A)$, which implies $\det(\text{adj}(A))=0$.