This is what I need to prove:

You have an invertible matrix $A \in M_3(\Bbb R^3)$. Prove that $\operatorname{adj}(\operatorname{adj}(A))=\det{(A)}^{n-2}A$

The proof goes as follows:

$= \det{(\det{(A)}A^{-1})}(\det{(A)}A^{-1})^{-1}$
$= \det{(A)}^n\det{(A^{-1})}\det{(A)}^{-1}A$

I'm having trouble going from the second to the last step (The proof isn't complete, but I understand everything except that part).

I have already proven that:
$\det{(A^{-1})} = \frac{1}{\det{A}}$
and that

So I think I need to use one of those but I can't seem to find anything useful. Thanks in advance!

  • $\begingroup$ If you rewrite the last, you get $\operatorname{adj} A = \det A \cdot A^{-1}$. So $\operatorname{adj} (\operatorname{adj} A) = \det (\operatorname{adj} A)\cdot (\operatorname{adj} A)^{-1}$. Can you go on from there? $\endgroup$ – Daniel Fischer Jan 12 '14 at 22:28

Everything boils down to realize that for any scalar $\;c\in\Bbb R\;$ ,we have $\;\det(cA)=c^n\det A\;$ , where $\;n=\;$ the number of rows in the square matrix $\;A\;$ , and thus

$$\det\left(\det(A)\cdot A^{-1}\right)\left(\det(A)\cdot A^{-1}\right)^{-1}=\frac{(\det A)^n\det\left(A^{-1}\right)}{\det(A)}A$$

and now reach the end.

  • $\begingroup$ Thanks! I remembered a formula like that with a scalar but I just couldn't find it anywhere in my notes so thanks a bunch! $\endgroup$ – Joshua Jan 12 '14 at 22:52

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