Maximal order with primitive determinant in $\operatorname{GL}_n(\mathbb{F}_q)$ The following question has come up in a facet of a current project.  Having an answer (hopefully affirmative) will help me design and test some computational simulations.
$\mathbb{F}_q$ denotes the finite field of $q$ elements and $\mathbb{F}_q^*$ its group of units.  Consider the general linear group $\operatorname{GL}_n(\mathbb{F}_q)$.  It is known that the maximal possible order of an element from this group is $q^n-1$, realized by Singer cycles.  (That this is an upper bound is a not-so-hard argument with Cayley-Hamilton; that it is actually achievable is where Singer cycles enter.  Thanks to Derek Holt for teaching me this.)
Given a matrix $A \in \operatorname{GL}_n(\mathbb{F}_q)$, we can ask if its determinant is a primitive element of $\mathbb{F}_q$.  In case it's not, it is not hard to "modify" $A$ to make its determinant generate $\mathbb{F}_q^*$:  take your favorite primitive $\alpha$ and scale the first row of $A$ by $\frac{\alpha}{\det(A)}$.  This, of course, likely alters the order of the matrix:  there is no reason this new matrix should have the same order as $A$.  This approach is too naive for what I need.
My question is how to simultaneously optimize both situations.  That is,

Given $n$ and $q$, must there exist a matrix $A \in \operatorname{GL}_n(\mathbb{F}_q)$ for which (1) $A$ has order $q^n-1$ and (2) $\det(A)$ is a primitive element of $\mathbb{F}_q$?

Some lame trial-and-error in a few small cases (low $n$, low $q$) does not exclude "yes" as a possible answer.  I've tried to argue by counting/pigeonholing, but I cannot prove this.  Here's what I tried.  The map
$$ 
\det: \operatorname{GL}_n(\mathbb{F}_q) \to \mathbb{F}_q^*
$$
is surjective with rather large kernel $\operatorname{SL}_n(\mathbb{F}_q)$.  Hence each $d \in \mathbb{F}_q^*$ has $|\operatorname{SL}_n(\mathbb{F}_q)|$ preimages.  At the same time, there are $\varphi(q-1)$ primitive elements in $\mathbb{F}_q$, which is also proportionally "high," each primitive having  $|\operatorname{SL}_n(\mathbb{F}_q)|$ preimages.  Thus we have a pool of
$$
\varphi(q-1) \cdot |\operatorname{SL}_n(\mathbb{F}_q)| = \frac{\varphi(q-1)}{q-1} \cdot |\operatorname{GL}_n(\mathbb{F}_q)|
$$
matrices in $\operatorname{GL}_n(\mathbb{F}_q)$ whose determinants are primitive. Given the asymptotics of $\frac{\varphi(k)}{k}$, these seem to make up a decent portion of $\operatorname{GL}_n(\mathbb{F}_q)$, leading me to think this is always possible. Must one of these preimages have maximal order $q^n-1$?
Is this argument salvagable or is it too crude?  Or, is it simply false?  Any arguments or counterexamples are appreciated.
 A: *

*For $A\in GL_n(\Bbb{F}_q)$ if $order(A)=q^n-1$ then $A$'s minimal polynomial $P(x)\in \Bbb{F}_q[x]$ divides $x^{q^n-1}-1$ which is separable so that $A$ is diagonalizable (over the algebraic closure).
Expressing the order of $A$ in term of the lcm of the order of the eigenvalues, that is the roots of $P(x)$, you'll get that $order(A)=q^n-1$ iff the roots of $P$ have order $q^n-1$, which implies that $P$ is irreducible of degree $n$  and that the roots generate $\Bbb{F}_{q^n}^\times$.
Letting $a$ be a root then the others are $a^{q^m},m\in 1\ldots n-1$ (Galois theory, Frobenius automorphism).
Since $\deg(P)=n$ then $P$ is also the characteristic polynomial of $A$, therefore
$$\det(A)=  \prod_{m=0}^{n-1} a^{q^m} = a^{(q^n-1)/(q-1)}$$
It has order $q-1$ and it generates $\Bbb{F}_q^\times$.


*Conversely take $a\in \Bbb{F}_{q^n}^\times$ of order $q^n-1$, let $P(x)\in \Bbb{F}_q[x]$ be its $\Bbb{F}_q$-minimal polynomial and $A$ its companion matrix. Then $P$ is the characteristic and minimal polynomial of $A$ so that $a$ is an eigenvalue of $A$ which will have order $q^n-1$.
