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Prove that if $AC^T = |A|I \implies \det C = (\det A)^{n-1}$

Ran into trouble with a proof for linear algebra. $C$ is the cofactor matrix of $A \in \mathbb{R}^{n\times n}$, and I'm not sure how to even approach this problem. Any tips for starting? Not the entire proof, please.

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  • $\begingroup$ Prove that if $AC^T = |A|I \implies detC = (det A)^{n-1}$ then what? Is $n$ any arbitrary non-negative integer? $\endgroup$
    – Git Gud
    Apr 5, 2013 at 18:49
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    $\begingroup$ @GitGud I think it should be read without the "if", and only with the implication arrow. $\endgroup$
    – apnorton
    Apr 5, 2013 at 18:52
  • $\begingroup$ @GitGud: I suspect that the OP means "Prove that if $AC^T=|A|I$, then $\det C=(\det A)^{n-1}.$" $\endgroup$ Apr 5, 2013 at 18:52
  • $\begingroup$ I assumed so too. My comment was a hint to his mistake. $\endgroup$
    – Git Gud
    Apr 5, 2013 at 18:53

2 Answers 2

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Hint:$$\det(AC^T)=\det(A)\det(C^T) =\det(A)\det(C)$$

What is $\det(\alpha I)$, for any scalar $\alpha$?

If $\det(A)=0$, note that $AC^T=CA^T =0_{n\times n}$. What does this say about the nullspace of $C$?

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If $\det(A) = 0$, one entire row or column of the cofactor matrix must be zero by the definition of $\det(A)$ and thus $$\det(C) = 0 = 0^{n-1} = \det(A)^{n-1}.$$ If $\det(A) \neq 0$, then $$\det(A C^T) = \det(A) \det(C^T) = \det(A) \det(C)$$ and $$\det(|A|I) = |A|^n \det(I) = \det(A)^n.$$ By assumption these are equal and so dividing both sides by $\det(A)$ gives the desired result.

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  • $\begingroup$ What if $A=0$ and $n=1$? $\endgroup$
    – Git Gud
    Apr 5, 2013 at 18:55
  • $\begingroup$ If $A = 0$, then the cofactor matrix is also zero and so $\det(C) = 0 = 0^0 = \det(A)^{n-1}$. $\endgroup$
    – Suugaku
    Apr 5, 2013 at 18:58
  • $\begingroup$ I'm confused with $0=0^0$. $\endgroup$
    – Git Gud
    Apr 5, 2013 at 19:00
  • $\begingroup$ $0^0 = 0$ since $0$ to any power is always $0$. Also, please see my edit above. I realized that case $\det(A) = 0$ was not so obvious. $\endgroup$
    – Suugaku
    Apr 5, 2013 at 19:02
  • $\begingroup$ I would ask what your definition of 'power' is because I don't know why $0=0^0$, but until the OP clarifies the range of $n$, this discussion isn't suitable here. $\endgroup$
    – Git Gud
    Apr 5, 2013 at 19:10

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