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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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The key here is to avoid any formulation which uses $A^{-1}$. Move things to other side, and use Gerry's answer. Hint for clarity though: $\det(\det(A)I) = \det\pmatrix{\det(A) & \dots \\ \dots & \dots \\ \dots & \det(A)} = \underbrace{\det(A)\det(A)\cdots\det(A)}_{n\text{ times}}.$ That gives you $\det(\text{adj}(A)) =\det(A)^{n-1}.$ – user2468 Aug 22 '12 at 4:52
@Jennifer, you've cancelled a factor of $\det A$ from both sides, no? If $\det A=0$, that requires some justification. – Gerry Myerson Aug 22 '12 at 4:56
@GerryMyerson you're right. – user2468 Aug 22 '12 at 4:59
If $det( A ) = 0$, you can't write $A^{-1}$ either... – xavierm02 Aug 22 '12 at 11:23
@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. – Bill Dubuque Aug 22 '12 at 18:35
up vote 2 down vote accepted

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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"Continuity"? Why to introduce topology here? Perhaps you meant "inductively"? – DonAntonio Aug 22 '12 at 4:46
I assume it's the continuity of the determinant as a polynomial. – user2468 Aug 22 '12 at 4:48
+1, by the way. – DonAntonio Aug 22 '12 at 4:49
@JenniferDylan , so do I, but what for was my question. Perhaps there's something I'm missing, though I believe the equation $\,\det A'=\left(\det A\right)^{n-1}\,$ already proves what the OP wanted. – DonAntonio Aug 22 '12 at 4:50
@Don, that equation was derived under the hypothesis $\det A\ne0$, so it can't be used, without some justification, in the case $\det A=0$. Topology supplies the justification. – Gerry Myerson Aug 22 '12 at 4:54

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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Thanks. I've just noticed it & removed my comment. – user2468 Aug 22 '12 at 5:04
I think it is because $\,AB=\det A\cdot I = 0=\,$ the zero matrix, @JenniferDylan – DonAntonio Aug 22 '12 at 5:04

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