How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?
I have been trying to solve this but I can't use $$\det(A^{-1})=\det \Big(\frac{1}{\det(A)} \operatorname{adj}(A) \Big)$$ because $\det(A)=0$ and $\frac{1}{0}$ is not allowed.
Linear Algebra – How to Prove that if det(A)=0 then det(adj(A))=0
linear algebra
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Best Answer
Let $B$ denote the adjugate matrix of $A$. Suppose for the sake of contradiction that $\det(B) \neq 0$. Then $B$ is invertible. Since the equation $AB =(\det{A})I = 0$ is true, we have then $$AB\vec{v} = \vec{0}\ \ \forall \vec{v}$$ which implies $A$ is the zero matrix. But then the adjugate of the $0$ matrix is clearly $0$ itself which contradicts the fact that $B$ was invertible.