Let $\displaystyle A \in \mathbb{Z}^{m \times m}$ with $\displaystyle det (A) \not =0$. Then there exists unique matrix $\displaystyle B \in \mathbb{Q}^{m \times m}$ such that AB=BA=det(A)I and B has integer entries.

Help me to prove this.

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- Dec 5th 2008, 12:20 PMdimukmatrix question
Let $\displaystyle A \in \mathbb{Z}^{m \times m}$ with $\displaystyle det (A) \not =0$. Then there exists unique matrix $\displaystyle B \in \mathbb{Q}^{m \times m}$ such that AB=BA=det(A)I and B has integer entries.

Help me to prove this. - Dec 5th 2008, 03:33 PMNonCommAlg
the matrix $\displaystyle B$ is the adjugate of $\displaystyle A.$ the proof of $\displaystyle \text{adj}(A) A=A \ \text{adj}(A)=\det(A)I$ can be found in any linear algebra textbook. the only thing that i have to add here is that if $\displaystyle A \in \mathbb{Z}^{m \times m},$

then from the definition of the adjugate matrix, it's clear that $\displaystyle \text{adj}(A) \in \mathbb{Z}^{m \times m}$ as well. so i don't know why your problem says $\displaystyle \text{adj}(A) \in \mathbb{Q}^{m \times m}$? we also don't need to have $\displaystyle \det(A) \neq 0.$

**Edit**: just realized that we want a unique $\displaystyle B.$ in this case, the condition $\displaystyle \det(A) \neq 0$ will take care of that. - Dec 5th 2008, 09:46 PMdimukmatrix question
I think that is for uniqueness of B