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In a problem, I got $ AB=N^k $ where $\det(N)=1$. Can one show that there exists a matrix $M$ such that $BA=M^k$, where all $A,B$, and $M$ are $3 \times 3$ matrices with integer entries.

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I'm not entirely sure what the question is, but I presume it is "Suppose $A$ and $B$ are $3\times 3$ integers matrices with the property that $AB=N^k$, where $\det(N)=1$ . Prove that $BA=M^k$ where $M$ is a $3\times 3$ integer matrix".

As $A$ and $B$ are integer matrices, and $\det(A)\det(B)=\det(AB)=\det(N)^k=1$ then $\det(A)=\pm1$. Therefore $A^{-1}$ is also an integer matrix. Then $$BA=A^{-1}(AB)A=A^{-1}N^kA=(A^{-1}NA)^k=M^k$$ where $M=A^{-1}NA$ has integer entries.

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