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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.

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    @miguel Do you know how to turn Gerry's hint into a rigorous proof? If - as for many students - this continuity argument is not clear, then you should ask for further details before accepting an answer. When you are learning about such matters it is *crucial* that you understand the details of such arguments (esp. since there are many pitfalls in this area). Do not settle for handwaving - rigor is essential.2012-08-22

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Let's write $A'$ for the adjoint of $A$. $AA'=(\det A)I$, so $\det A\det A'=(\det A)^n$ (where $A$ is an $n\times n$ matrix). If $\det A\ne0$, this yields $\det A'=(\det A)^{n-1}$. By continuity, this last equation is true even when $\det A=0$, and you're done.

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    In this approach, the proof of continuity is the *crux* of the OP's problem, so I think if you propose this as an answer (vs. a comment), then you should at least sketch the proof, esp. since it often trips up many students (and even some teachers) in my experience.2012-08-22
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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.

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    I think it is because $\,AB=\det A\cdot I = 0=\,$ the zero matrix, @JenniferDylan2012-08-22