Proof that All Entries in an Inverse Matrix are Integers "Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^{-1}$ are integers."
I began by setting up the adjoint method for finding the inverse.
$A^{-1} = \cfrac {1}{det(A)} adj(A)$
given that $det(A)=1$, 
$A^{-1}=adj(A)$
At this point I'm stuck. I think I need to work with an example containing integers, but I'm not quite sure.
 A: The coefficient at place $(i,j)$ of $\operatorname{adj}A$ is the determinant of the matrix obtained by removing row $j$ and column $i$ from $A$, multiplied by $(-1)^{i+j}$.
The determinant of a matrix with integer coefficients is an integer.
A: Here is a rewording of the adjoint argument.
The inverse matrix is obtained by solving a matrix equation $AX=I$, which translates to $n$ linear systems. By Cramer's rule, the solution of each linear system has integer entries because $\det A=1$.
A: For a $2 \times 2$ matrix $A$, the adjoint $Adj(A)$ coincides with its matrix of cofactors, with (egreg refreshed my memory) a change of sign for the even entries, meaning those entries $a_{ij}$ with $i+j$ even.
And you can go further: an integer-valued matrix, i.e., a matrix with integer entries is invertible iff its determinant is $ \pm 1$: You know the determinant will be Integer-valued, and you need $Det(A)Det(A^{-1})=1$. So the determinant DetA will have to be a unit in the ring of integers , so that it must be $\pm 1$
