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I have the following proof that I would like to be walked through because I'm not intuitively seeing what to do:

If $A$ is $n\times n$, prove $\det\left(\operatorname{adj}(A)\right) = \det(A)^{n-1}$.

I know the property of $A\operatorname{adj}(A) = \det(A)I$ is important but I don't know how to apply it to get an answer. Any help is much appreciated.

Thanks,

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    $\begingroup$ All you need to use is that important property, and $\det(XY)=\det X\det Y$, and the formula for $\det cX$ when $c$ is a constant (and you need to know $\det I$). Try it! $\endgroup$ – Gerry Myerson Oct 6 '13 at 2:01
  • $\begingroup$ @GerryMyerson to lay some basic ground level knowledge, det I = 1, det cX = c^n(det X), correct? $\endgroup$ – n8sty Oct 6 '13 at 2:08
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – Gerry Myerson Oct 6 '13 at 2:11
  • $\begingroup$ @GerryMyerson I hope I understand it now intuitively: since the minor of A is used in the calculation of the cofactor of A there will be 1 less row, hence n-1. What I still don't understand is how to use the various theorems established above to prove this relationship. $\endgroup$ – n8sty Oct 6 '13 at 2:19
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    $\begingroup$ Take determinants on both sides of your $A({\rm adj}A)=\det(A)(I)$ formula. $\endgroup$ – Gerry Myerson Oct 6 '13 at 2:22
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A(adj A) = |A|(I)

|A(adj A)| = |(|A| I)|

|A| |adj A| = $|A|^n * |I|$

|A| |adj A| = $|A|^n $

case 1: if |A|$\neq0$

Then we get ,|adj A| = $|A|^{n-1} $

case 2: if |A|$=0$

Then,|adj A|$=0$

And, we again get |adj A| = $|A|^{n-1} $

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  • $\begingroup$ Wait a minute --- what does $||A||$ mean? Does it mean $\det(\det A)$? But $\det A$ is a number, not a matrix, so what does $\det(\det A)$ mean, and why is it $(\det A)^n$? And how do you go from the next-to-last line to the last line? Did you divide by $\det A$? What if $\det A=0$? And why didn't you let OP write it out? We were so close.... $\endgroup$ – Gerry Myerson Oct 6 '13 at 8:50
  • $\begingroup$ OK, you took care of the $\det A=0$ problem --- but I had some other questions.... $\endgroup$ – Gerry Myerson Oct 6 '13 at 12:05
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    $\begingroup$ Sorry, but $\det(\det A)=(\det A)^n$ is nonsense. What's true is $\det((\det A)I)=(\det A)^n\det I=(\det A)^n$. We take determinants of matrices, not of numbers. $\endgroup$ – Gerry Myerson Oct 6 '13 at 22:06
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    $\begingroup$ How can you know that if |A| =0 then |adj(A)| =0? $\endgroup$ – user124697 Jan 21 '16 at 14:16
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    $\begingroup$ @user124697: math.stackexchange.com/questions/881654/… $\endgroup$ – Hans Lundmark Apr 27 '16 at 8:00
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We have the relation $$A \operatorname{Adj}(A)=\det(A)I $$ Now take determinant on both the sides we get, $$\det(A)\det(\operatorname{Adj}(A))=\det(\det(A)I) \tag1$$ use the relation that, for a matrix $A$ that is $n \times n$,

$$\det(kA)=k^n (\det(A))$$ where k is a numerical constant. Hence we have

$$\det(\det(A)I)=\det(A)^n \det (I)=\det(A)^n \tag2$$ and from equations $(1)$ and $(2)$ it follows that $$\det(A)\det(\operatorname{Adj}(A))=\det(A)^n ,$$ which implies $$\det(\operatorname{Adj}(A)) = \det(A)^{n-1} .$$

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  • $\begingroup$ This is very difficult to read. MathJax would certainly help as well as some other formatting. Regards $\endgroup$ – Amzoti Oct 28 '14 at 14:45
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[Case I] det (Adj(A))=0

[Case II] det (Adj(A)) = nonzero, so Adj(A) is invertible.

Let (Adj(A))^{-1} =B. From A Adj(A)=det(A)I, A Adj(A) B= det(A)I B.

So A = B det(A)I.

Suppose that det(A)=0. Then A = 0. So Adj A =0 implies det (Adj A)=0, a contradiction.

Therefore det (A) = nonzero. Now we have det (Adj(A))=det (A)^{n-1}.

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    $\begingroup$ Welcome to Math.SE. You chose to respond to a Question that is three years old, and you've given no indication what you think is new information, relative to older Answers, in your post. Although distinguishing the cases $\det(Adj(A))= 0$ and $\det(Adj(A))\neq 0$ may be a useful tactic, there are some details you omitted in the proof or calculation. See this introduction to posting mathematical expressions. $\endgroup$ – hardmath Apr 5 '17 at 12:38

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