# Can a Gaussian integer matrix have an inverse with Gaussian integer entries?

Is there any way to characterize the set of complex matrices with Gaussian integer entries whose inverses also have Gaussian integer entries? I'm aware of the numerous examples of integer matrices whose inverses also have integer entries (usually involving binomial coefficients), but I'm wondering if those constructions can be generalized to Gaussian integers.

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If $A$ is a matrix over any commutative ring such that $A^{-1}$ also has entries in that ring, then from the equation $AA^{-1} = I$ we see that the determinant of $A$ must be a unit. And conversely, if $\det(A)$ is a unit, then $A^{-1}$ will have entries in the given ring (using the formula for adjugate matrix). So in particular, since the units in the ring of Gaussian integers are $\pm 1, \pm i$, it follows that a square matrix has inverse whose entries are also Gaussian integers if and only if its determinant is equal to $\pm 1, \pm i$.
If $R$ is a subring of $\Bbb C$, the group ${\rm GL}_n(R)$ of the invertible matrices with coefficients in $R$ consists of the matrices with determinant in $R^\times$ (the group of invertible elements in $R$).
Thus, if $R={\Bbb Z}[i]$ is the ring of Gaussian integers, the group ${\rm GL}_n(R)$ consists of the matrices $M$ with coefficients in $R$ such that $$\det(M)\in\{\pm1,\pm i\}.$$
Why do you need that $R$ is a subring of $\mathbb{C}$? This is true for any commutative ring. –  t.b. Jul 6 '11 at 8:56