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Is this statement true? If $A$ is an $n\times n$ matrix and $A^m=I_n$ for some $m\in \Bbb{N}$, then $A$ is invertible.

My trial

Let $n\in \Bbb{N}$ be fixed. Then, $$[\det(A)]^m=\det(A^m)=I_n=1.$$ Hence, $$\det(A)=1\neq 0.$$ Thus, $A$ is invertible since $\det(A)\neq 0.$. I'm I right or is there a counter-example?

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    $\begingroup$ Looks good. Alternatively, $A(A^{m-1})=I=(A^{m-1})A$ so $A^{-1}=A^{m-1}$. $\endgroup$ – Chrystomath Mar 18 at 11:32
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The right conclusion is $\det(A) \ne 0$, hence it is invertible.

Notice that we can't conclude that $\det(A)=1$. After all, it can take value $-1$.

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  • $\begingroup$ Sorry, I'll edit! $\endgroup$ – Omojola Micheal Mar 18 at 11:33
  • $\begingroup$ Thanks for the prompt reply. $\endgroup$ – Omojola Micheal Mar 18 at 11:34
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I think it is easier to see it this way: what is $AA^{m-1} = A^{m-1}A$?

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It's invertible since it has a inverse, $A^{m-1}$ that is. By the way, $\det(A)$ could also be -1.

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